Electric Potential at a Point:

The electric potential at any point in the electric field is defined as the work that must be done by the external force to move unit positive charge from infinity to that point without acceleration.

If W is the work done in moving the charge q_o from infinity to the point in the electric field. Then the potential at that point is given by

The S.I. unit of potential difference is volt (V).

The electric potential at a point in electric field is said to be 1 volt, if 1 joule of work is done, against an electric intensity, to bring a charge of 1 coulomb from infinity to that point.

The Concept of Potential Difference:   

The electric potential difference between two points A and B in an electric field is work done in moving a unit positive charge without acceleration from A to B.

Explanation:

Consider charge (+q) kept in the medium of dielectric constant K.  It will create an electric field around it.  If another charge is kept in this field then it will always experience a force.  Hence to move the charge in the electric field from one point to another some work has to be performed. The potential difference between points A & B is defined as work done (WAB) in moving unit positive charge from point A to point B without acceleration.  Let q0 be the +ve charge kept in the electric field created by charge +q.  The potential difference between two points A & B is given by

The potential difference between two points in the electric field is said to be 1 volt if 1 joule of work is done in moving a charge of 1 coulomb from a point at lower potential to a point at higher potential without acceleration.

The Expression for Electric Potential at a Point:

Consider charge (+q) kept in the medium of dielectric constant k.  It will create an electric field around it.  If another charge is kept in this field then it will always experience a force.  Hence to move the charge in the electric field from one point to another some work has to be performed. The potential difference between points A & B is defined as work done (W AB) in moving unit positive charge from point A to point B without acceleration.  Let q0 be the +ve charge kept in the electric field created by charge +q.  The potential difference between two points A & B is given by

Let P be the point at a distance ‘r’ from charge +q.  Let q₀ be the charge kept at point P.  The force F acting on charge q₀ is away from charge +q and along vector OP. The electric intensity at point P is given by  

By force F  the charge q₀ is repelled. To move charge from A to B we have to apply equal and opposite force which to given by

The work done by this force against repulsive force to move charge from very small distance   is given by

Total work done in moving the charge from A to B can be found by integrating the above equation.

This is an expression for the potential difference between two points A & B in the electric field.

Now, potential at a point is defined as work done in moving unit +Ve charge from infinity to that point without acceleration

∴ r_A = ∞

 let,  r_B = r

∴ V_B = V and V_A = 0

This is an expression for potential at a point in the electric field.

Electron Volt:

Electron volt is a unit of energy used in atomic and nuclear physics. The kinetic energy acquired by the an electron, when it is accelerated through a potential difference of 1 volt in vacuum is called one electron volt (1 eV).

Relation Between electron volt and joule:

When a charge moves through the electric field work is done which is given by

Work done = charge x potential difference

This work done is converted into kinetic energy of charge

kinetic energy of charge = charge x potential difference

   1 electron volt = Charge on one electron x 1 volt

     1 eV = 1.6 x 10^-19  joule 

This is the relation between the electron volt and joule.

Relation Between Electric Intensity and Potential:

Let us consider uniform electric field of intensity . In uniform electric field the lines of force are equispaced and parallel to each other. Let A and B two points situated in this electric field such that line joining the two points is parallel to the direction of electric field. Let d be the distance between the two points.

Let us consider a test charge q0 kept at point O. Then the force acting on this charge in electric field is given by. The direction of this force is same as that of the electric field i.e. downwards.  To move the charge from A to B without acceleration in the electric field we have to apply equal and opposite force given by. Thus the direction of this force is upward.

Let us assume this force moves the charge q0 from A to B through a small distance. Then, the work done is given by

The total work done can be calculated by integrating above equation.

This is the relation between electric intensity and potential in uniform electric field.

Electric Dipole and Electrical Dipole Moment:

Consider two equal and opposite charges say +q & -q separated by some finite distance ‘2a’ then the two charges are said to form what is called as an electric dipole. The dipole consists of two equal and opposite charges, hence the total charge of the dipole is zero, but it still exhibits electrical properties due to the separation of the charges.

The electric dipole movement is given by

Considering magnitude only,  

p = q. 2a

Electric dipole movement is the vector quantity whose direction is from negative charge to positive charge.

Electric Flux.

1. The total number of electric lines of force passing normally though a given area in an electric field is called the electric flux though that area.
2. Consider an infinitesimal area (very small area) ds drawn in an electric field. Let E be the electric intensity at the centre of this area. The area ds is so small that the intensity at every point of this area can be assumed to be the same as E. The area ds can be represented by a vector drawn perpendicular to it. Let q be the angle between E and ds.

Then the electric flux (Φ) through the area ds is given by the relation.

Φ = E ds cos θ

Unit of Electric Flux

We have Φ = E ds cos θ

Thus, Unit of Φ  =  unit of E x unit of ds

Unit of Φ=  V/m  x m²

Unit of Φ =  Vm

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When a charge is kept in a medium it sets around it, what is called an electric field. It can be represented by drawing what is called lines of force. If another charge is brought into this region it experiences a force. Due to the field, the charges experience the force of attraction or repulsion. The region around a charge or system of charges within which other charged particles experience an electrostatic force is called an electric field.

Lines of Force and Line of Induction:

The electric field can be represented by drawing what is called the lines of force. They can be straight or curved. A line of force is a locus of all the points through which a free unit positive charge (test charge) will move when kept in the field. If the medium is other than air or vacuum, the lines of force are called lines of induction.

The direction of the electrical field at a point can be obtained by drawing a tangent to the line of force passing through that point.

Characteristics of Electric Lines of Force:

• Lines of force are imaginary lines or hypothetical lines.
• Lines of force always emerge out from a positive charge and terminate on a negative charge. Thus the lines of force coming out from positive charge are supposed to end at a negative induced charge on the surrounding and the lines of force coming towards negative charge are supposed to start from a positive induced charge on the surrounding. 

In the following photograph a student with long hair is touching electrode of Van de Graaff generator, we can see that the hair follow electric lines of force.

• Lines of force always emerge out or terminate at right angles to the surface (normally).
• When two similar charges are kept near to each other the lines of force push each other and thus there is repulsion between the two similar charges.

• When two dissimilar charges are kept near to each other the lines of force will try to come near to each other and hence there is an attraction between two dissimilar charges. The lines of force start from a positive charge and end on a negative charge. So the lines of force do not form closed loops.

