The electrostatic potential on the surface of | vedclass

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

Question

(A) For a uniformly charged spherical shell,the electrostatic potential inside the shell is constant and equal to the potential on its surface.Given:

Vedclass · Accepted Answer

$120 \ V, 120 \ V, 80 \ V$

Vedclass · Answer

(A) For a uniformly charged spherical shell,the electrostatic potential inside the shell is constant and equal to the potential on its surface.Given: Potential on the surface $V_{surface} = 120 \ V$ and radius $R = 10 \ cm$.$1$. At the centre $(r = 0 \ cm)$: Since $r $2$. At a distance $r = 5 \ cm$ from the centre: Since $r $3$. At a distance $r = 15 \ cm$ from the centre: Since $r > R$,the shell acts as a point charge. The potential is given by $V = \frac{kQ}{r}$.We know $V_{surface} = \frac{kQ}{R} = 120 \ V$,so $kQ = 120 	imes 10 = 1200 \ V \cdot cm$.Thus,$V_{r=15cm} = \frac{1200}{15} = 80 \ V$.Therefore,the potentials are $120 \ V, 120 \ V, 80 \ V$.

The electrostatic potential on the surface of a uniformly charged spherical shell of radius $R = 10 \ cm$ is $120 \ V$. The potential at the centre of the shell,at a distance $r = 5 \ cm$ from the centre,and at a distance $r = 15 \ cm$ from the centre of the shell respectively,are:

Similar Questions

$A$ pellet carrying a charge of $0.5\, C$ is accelerated through a potential difference of $2,000\, V$. What is the kinetic energy attained by the pellet?

An electric charge $10^{-6} \mu C$ is placed at the origin $(0,0) \text{ m}$ of an $X-Y$ coordinate system. Two points $P$ and $Q$ are situated at $(\sqrt{3}, \sqrt{3}) \text{ m}$ and $(\sqrt{6}, 0) \text{ m}$ respectively. The potential difference between the points $P$ and $Q$ will be:

Statement-$1$: For a charged particle moving from point $P$ to point $Q$,the net work done by the electrostatic force is independent of the path connecting point $P$ to point $Q$.
Statement-$2$: The net work done by a conservative force on an object in a closed loop is zero.

For a uniformly charged thin spherical shell,the electric potential $(V)$ radially away from the center $(O)$ of the shell can be graphically represented as:

Particles $A$ and $B$ have charges $+q$ and $+4q$ respectively. Both have the same mass $m$. If they are released from rest and accelerated through the same potential difference $V$,the ratio of their velocities $v_A/v_B$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module