Question

When the positively charged hanging pendulum bob is made fixed, the work done in slowly shifting a unit positive charge from infinity to $P$ is $V$. If the pendulum is free to move, the corresponding work done is $V^{{}^{\prime }}$. Then :

• A. $V=V^{{}^{\prime }}$
• B. $V>V^{{}^{\prime }}$
• C. $V<V^{{}^{\prime }}$
• D. $V\le V^{{}^{\prime }}$

Answer 1

The correct option is B $V>V^{{}^{\prime }}$

If bob is positively charged, it will be repelled by unit positive charge and in second case, distance between charges may increase. Thus final energy in second case may be less than in first case. Hence, less work may be done in second case.
$\Rightarrow V>V^{{}^{\prime }}$