• The lines of force exert a lateral force on each other. i.e. they try to repel each other. This explains repulsion between two like charges.
• The lines of force possess a longitudinal tension hence they have a tendency to shrink i.e. to contract lengthwise. This explains the attraction between unlike charges.
• Lines of force never intersect with each other. If they do so, it means that at the point of intersection of the two lines the electric field has two directions which is not possible. Having two directions means the test charge at that point can move in two different directions simultaneously.
• It is possible for us to draw lines of force through each and every point in the medium. But it will result in overcrowding of the lines. And thus it is not possible for us to study them. Hence lines of forces are grouped into what is called ‘tubes of force.
• If the tubes of force are crowded together then the field is strong.
• If the tubes are equally spaced then the field is uniform.

Types of Electric Field:

Uniform Electric Field:

When magnitude and direction of electric intensity are the same at all the points in the electric field, then it is called a uniform electric field. It is represented by drawing equidistant parallel straight lines in the direction of the field.

Non-Uniform Electric Field:

When magnitude and direction of electric intensity are not the same at all the points in the electric field, then it is called a non-uniform electric field.

Radial Electric Field:

When electric intensity E at any point in the electric field is directed towards or away from the same fixed point, then the field is called a radial electric field.

Electric Field Intensity:

Electric intensity at a point in an electric field is defined as the force acting on a unit positive charge when placed at that point.

It is denoted by the letter ‘E’ and its S.I. unit is N/C or V/m. Mathematically

Electric intensity at a point gives us the idea of the strength of the electric field at that point. Stronger the field at a point greater is the magnitude of the electric intensity.

Electric intensity is a vector quantity and its direction at a point is the same as the direction of force acting on positive test charge kept at that point.

Expression for Electric Intensity at a Point Due to Point Charge:

Consider a charge ‘q’ a placed at point O as shown. Let P be the point at a distance of ‘r’ from O where electric intensity is to be found.

Let us consider a positive charge q_o kept at point P. The force acting on q_o, due to q is given by Coulombs Law.

By definition of Electric Intensity,

This is an expression for electric intensity at a point at a distance of ‘r’ from a point charge. The direction of electric intensity is the same as that of force at that point.

The Expression for Electric Intensity in a Vector Form:

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In this article, we shall study the COulomb’s law of electrostatics, its explanation, uses, and limitations.

Coulomb’s Law of Electrostatics:

Statement:

The force of attraction or repulsion between two electric point charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between the two charges. The force acts along the line joining the two charges.

Explanation:

Let q₁ and q₂ be the two-point charges separated by distance ‘r’ in a medium with dielectric constant ‘k’  Then,

F  ∝ q₁ . q₂   ………..  (1)

F  ∝ 1/r²…………(2)

From statement (1) & (2) we get

Where F   = force between the two charges,

∈_o  = electrical permittivity of free space. = 8.8542 × 10^-12 C² N^-1m^-2 or F m^-1.

k  = dielectric constant of the medium.

This expression is called a mathematical statement of Coulomb’s Law.

The value of the constant

Hence the equation of Coulomb’s law can be written as

In cgs system the value of 1/4πε_o = 1

Coulomb’s force is not affected by the presence or absence of other charges.

Coulomb’s Law in a Vector Form:

Where  F ₂₁ = Force on the second charge due to the first charge.

r ₁₂ = Unit vector from the first charge due to the second charge

Where  F ₁₂ = Force on the second charge due to the first charge.

r ₂₁ = Unit vector from the first charge due to the second charge

as unit vector r ₁₂ = – unit vector r ₂₁

 F ₂₁ = – F ₁₂

Thus the forces are equal but oppositely directed.

If we consider the signs of the charges (positive and negative) then the vector form of coulomb’s law can be written as

Coulomb’s Law in Terms of Position Vector:

Let r₁ and r₂  be the position vectors of charges q₁ and q₂ situated at point A and B respectively w.r.t. origin O.

By triangle law of vector addition

By Coulomb’s law, the force on charge q₁ due to charge q₂ is given by

This is the vector form of Coulomb’s law in terms of position vectors of charges.

Limitations of Coulomb’s law:

• Coulomb’s law is applicable to point charges only and is not applicable to extended bodies.
• It can be applied to extended bodies by assuming them to be point charges but the distance between them should be sufficiently large,
• The law holds good only for stationary charges.
• Coulomb’s law is verified for distances from 10^-15 m to several kilometres. We are not sure about its validity for a distance less than 10^-15 m (nuclear dimensions).

Notes:

• The charges possessing the same nature of charge are called like charges. Like charges either all positive or all negative. Like charges always repel each other.
• The charges possessing opposite nature of charge are called, unlike charges. Unlike charges always repel each other.

Dimensions of Electrical Permittivity:

Hence dimensions of universal electrical permittivity are [L^-3M^-1T⁴I²]

Units of charge:

Coulomb: SI unit of charge is coulomb.

Definition of Coulomb (Using Coulomb’s Law):

By Coulomb’s Law

If q₁ = q₂ = q, r = 1 m and F = 9 × 10⁹ N, k =1, then

q² = 1 and q = ± 1 C

The charge of 1 coulomb is that charge which, when placed in free space at a distance of 1 metre from an equal and similar charge, repels it with a force of  9 × 10⁹ N

Definition of Coulomb (Using Electric Current):

The electric current flowing through any section of a conductor is given by

I = q/t

Where q   =   electric charge flowing through any section of the conductor

t    =   time for which the charge is flowing

q = I × t

unit of charge = unit of a current × unit of time

1 coulomb = 1 ampere × 1 second

Hence a charge of 1 coulomb is defined as the quantity of a charge which flows per second through any cross-section of a conductor when there is a steady current of 1 ampere in the conductor.

statcoulomb or electrostatic unit (esu) of charge:

statcoulomb or esu of charge is cgs unit of charge. In cgs system the value of 1/4πε_o = 1

If q₁ = q₂ = q, r = 1 cm and F = 1 dyne, then

q² = 1 and q = ± 1 C

Hence one statcoulomb of a charge is that charge which, when placed in free space at a distance of 1 cm from an equal and similar charge, repels it with a force of 1 dyne. In cgs system unit of electromagnetic charge is abcoulomb or emu of charge.

Conversions:

Note:

These definitions are for convenience and not standard. When defining S.I. Unit coulomb in standard conditions help of the magnetic effect of electric current is considered.

Dielectric Constant of Medium:

Definition:

The ratio of the electrical permittivity of the medium to the electrical permittivity of free space is called the dielectric constant of the medium.

k = ∈ / ∈_o 

The dielectric constant of e medium is denoted by letter ‘k’ or ‘∈_r‘. As it is a pure ratio it has no unit. For air or vacuum k = 1 for any other medium k > 1.

