There are two equipotential surface as shown in figure. The distance between them is r. The charge of -q coulomb is taken from the surface A to B, the resultant work done will be

Question

There are two equipotential surface as shown in figure. The distance between them is r. The charge of -q coulomb is taken from the surface A to B, the resultant work done will be (A) W=(1/4 π εo) (q/r) (B) W=(1/4 π ε0) (q/r2) (C) W=-(1/4 π ε0) (q/r2) (D) W= zero

Tardigrade Admin · Accepted Answer

The work done is given by =q(V2-V1)=0

Q. There are two equipotential surface as shown in figure. The distance between them is $r$ . The charge of $- q$ coulomb is taken from the surface $A$ to $B$ , the resultant work done will be

$W = \frac{1}{4 π ε _{o}} \frac{q}{r}$

$W = \frac{1}{4 π ε _{0}} \frac{q}{r ^{2}}$

$W = - \frac{1}{4 π ε _{0}} \frac{q}{r ^{2}}$

$W =$ zero

The work done is given by $= q (V_{2} - V_{1}) = 0$

Q. There are two equipotential surface as shown in figure. The distance between them is r. The charge of −q coulomb is taken from the surface A to B, the resultant work done will be

W=4πεo​1​rq​

W=4πε0​1​r2q​

W=−4πε0​1​r2q​

W= zero