A spherically symmetric charge system is centered at the origin. Given electric potential

$V = \frac{Q}{4 π ε_{0} R_{0}} (r \leq R_{0}),$ $V = \frac{Q}{4 π ε_{0} r} (r > R_{0}) .$ [2006]

Question

A spherically symmetric charge system is centered at the origin. Given electric potential

$V = \frac{Q}{4 π ε_{0} R_{0}} (r \leq R_{0}),$ $V = \frac{Q}{4 π ε_{0} r} (r > R_{0}) .$ [2006]

Smart Achievers · Accepted Answer

(1, 2, 4)

The given $V - r$ graph is of a charged conducting sphere of radius $R_{0}$

(1) The whole charge $Q$ will be enclosed in a sphere of diameter $2 R_{0}$ .

(2) Electric field $E = 0$ inside the sphere. Hence electric field is discontinued at $r = R_{0}$ .

(3) Changes in $V$ and $E$ are continuously present for $r > R_{0}$ .

Option (3) is incorrect.

(4) For $r < R_{0}$ , the potential $V$ is constant and the electric intensity is zero. Obviously the electrostatic energy is zero for $r < R_{0}$

Want Us to Email you About Special Offers & Updates?