Let q₁ and q₂ be the two point charges separated by distance ‘r’ in a medium in a vacuum, and then the same charges separated by the same distance in a medium with dielectric constant ‘∈_r‘.

Thus the dielectric constant or relative permittivity of a medium can be defined as the ratio of the electrostatic force between two charges separated by a certain distance in vacuum to the electrostatic force between the same two charges separated by the same distance in the medium. The presence of dielectric reduces electrostatic force between two charges which implies that the electrostatic force between two charges is maximum in a vacuum.

Comparison Between Electrostatic Force and Gravitational Force:

Points of Similarity:

• Both the forces obey the inverse square law of distances.
• Work done by both the forces does not depend on the path followed. Hence both the forces are conservative forces.
• Both the forces act along the line joining the centres of interacting bodies. Hence both the forces are central forces.
• Both the forces operate even in the vacuum.

Point of Dissimilarity:

• Gravitational force is always that of attraction while the electrostatic force may be attractive or repulsive depending upon the nature of interacting charges.
• Gravitational force is unaffected by the intervening medium. The electrostatic force depends on the dielectric constant of the intervening medium.
• The electrostatic force is very much stronger than the gravitational force. Let us consider an electron-proton system in an atom. The electrostatic force F_e and gravitational force F_g between them are given by

• We can see that the electrostatic force is 1039 times stronger than the gravitational force. But gravitational force plays a more significant role than the electrostatic force because most of the bodies in the universe are electrically neutral. Similarly, due to the attractive or repulsive nature of electrostatic force, they can cancel each other while gravitational force is only of attraction hence there is no question of cancellation of the gravitational force. Thus gravitational force is weak but has the dominating effect.
• Electrostatic forces are weaker than nuclear forces.

The principle of Superposition of Forces:

• This principle is used for calculation of net force acting on an individual charge due to other when more than two charges are interacting with each other.
• When a number of charges are interacting, the total force on a given charge is the vector sum of individual forces exerted on a given charge by all other charges.
• Let us consider a  system of point charges q1, q2, q3, q4, ……..qn.  Let F12, F13, F14, F15, ……..F1n be the forces acting on charge q1 due to charges q₂, q₃, q₄, ……..q_n respectively.
•  Then according to the principle of superposition of forces, the force acting on charge q₁ is given by

 F₁ =  F₁₂ + F₁₃ + F₁₄ +  ……+ F_1n

By vector representation

Applying the same logic to all such pairs we have

• The same procedure can be adopted for finding the force on any other charge due to the remaining charges.

Numerical Problems:

Example – 01:

The distance between electron and proton in a hydrogen atom is 5.3 × 10^-11 m. What is the magnitude of the electric force between them?

Given: Distance between proton and electron = r =  5.3 × 10^-11 m, The magnitude of the charge on proton = q₁ = 1.6 × 10^-19 C, Magnitude of the charge on electron = q₂ = 1.6 × 10^-19 C.

To find: Force =?

Solution:

By Coulomb’s law

Ans: The electric force between electron and proton is 8.202 × 10^-8 N.

Example – 02:

Two charges of magnitudes 1 μC and 2 μC are separated by a medium of dielectric constant 2. What is the magnitude of the electric force between them?

Given: Distance between proton and electron = r =  10 cm = 0.1 m, The magnitude of the first charge = q₁ = 1 μC = 1 × 10^-6 C, Magnitude of the charge on second charge = q₂ = 2 μC = 2 × 10^-6 C, dielectric constant of medium = k = 2.

To find: Force =?

Solution:

By Coulomb’s law

Ans: The electric force between the charges is 0.9 N.

Example – 03:

Two small spheres are charged positively. The combined charge being 5.0 x 10-5 C. If each sphere is repelled from other by a force of 1.0 N when they are 2.0 m apart. How is the total charge distributed between two spheres?

Given: Distance between two charges = r =  2.0 m, Sum of the cahrges = 5.0 × 10^-5 C =  50 × 10^-6 C, Let , The magnitude of the first charge = q₁ = q μC = q × 10^-6 C, Magnitude of the other charge = q₂ =(50 – q)  × 10^-6 C, dielectric constant of medium = k = 1. Force = F = 1.0 N

To find: charges = ?

Solution:

By Coulomb’s law

∴ q₁= 38.43 m C = 38.43 × 10^-6 C = 3.84 × 10^-5 C

∴ q₂= 11.57 m C = 11.57 × 10^-6 C = 1.16 × 10^-5 C

Ans: The distribution of charge is 3.84 × 10^-5 C and 1.16 × 10^-5 C

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In this article, we shall study the concept of normal electric induction, electric field, and electric flux.

Normal Electrical Induction:

The number of tubes of induction passing normally through a unit area in an electric field is called Normal electric induction.

Total Normal Electrical Induction:

The total number of tubes of induction passing normally through a given surface in an electric field is called the total normal electric induction

Electric Flux:

The number of tubes of force passing normally through a given surface in an electric field is called electric flux.

Electric Field:

When an electric charge is kept in a medium, it creates around it, what is called an electric field. If another charge is kept in this field it experiences a force of attraction or repulsion. Thus electrical charge modifies properties of surrounding space by producing an electric field. Hence we can consider the electric field is a characteristic property of the system of charges.

Electric Lines of force:

The electric field near a charge is represented by drawing a line of force. A line of force in an electric field is an imaginary line drawn in such a way that the direction of the line of force at any point is the same as the direction of the field at that point.

An electric line of force is defined as the path along with a free positive charge, moves when it is placed in an electric. However, since the direction of the field varies from point to point, the lines of force are usually curves.

Characteristics of Electric Lines of Force:

• Lines of force are imaginary. They start from a positive charge and end on a negative charge.
• A tangent drawn to a line of force at any point shows the direction of the electric field at that point.
• Two lines of force never intersect each other. Only one line of force can pass through one point in the electric field.
• The number of lines of force passing normally through a unit area in the direction of the field represents the magnitude of electric intensity.
• The line of force always originates perpendicular to the surface of a charged conductor.
• Lines of force do not pass through the conductor. They can pass through a dielectric.
• Lines of force have a tendency to contract in length. This explains electrostatic attraction between unlike charges.
• Lines of force exert lateral pressure on each other. This explains electrostatic repulsion between like charges.

