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

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Next Topic: Mechanical Force Per Unit Area of Charged Conductor

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

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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)

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

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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

Science > Physics > Electrostatics > Dielectrics

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