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Home » A spherical conductor of radius has a charge of distributed uniformly on its surface. What is the electric field (1) inside the sphere. (2) just outside the sphere. (3) at a point from the centre of sphere.
A spherical conductor of radius 12 \mathrm{~cm} has a charge of 1.6 \times 10^{-7} \mathrm{C} distributed uniformly on its surface. What is the electric field
(1) inside the sphere.
(2) just outside the sphere.
(3) at a point 18 \mathrm{~cm} from the centre of sphere.
CBSE | Class 12 | Electrostatic Potential and Capacitance | NCERT | Physics
A spherical conductor of radius 12 \mathrm{~cm} has a charge of 1.6 \times 10^{-7} \mathrm{C} distributed uniformly on its surface. What is the electric field
(1) inside the sphere.
(2) just outside the sphere.
(3) at a point 18 \mathrm{~cm} from the centre of sphere.

Solution:

(1) Given:

Radius of spherical conductor, r=12 \mathrm{~cm}=0.12 \mathrm{~m}

Charge is spread uniformly across the surface of the material. q=1.6 \times 10^{-7} \mathrm{C}.

When an electric field is created inside a spherical conductor, it is equal to zero.

(2) The relationship between the electric field E just outside the conductor and the conductor itself is described by the equation

\mathrm{E}=\frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q}{r^{2}}

Here,

\varepsilon_{0}= permittivity of free space and \frac{1}{4 \pi \epsilon_{o}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}

Therefore,

\mathrm{E}=\frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{(0.12)^{2}}=10^{5} \mathrm{NC}^{-1}

Therefore, just outside the sphere the electric field is 10^{5} \mathrm{NC}^{-1}.

(3) From the centre of the sphere the electric field at a point 18 \mathrm{~m}=\mathrm{E}_{1}.

From the centre of the sphere, the distance of point d=18 \mathrm{~cm}=0.18 \mathrm{~m}.

\mathrm{E}<em>{1}=\frac{1}{4 \pi \epsilon</em>{0}} \cdot \frac{q}{d^{2}}=\frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{\left(1.8 \times 10^{-2}\right)^{2}}=4.4 \times 10^{4} \mathrm{NC}^{-1}

So, from the centre of sphere the electric field at a point 18 \mathrm{~cm} away is 4.4 \times 10^{4} \mathrm{NC}^{-1}.

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