Concept of Tubes of Force:

Imagine a small area of the surface of a positively charged conductor and suppose that lines of force are drawn from every point on the boundary of this area. These lines enclose a tube of force. The sides of a tube of force are formed by lines of force and the tube of force has the same properties as lines of force.

In order to obtain a relation for the tube of force, it is assumed that, in a medium of permittivity ε = ε_ok , the number of tubes of force originating from a unit positive charge is  (1/ε_ok). Therefore the number of tubes of force originating from a charge +q is (q/ε_ok). This is also an expression for electric flux from charge +q.

Suppose that charge +q is situated at the centre of a sphere of radius ‘r’.  The total number of tubes force is (q/ε_ok) passes normally from the surface of the sphere of normally through unit area of the surface is 4πr².

Therefore, the number of tubes of force passing normally through the unit area of the surface is

This value is also equal; to the magnitude of electric intensity (E) at a distance `r’ from a charge +q.

Concept of Electric Flux:

The number of tubes of force passing normally through a given surface in an electric field is called electric flux. (∅). As the number of tubes of force passing normally per unit area is the electric intensity (E), it is also the electric flux per unit area. S.I. unit of electric flux is Nm²/C.

Consider a small area dS in an electric field of intensity E.  Let θ be the angle made by the normal drawn to area vector dS and the direction of E.

The component of electric intensity i.e. E cosθ is parallel to dS.

Electric flux per unit area  = E cosθ

and Electric flux through dS area = ∅ = E cosθ . dS ……….. (1)

Also, electric flux from change q is given by  ∅ = q/ε_ok ………(2)

From equation (1) and (2) we get

Concept of Tubes of Induction:

A tube of force in any medium other than a vacuum is called tube of induction. The number of tubes of force originating from a charge depends on the permittivity of the medium and is therefore different for different media.

In order that this number does not depend on the nature of the medium, the concept of tubes of induction was introduced. According to this concept, only one tube originates from a unit positive charge, whatever be the medium surrounding this charge. Such a tube is called a tube of induction. The number of tubes of induction originating from charge +q is q.

Concept of Normal Electric Induction:

The number of tubes of induction originating from charge +q is q. If this charge is situated at the centre of a sphere of radius ‘r’, then the number of tubes of induction passing normally through unit area 4πr² of the sphere is q/4πr².

The number of tubes of induction passing normally through the unit area in the electric field is called Normal electric induction.

Normal Electric Induction is also called an electric displacement vector represented by

Concept of Total Normal Electric Induction:

The total number of tubes of induction passing normally through a given surface in an electric field is called the normal electric induction (T. N. E. I.)

If E cos θ is the component of electric intensity E parallel to the area vector of dS, then

T.N.E.I. through area dS  = Normal Electric Induction × area dS.

Previous Topic: Coulomb’s Law

Next Topic: Gauss’s Theorem and its Applications

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In this article, we shall study the gauss’s theorem and its applications to find electric intensity at a point outside charged bodies of different shapes.

Gauss’s Theorem:

Statement: 

Total Normal Electric Induction over any closed surface of any shape in an electric field is equal to the algebraic sum of electric charges enclosed by that surface.

Explanation: 

Consider a closed surface enclosing number of charges such as +q₁, + q₂, – q₃ …… then

(TNEI)_S = q₁ + q₂ – q₃ + ……= ∑ q.

Proof:

Consider a charge +q situated at a point ‘O’ inside a closed surface of any shape. Such a surface is called a Gaussian surface. Consider a small element dS on its surface. at a distance of ‘r’ from the charge ‘+q’.

The electric intensity  at any point over element dS is given by

The normal drawn to the area dS makes an angle θ with the direction of electric intensity E.

Therefore, the total normal electric induction over area dS is given by

The total Normal Electric Induction over the whole Gaussian surface can be obtained by integrating the above expression.

The same argument is true for any other charge present inside the closed surface. Thus if the closed surface enclosed a number of charges such as +q₁, + q₂, – q₃ …… then

(TNEI)_S = q₁ + q₂ – q₃ + ……= ∑ q

Thus Gauss’s theorem is proved.

Electric Intensity at a Point Outside a Charged Sphere: (Application of Gauss’s Theorem):

Consider a conducting sphere of radius R on which a charge +q is deposited. The deposited charge gets distributed uniformly over the surface of the sphere. The lines of induction will be normal to the surface, radially outwards. Consider a point P at a distance r from the centre of the sphere (r > R) at which electric intensity is to be found.

To find the electric intensity at point P, construct an imaginary Gaussian spherical surface of radius ‘r’ through P.

The total Normal electric induction over element dS is given by

(TENI)_dS  = k E ε_o cos θ dS

As electric intensity vector and area vector are parallel to each other,

θ = 0° and hence cos θ  = cos 0° = 1

∴  (TENI)_dS  = k E ε_o dS.

Total Normal Induction over whole Gaussian surface is

This is the expression for an electric intensity at a point outside a charged sphere. This relation is the same as the expression electric intensity at distance ‘r’ due to a point charge q. Hence a charged sphere behaves as if the entire charge is concentrated at its centre.

Electric Intensity at a Point Outside a Charged Sphere in Terms of Surface Charge Density of the Sphere:

The electric intensity at a point outside a charged sphere is given by

The surface area of charged sphere = A = 4πR²

Let σ be the charge per unit area of the sphere (surface charge density)

By definition of surface charge density

σ = q /A = q / 4πR²

q = σ × 4πR²

Substituting for q, in equation (1) we get

Where R = Radius of the sphere

r  =  distance of the point from the centre of the sphere.

This is the expression for an electric intensity at a point outside a charged sphere in terms of surface charge density.

Electric Intensity at a Point on the Surface of a Charged Sphere:

The expression for an electric intensity at a point outside a charged sphere in terms of surface charge density is

This is the expression for an electric intensity at a point on the surface of a charged sphere.

Expression for an Electric Intensity at Point Outside a Charged Cylinder: (Application of Gauss’s Theorem)

Consider a long uniform cylinder of radius ‘R’ carrying charge +q per unit length. Let ‘P’ be a point at a perpendicular distance ‘r’ from the axis (r > R) at which electric intensity is to be found.

To find the electric intensity at point P, let us consider a coaxial imaginary Gaussian cylindrical surface of radius ‘r’ and height ‘l ‘ passing through the point P.

If we consider an element dS on the flat face of the Gaussian cylinder, for this element the electric intensity vector and the area vector are perpendicular to each other. Hence θ = 90°, cos 90° = 0.

