Question

A spherical conductor of the radius $ 12\;cm $ has a charge of $ 1.6 \times {10^{ - 7}}C $ distributed uniformly on its surface. What is the electric field
(a) Inside the surface
(b) Just outside the surface
(c) At a point $ 18\;cm $ from the center of the sphere?

Answer 1

Hint : The electric field at a point depends on the charge and the distance of the point from the center outside the surface of the conductor. The formula to calculate the value of an electric field is $ E = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{Q}{{{R^2}}} $ , where R is the distance from the center.

Complete Step By Step Answer:
From the given data, we can write
Charge $ Q = 1.6 \times {10^{ - 7}}C $
Radius of conductor $ R = 12cm = 0.12m $
The electric field can be calculated by the formula
 $ E = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{Q}{{{R^2}}} $
Here, we know that the value of the constant is $ \dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}} $
Let us consider each case at a time
Case (a): Inside the surface
We know that the charge of a conductor resides only on the surface of the conductor. As we are given that the charge is uniformly distributed, for the points inside the conductor the charges are equal and opposite in direction. Hence they cancel each other.
Therefore, the electric field inside the surface is zero.
Case (b): Just outside the surface
For the points outside the surface, the electric field is not zero and can be calculated by the given formula.
For just outside the surface, the distance from the center to the point is $ R = 0.12m $
Hence, the electric field is
 $ \therefore E = 9 \times {10^9}N{m^2}{C^{ - 2}} \times \dfrac{{1.6 \times {{10}^{ - 7}}C}}{{{{(12 \times {{10}^{ - 2}}m)}^2}}} $
Without considering the units,
 $ \therefore E = 9 \times {10^9} \times \dfrac{{1.6 \times {{10}^{ - 7}}}}{{144 \times {{10}^{ - 4}}}} $
 $ \therefore E = 9 \times {10^9} \times \dfrac{{16 \times {{10}^{ - 8}}}}{{144 \times {{10}^{ - 4}}}} $
Multiplying the powers and the values,
 $ \therefore E = \dfrac{{9 \times {{10}^9} \times 16 \times {{10}^{ - 8}} \times {{10}^4}}}{{144}} $
 $ \therefore E = {10^5}N{C^{ - 1}} $
Thus, the electric field just outside the surface is $ {10^5}N{C^{ - 1}} $ .
Case (c): At a point $ 18\;cm $ from the center of the sphere
For the point outside the surface, the radius R is considered as the distance of the point from the center.
Thus here, $ R = 18cm = 0.18m $
Hence, the electric field is
 $ \therefore E = 9 \times {10^9}N{m^2}{C^{ - 2}} \times \dfrac{{1.6 \times {{10}^{ - 7}}C}}{{{{(18 \times {{10}^{ - 2}}m)}^2}}} $
Without considering the units,
 $ \therefore E = 9 \times {10^9} \times \dfrac{{1.6 \times {{10}^{ - 7}}}}{{324 \times {{10}^{ - 4}}}} $
 $ \therefore E = 9 \times {10^9} \times \dfrac{{16 \times {{10}^{ - 8}}}}{{324 \times {{10}^{ - 4}}}} $
Multiplying the powers and the values,
 $ \therefore E = \dfrac{{9 \times {{10}^9} \times 16 \times {{10}^{ - 8}} \times {{10}^4}}}{{324}} $
 $ \therefore E = 0.444 \times {10^5}N{C^{ - 1}} $
Thus, the electric field just outside the surface is $ 0.444 \times {10^5}N{C^{ - 1}} $ .

Note :
The point to note here is that the electric field inside a uniformly charged surface is always zero. The electric field is also zero on the surface of the body. But an electric field exists for a point just outside the surface.