A solid conducting sphere with charge Q is su | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the radius of the inner solid sphere be $r$ and the radius of the outer shell be $R$.According to Gauss's law,the electric field $E$ in the re

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let the radius of the inner solid sphere be $r$ and the radius of the outer shell be $R$.According to Gauss's law,the electric field $E$ in the region between the sphere and the shell $(r The charge enclosed is the charge on the inner sphere,which is $Q$.Thus,$E = \frac{Q}{4 \pi \epsilon_0 x^2}$.The potential difference $V$ between the surface of the inner sphere and the surface of the outer shell is given by $V = \int_{r}^{R} E \, dx = \int_{r}^{R} \frac{Q}{4 \pi \epsilon_0 x^2} \, dx$.Evaluating the integral,$V = \frac{Q}{4 \pi \epsilon_0} \left[ -\frac{1}{x} \right]_{r}^{R} = \frac{Q}{4 \pi \epsilon_0} \left( \frac{1}{r} - \frac{1}{R} \right)$.Since the potential difference $V$ depends only on the charge $Q$ of the inner sphere and the radii $r$ and $R$,changing the charge on the outer shell does not affect the potential difference between the two surfaces.Therefore,the new potential difference remains $V$.

Similar Questions

$A$ charge of total amount $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R$ $(R > r)$ such that the surface charge densities on the two spheres are equal. The electric potential at the common centre is

$A$ hollow metal sphere of radius $15 \,cm$ is charged such that the potential on its surface is $20 \,V$. Then, the potential at the centre of the sphere is (in $V$)

Which of the following is a volt?

Energy of electrons can be increased by allowing them

Vedclass Test Series

Exam Paper Generator

Online Exam Module