Hence TNEI over flat face = 0

Therefore, the Total Normal Electric Induction over the whole surface is the same as the TNEI over a curved surface. Consider an element dS over the curved surface of the Gaussian cylinder. Over this element,

(TENI)_dS  = k E ε_o cos θ dS

As electric intensity vector and area vector are parallel to each other,

θ = 0° and hence cos θ  = cos 0° = 1

∴  (TENI)_dS  = k E ε_o dS.

Total Normal Induction over whole Gaussian surface is

As ‘q’ is the charge per unit length of the cylinder, the charge enclosed by the Gaussian cylinder is ‘ql’.

Applying Gauss’s theorem,

This is the expression for an electric intensity at a point outside a charged cylinder.

Electric Intensity at a Point Outside a Charged Sphere in Terms of Surface Charge Density of the Cylinder:

The electric intensity at a point outside a charged cylinder is given by

Let σ be the charge per unit area of the cylinder (surface charge density)

Let, q  =  Charge per unit length of the cylinder

∴   q      =  Charge per unit area × area of unit length.

R = radius of the cylinder

r = perpendicular distance of the point from the axis.

This is the expression for an electric intensity at a point outside a charged cylinder in terms of surface charge density.

Electric Intensity at a Point on the Surface of a Charged Cylinder:

The expression for an electric intensity at a point outside a charged cylinder in terms of surface charge density is

This is the expression for an electric intensity at a point on the surface of a charged cylinder.

Expression for an Electric Intensity at a Point Near a Charged Conductor: (Application of Gauss’s Theorem)

Consider a uniform charged conductor of any shape. Let σ be the surface charge density or charge per unit area of its surface. Consider a point P very close to the surface of the conductor.

Consider an element dS near point P. Construct an imaginary Gaussian surface of cross-sectional area dS with its axis normal to the surface, partly inside and partly outside the conductor. As the electric Intensity inside a conductor is zero, there is no TNEI through cross-sectional area dS outside the conductor. Therefore TNEI through cross-sectional area dS outside the conductor is

(TENI)_dS  = k E ε_o cos θ dS

As electric intensity vector and area vector are parallel to each other,

θ = 0° and hence cos θ  = cos 0° = 1

∴  (TENI)_dS  = k E ε_o dS.    ….  (1)

Since is the charge density, the total charge enclosed by Gaussian surface over area dS is dS. Hence by Gauss’s theorem,

(TNEI) ds  = σ  dS    ….  (2)

From (1) and (2)

k E ε_o dS    =  σ  dS

∴   k E ε_o    =  σ

∴  E =  σ / k ε_o

This is the expression for an electric intensity at a point on near a charged conductor.

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In this article, we shall study to solve problems to find electric intensity at a point due to a charged sphere.

Example – 01:

A charge of 0.002 µC is given to an isolated conducting sphere of radius 0.5 m. Calculate the electric intensity (i) at a point on the surface of the sphere and (ii) at a point 1.5 m away from its centre. (iii) at the centre of the sphere.

Given: Charge = 0.002 µC = 0.002 x 10^-6 C = 2 x 10^-9 C, radius of sphere = R = 0.5 m, k = 1, ε_o = 8.85 x 10^-12 C²/Nm²

To Find: Electric intensity (i) at a point on the surface of the sphere and (ii) at a point r = 1.5 m

Solution:

Surface charge density is given by

Electric intensity on the surface of a charged sphere is given by

Electric intensity at a point outside charged sphere is given by

Electric intensity at any point inside a charged conductor is zero. Hence electric intensity at the centre of the sphere is zero.

Ans: Electric intensity at a point on the surface of the sphere is 71.93 V/m, Electric intensity at a point at a distance of 1.5 m from centre of the charged sphere is 7.992 V/m and electric intensity at the centre of the sphere is zero.

Example – 02:

An isolated conducting sphere of radius 0.1 m placed in vacuum carries a positive charge of 0.l µC. Find the electric intensity at a point at a distance of 0.2 m from the centre.

Given: Charge = 0.1 µC = 0.1 x 10^-6 C = 1 x 10^-7 C, radius of sphere = R = 0.1 m, distance of point from centre = r = 0.2 m, k = 1, ε_o = 8.85 x 10^-12 C²/Nm²

To Find: Electric intensity at a point r = 0.2 m =?

Solution:

Surface charge density is given by

Electric intensity at appoint outside the charged sphere is given by

Ans: Electric intensity at a point at a distance of 0.2 m from the centre of the charged sphere is 2.248 x 10⁴ V/m

Example – 03:

The electric intensity at a point at a distance of 1 m from the centre of a sphere of radius 25 cm is 10⁴ N/C. Find the surface density of charge on the surface of the sphere; The sphere is situated in air.

Given: Radius of sphere = R = 25 cm = 0.25 m, distance of point from centre = r = 1 m, k = 1, ε_o = 8.85 x 10^-12 C²/Nm², Electric intensity = E = 10⁴ N/C

To Find: Surface charge density = σ =?

Solution:

Electric intensity at appoint outside charged sphere is given by

Ans: The surface charge density is 1.416 x 10^-6 C/m²

Example – 04:

A metal sphere of radius 20 cm is charged with 12.57 µC situated in air. Find the surface density of charge. Calculate the distance of point from centre of sphere where electric intensity is 1.13 x 10⁵ N/C

Given: Radius of sphere = R = 20 cm = 0.20 m, Charge = 12.57 µC = 12.57 x 10^-6 C, k = 1, ε_o = 8.85 x 10^-12 C²/Nm², Electric intensity = E = 1.13 x 10⁵ N/C

To Find: Surface charge density = σ = ?, distance of point from centre = r =?

Solution:

Surface charge density is given by

Electric intensity at appoint outside the charged sphere is given by

Ans: Surface charge density = σ = 2.5 x 10^-5 C/m² and the distance of the point from the centre of the sphere where the electric intensity is 1.13 x 10⁵ N/C is 1 m

Example – 05:

A metal sphere of radius 1 cm is charged with 3.14 µC. Find the electric intensity at a point situated at a distance of 1 m from centre of metal sphere.

Given: Charge = 3.14 µC = 3.14 x 10^-6 C, radius of sphere = R = 1 cm = 0.01 m, k = 1, ε_o = 8.85 x 10^-12 C²/Nm²

To Find: Electric intensity at a point r = 1 m

Solution:

Surface charge density is given by

Electric intensity at appoint outside charged sphere is given by

Ans: Electric intensity at a point 1 m from centre of the charged sphere is 2.825 x 10⁴ N/C

Example – 06:

A uniformly charged metal sphere of radius 1.2 m has a surface charge density of 16 µC/m². Find the charge on the sphere. What is the electric flux emanating from the sphere?

