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

Science > Physics > Electrostatics > Electric Potential

The post Electric Potential appeared first on The Fact Factor.

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

Science > Physics > Electrostatics > Normal Electric Induction

The post Normal Electric Induction appeared first on The Fact Factor.

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.

Previous Topic: Normal Electric Induction

Next Topic: Numerical Problems on Electric Intensity Due to Charged Sphere

Science > Physics > Electrostatics > Gauss’s Theorem and its Applications

The post Gauss’s Theorem and its Applications appeared first on The Fact Factor.

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

Science > Physics > Electrostatics > Electric Intensity Due to Charged Sphere

The post Numerical Problems on Electric Intensity Due to Charged Sphere appeared first on The Fact Factor.

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:

Previous Topic: Numerical Problems on Electrical Intensity Due to Charged Sphere

Next Topic: Dielectrics (Dielectric Materials)

Science > Physics > Electrostatics > Mechanical force per Unit Area of Charge Conductor

The post Mechanical force per Unit Area of Charge Conductor appeared first on The Fact Factor.

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.

Previous Topic: Mechanical Force per Unit Area of Charged Conductor

Next Topic: Concept of Capacity of a Conductor

Science > Physics > Electrostatics > Dielectrics

The post Dielectrics appeared first on The Fact Factor.