Electric Charges and Fields - Result Question 14

14. A hollow metal sphere of radius $R$ is uniformly charged. The electric field due to the sphere at a distance $r$ from the centre :

[2019]

(a) increases as $r$ increases for $r<R$ and for $r>R$

(b) zero as $r$ increases for $r<R$, decreases as $r$ increases for $r>R$

(c) zero as $r$ increases for $r<R$, increases as $r$ increases for $r>R$

(d) decreases as $r$ increases for $r<R$ and for $r>R$

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Answer:

Correct Answer: 14. (b)

Solution:

(b) Charge $Q$ will be distributed over the surface of hollow metal sphere.

(i) For $r<R$ (inside)

[At a point inside the hollow sphere]

By Gauss’s law, $\oint \overrightarrow{{}E} _{in} \cdot \overrightarrow{{}dS}=\frac{q _{en}}{\varepsilon_0}=0$

As enclosed charge is $=0$

So, $E _{\text{in }}=0$ the electric field inside the hollow sphere is always zero.

(ii) For $r>R$ (outside)

[At a point outside hollow sphere]

By Gauss’s law, $\oint \overrightarrow{{}E}_0 \cdot \overrightarrow{{}dS}=\frac{q _{en}}{\varepsilon_0} \quad(Q q _{en}=Q)$

$ \begin{aligned} & \therefore E_0 4 \pi r^{2}=\frac{Q}{\varepsilon_0} \\ & \therefore E_0 \propto \frac{1}{r^{2}} \end{aligned} $



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