Given: Surface charge density = 16 µC/m²= 16 x 10^-6 C/m², radius of sphere = R = 1.2 m, k = 1, ε_o = 8.85 x 10^-12 C²/Nm²

To Find: Charge on sphere = q = ?, electric intensity flux = ϕ = ?

Solution:

Surface charge density is given by

Electric flux is given by

Ans: Charge on the sphere is 290 µC and electric flux is 3.277 x 10⁷ Nm²/C

Example – 07:

A metal sphere of diameter 20 cm is charged with 4π µC. Find the surface density of charge on the sphere and the distance of a point from the centre of the sphere where the electric intensity is 2.26 x 10⁵ V/C.

Given: Diameter of sphere = 20 cm, radius of sphere = R = 20/2 = 10 cm = 0.10 m, Charge = 4π µC = 4π x 10^-6 C, k = 1, ε_o = 8.85 x 10^-12 C²/Nm², Electric intensity = E = 2.26 x 10⁵ N/C

To Find: distance of point from centre = r =?

Solution:

Surface charge density is given by

Electric intensity at a point outside charged sphere is given by

Ans: Surface charge density = σ = 10^-4 C/m² and the distance of the point from the centre of the sphere where the electric intensity is 2.26 x 10⁵ N/C is 7.07 m

Example – 08:

The electric flux due to a point charge q passing through a sphere of radius 15 cm is 12 x 10³ Nm²/C. What would be the flux due to the charge through a sphere of radius 18 cm? Find q.

Given: Radius of sphere = R = 15 cm = 0.15 m, Electric flux = ϕ = 12 x 10³ Nm²/C, k = 1, ε_o = 8.85 x 10^-12 C²/Nm²

To Find: Electric flux when radius of sphere is 18 cm and charge = q =?

Solution:

The charge on the surface of sphere behaves like it is concentrated at the centre of the sphere. Hence the electric flux is independent of the radius of the sphere. Hence I case of a sphere of radius 18 cm, the electric flux will remain the same. 12 x 10³ Nm²/C.

Ans: the flux due to the charge through a sphere of radius 18 cm is also 12 x 10³ Nm²/C, and charge on the sphere is 1.062 x 10^-7 C

Example – 09:

A point charge is enclosed by spherical Gaussian surface of radius 5 cm. If electrical flux passing through it is 5 x 10³ Nm²/C. Find the charge and flux density over the surface of the sphere.

Given: Radius of sphere = R = 5 cm = 0.05 m, Electric flux = ϕ = 5 x 10³ Nm²/C, k = 1, ε_o = 8.85 x 10^-12 C²/Nm²

To Find: charge = q =?, flux density over the surface of sphere = ?

Solution:

Electric flux is given by

Surface charge density is given by

Electric flux density over the surface of a sphere is given by

Ans: The charge is 1.408 x 10^-8 C and flux density on the surface of the sphere is 1.591 x 10⁵ N/C.

Example – 10:

A hollow metal ball 10 cm in diameter is given a charge of 0.01 C. What is the intensity of the electric field at a point 20 cm from the centre of the ball.

Given: Charge = 0.01 C, radius of sphere = R = 10 cm = 0.1 m, k = 1, ε_o = 8.85 x 10^-12 C²/Nm²

To Find: Electric intensity at a point r = 20 cm = 0.2 m

Solution:

Surface charge density is given by

Electric intensity at a point outside charged sphere is given by

Ans: Electric intensity at a point at a distance of 0.2 m from centre of cthe harged sphere is 2.248 x 10⁹ N/C

Previous Topic: Gauss’s Theorem and its Applications

Next Topic: Mechanical Force Per Unit Area of Charged Conductor

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In this article, we shall study the cause of mechanical force acting on a charged conductor and hence shall derive an expression for mechanical force per unit area of the conductor.

Expression for Mechanical Force per Unit Area of a Charged Conductor:

Every element of a charged conductor experiences a normal outward mechanical force. This is the result of repulsive force from similar charges present on the rest of the surface of the conductor.

Consider a small element, dS on the surface of a charged conductor. If σ is the surface charge density and the charge carried by the element dS be dq.

dq =  σ .dS     ….(1)

Consider a point P just outside the surface near the element dS. The electric intensity at a point P is given by

E = σ /ε_ok       ….(2)

The direction of the intensity is normally outwards.

Now, consider a point Q inside the conductor very near to element dS.

Here E₂ is the electric intensity due to the charge on the rest of the conductor. Hence repulsive force experienced by element dS carrying charge dq due to is given by

This is the mechanical force over dS area of a charged conductor. S.I. unit of force per unit area is  N/m².

Expression for Energy per Unit Volume (Energy Density) of a Medium:

During the process of charging a conductor, work has to be done to bring a charge on the surface of the conductor. This work is stored in the electric field surrounding the conductor in the form of electrostatic energy. A charged conductor is in an electric field.

The mechanical force acting on it over dS area is given by

The force is directed normally outwards.

Under the action of force, of the element dS moves outward through a distance dl, then the work done by the force is given by,

The work done dW is stored in the medium as electric potential energy dU. i.e. dW = dU. Then the energy per unit volume or energy density is given by

This is an expression for energy per unit volume (energy density) of a medium. Its S.I. unit is joule per cubic metre (J/m³)

Note:

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Next Topic: Dielectrics (Dielectric Materials)

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In this article, we shall study dielectric materials (dielectrics), their working and types.

Dielectrics are non-conducting substances. In these materials, free-moving electrons are not present. Thus there is no question of the movement of charge. When the dielectric material is kept in an external electric field, a dipole movement is introduced in the dielectric due to the stretching and reorientation of the molecules of the dielectric. Due to this molecular dipoles, a charge is created on the surface of the dielectric. Which produces an electric field that opposes the external electric field.

• Examples of solid dielectrics: Ceramics, glasses, plastics, rubber, mica asbestose.
• Examples of liquid dielectrics: Mineral oil, silicone oil, magnesia.

Types of Dielectrics:

Polar Dielectrics:

A polar molecule is one in which the centre of gravity of positive nuclei and revolving electrons do not coincide. Examples: HCl, H₂O, N₂O molecules.

Polar molecules have a permanent dipole moment. Thus they behave like a tiny electric dipole. In the absence of an external electric field, the tiny molecular electric dipoles are randomly arranged due to thermal agitation. Thus the net electric dipole moment of the polar dielectric is zero. In presence of an external electric field, these tiny molecular electric dipoles align themselves in the direction of the external electric field. Thus there is net dipole moment in the direction of the field.

Non-Polar Dielectrics:

A non-polar molecule is one in which the centre of gravity of positive nuclei and revolving electrons coincide. Examples: O₂, H₂, CO₂, Polyethene, polystyrene.

Due to symmetry non-polar molecules do not have a permanent dipole moment. When non-polar molecules are subjected to an external electric field, the positive and negative charges in the molecules are displaced in the opposite direction. This displacement continues until the external force on the charges is balanced by restoring force due to the internal molecular field. Thus nonpolar molecules acquire induced dipole moment in the external electric field and are said to be polarised in the external electric field. The induced electric moments of different molecules add up and give rise to the net dipole moment.

Polarization:

When polar or a non-polar dielectric are kept in the external electric field, their molecules acquire induced dipole moment and the dielectric is said to be polarized in the external electric field. The phenomenon is known as polarization. Polarization is defined as dipole moment per unit volume and is given by

This relation is true for linear isotropic dielectrics.

Linear isotropic dielectrics are those dielectrics in which induced dipole moment is induced in the same direction of the external field and is proportional to the field strength.

Behaviour of Dielectric Slab Which is Subjected to External Electric Field:

Consider a thin slab of the dielectric of permittivity placed in an external uniform electric field. Irrespective of the nature of their molecules (polar or non-polar) the dielectric gets polarised. Due to polarization molecules are oriented such that the negative charges are on the left side and positive charges on the right side. The net electric charge on the dielectric is zero. The charges so obtained on the surface of the dielectric slab are called polarization charges.

Thus polarized dielectric is equivalent to two charged surfaces with polarization charges. These charges oppose the external electric field and thereby weaken the original field within the dielectric.

Polarization on the dielectric slab can be defined as the amount of induced surface charge per unit area (area right angle to the external electric field) of the surface.

Where, P = Polarization

q_P = Polarization charges

σ_P = Charge density of polarization charges

A = Area of cross-section of dielectric

P is a vector quantity and is directed from negative induced charges to positive induced charges.

Proof:

Polarization is defined as dipole moment per unit volume

Where,Q = nq = Net charge on all dipoles

n = number of molecular dipoles

q = charge of each dipole

l = length

A = area of cross-section of dipole

Thus polarization can be also defined as the amount of induced surface charge per unit area of the surface.

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Next Topic: Concept of Capacity of a Conductor

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In this article, we shall learn numerical problems to calculate the gravitational force of attraction between two bodies.

Example – 01:

Calculate the gravitational force of attraction between two metal spheres each of mass 90 kg, if the distance between their centres is 40 cm. Given G = 6.67 x 10^-11 N m²/kg². Will the force of attraction be different if the same bodies are taken on the moon, their separation remaining the same?

Given: Mass of first body = m₁ = 90 kg, mass of second body = m₂ = 90 kg, Distance between masses = r = 40 cm = 40 x 10^-2 m, G = 6.67 x 10^-11 N m²/kg² .

To Find: Force of attraction = F =?

Solution:

By Newton’s law of gravitation

If the same bodies are taken on the moon, their separation remaining the same, the force of attraction between the two bodies will remain the same, because the force of attraction between two bodies is unaffected by the presence of the third body and medium between the two bodies.

Ans: The force of attraction between two metal spheres is 3.377 x 10^-6 N

The force of attraction between two bodies remains the same

Example – 02:

Find the gravitational force of attraction between the moon and the earth if the mass of the moon is 1/81 times the mass of earth. G = 6.67 x 10^-11 N m²/kg², radius of moon’s orbit is 3.58 x 10⁵ km. Mass of the earth = 6 x 10²⁴ Kg.

Given: Mass of Moon =  1/81 times the mass of earth, m_m = 1/81 m_e ,Distance between the moon and earth  = r = 3.58 x 10⁵ km = 3.58 x 10⁸ m, G = 6.67 x 10^-11 N m²/kg² . Mass of earth = M_e = 6 x 10²⁴ Kg

To Find: Force of attraction = F =?

Solution:

By Newton’s law of gravitation

Ans: The gravitational force of attraction between the moon and the earth is 2.213 x 10²⁰ N

Example – 03:

Two bodies of masses 5 kg and 6 x 10²⁴ kg are placed with their centres 6.4 x 10⁶ m apart. Calculate the gravitational force of attraction between the two masses. Also, find the initial acceleration of two masses assuming no other forces act on them.

Given: Mass of first body = m₁ = 5 kg, mass of second body = m₂ = 6 x 10²⁴ kg, Distance between masses = r = 6.4 x 10⁶ m, G = 6.67 x 10^-11 N m²/kg² .

To Find: Force of attraction between two masses = F =? Initial accelerations of the two masses =?

Solution:

By Newton’s law of gravitation

Initial acceleration of body of mass 5 kg

By Newton’s second law of motion F = ma

Thus a = F/m = 48.85 / 5 = 9.77 m/s²

Initial acceleration of body of mass 6 x 10²⁴ kg

By Newton’s second law of motion F = ma

Thus a = F/m = 48.85 / 6 x 10²⁴ = 8.142 x 10^-24 m/s²

Ans: The force of attraction between the two masses = 48.85 N

The Initial acceleration of body of mass 5 kg is 9.77 m/s² and

That of body of mass 6 x 10²⁴ kg is 8.142 x 10^-24 m/s².

Example – 04:

A sphere of mass 40 kg is attracted by another spherical mass of 15 kg by a force of 9.8 x 10^-7 N when the distance between their centres is 0.2 m. Find G.

Given: Mass of first body = m₁ = 40 kg, mass of second body = m₂ = 15 kg, force between them = F = 9.8 x 10^-7 N, Distance between the masses = r = 0.2 m.

To Find: Universal gravitation constant = G =?

Solution:

By Newton’s law of gravitation

Ans: The value of universal gravitation constant is 6.533 x 10^-11 N m²/kg².

Example – 05:

A sphere of mass 100 kg is attracted by another spherical mass of 11.75 kg by a force of 19.6 x 10^-7 N when the distance between their centres is 0.2 m. Find G.

Given: Mass of first body = m₁ = 100 kg, mass of second body = m₂ = 11.75 kg, distance between masses = r = 0.2 m, force between them = F =  19.6 x 10^-7 N,

To Find: Universal gravitation constant = G =?

Solution:

By Newton’s law of gravitation

Ans: The value of universal gravitation constant is 6.672 x 10^-11 N m²/kg².

Example – 06:

The distance of a planet from the earth is 2.5 x 10⁷ km and the gravitational force between them is 3.82 x 10¹⁸ N. Mass of the planet and earth are equal, each being 5.98 x 10²⁴ kg. Calculate the universal gravitation constant.

Given: Mass of Planet = m₁ = 5.98 x 10²⁴ kg, mass of earth = m₂ = 5.98 x 10²⁴ kg, distance between them = r = 2.5 x 10⁷ km = 2.5 x 10¹⁰ m, force between them = F =  19.6 x 10^-7 N,

To Find: Universal gravitation constant = G =?

Solution:

By Newton’s law of gravitation

Ans: The value of universal gravitation constant is 6.676 x 10^-11 N m²/kg².

Example – 7:

Three 5 kg masses are kept at the vertices of an equilateral triangle each of side of 0.25 m. Find the resultant gravitational force on any one mass. G = 6.67 x 10^-11 S.I. units.

Given:  m₁ = 5 kg, m₂ = 5 kg, m₃ = 5 kg, r = 0.25 m, G = 6.67 x 10^-11 N m²/kg².

To find: Force on m₁ =?

Solution:

By Newton’s law of gravitation, the force on mass m₁ due to mass m₂.

By Newton’s law of gravitation, the force on mass m₁ due to mass m₃.

The angle between F₁₂ and F₁₃ is 60°. (Angle of an equilateral triangle). The net force on m₁ is given by

The two forces are equal, hence their resultant act along angle bisector towards centroid.

Ans:  Force on any mass is 4.621 x 10^-8 N towards the centroid

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In this article, we shall study Newton’s law of gravitation, its universality, and universal gravitation constant.

Newton’s Law of Gravitation:

Statement:

Every particle of matter in the universe attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Explanation:

Let ‘m₁’ and ‘m₂’ be the masses of two particles separated by a distance r as shown. According to Newton’s Law of gravitation, these particles will attract each other by a force ‘F’ such that

Where ‘G’ is a constant of proportionality and known as Universal gravitation constant. The value of ‘G’ in S.I. system is 6.673 10^-11 N m² kg^-2 and in c.g.s. system is 6.673 10^-8 dyne cm² g^-2.

The mathematical expression for the law of gravitation is sometimes written as

The negative sign indicates the force of attraction.

Newton’s Law of Gravitation in Vector Form:

Force of Gravitation:

The force of attraction between two material bodies in the universe is known as the force of gravitation.

If one of the body is the earth or some other planet or natural satellite then the force of gravitation is called the force of gravity.

Characteristics of Gravitational Force:

• The gravitational force between two bodies forms the action-reaction pair. The gravitational force between two masses is always that of attraction. If the first body attracts the second body with force F (direction of force from the second body to the first body), then the second body attracts the first body with equal force F  (direction of force from the first body to the second body).
• The gravitational force between two masses is always acting along the line joining the centre of the two masses. Hence it is a central force.
• The gravitational force between two masses is independent of the medium between the two masses. It means the gravitational force between two masses is the same when they are kept in a vacuum or in water or in the air. This fact rules out the possibility of making gravity screens.
• The gravitational force between two masses is independent of their sizes or distribution of mass of the bodies.
• The gravitational force between two bodies does not depend upon the presence or the absence of other bodies.
• If the masses of the body are small, the gravitational force between them is negligible. If the masses are large like that of the sun and the earth, the gravitational force of attraction is considerable.
• The gravitational force between two bodies is called action – at – a distance type of interaction, because the two particles interact even though they are not in contact with each other. Thus gravitational force is a non-contact force.
• Gravitational force is a conservative force because the work done by the gravitational force is independent of the path between the initial and final position.

The universality of Newton’s Law of Gravitation:

Newton’s law of gravitation is also called as the universal law of gravitation because

• It is applicable to all material bodies irrespective of their sizes. It is applicable to very minute particles like atoms, electrons at the same time it is applicable to heavenly bodies like planets, stars etc.
• The law is applicable to all material bodies irrespective of the distance between them. It is applicable to interatomic distances at the same time it is applicable to stellar distances i.e. the distance between stars.

Evidence Supporting Newton’s Law of Gravitation:

• The Earth moves around the Sun under the gravitational influence of the Sun on the Earth.
• The Moon moves around the Earth under the gravitational influence of the Earth on the Moon.
• The high tide and low tide are caused due to the gravitational influence of the Moon on the Earth.
• The times of lunar eclipses and solar eclipses calculated on the basis of Newton’s law of gravitation are found to be approximately correct.

Difference Between Gravitation and Gravitational Force:

Gravitation is a natural phenomenon by which material objects attract one another. While the gravitational force is the force of attraction which keeps two bodies in the universe bonded together.

Factors Affecting the Gravitational Force Between Two Bodies:

• The gravitational force of attraction between two bodies is directly proportional to the product of masses. If the distance between two masses is constant, then an increase in the mass of one of the two or of both increases the gravitational force of attraction between the two bodies.
• The gravitational force of attraction between two bodies is inversely proportional to the square distance between the two bodies. If the masses are kept constant, then the increase in distance between the two bodies decreases the gravitational force of attraction between the two bodies.
• The gravitational force between two masses is independent of the medium between the two masses.
• The gravitational force between two bodies does not depend upon the presence or the absence of other bodies.

S.I. Unit of G:

By Newton’s law of gravitation

The SI unit of constant of gravitation is N m² kg^-2 and c.g.s. unit is dyne cm² g^-2.

Dimensions of G:

By Newton’s law of gravitation

Hence the dimensions of universal gravitation constant are [M^-1 L³ T^-2]

Definition of G:

The gravitational force between two point masses ‘m₁’ and ‘m₂’ separated by distance ‘r; is given by

Let r = 1 unit, m₁ = m₂ = 1 unit, then G = F

Hence, the universal gravitational constant is the numerical value of the force between two unit masses kept at a unit distance from each other.

The value of G is very small and gravitational forces are small unless the masses of the two attracting bodies are large. If the value of G becomes 100 times its present value, then the earth’s attraction would be so large that we would be crushed to the earth. If the value of G becomes 1/100 times its present value, then we would be able to jump from a multi-story building.

Principle of Superposition of Forces:

Next Topic: Numerical Problems on Newton’s Law of Gravitation

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