Question

If $4\times {10}^{-3}J$ of work is done in moving a particle carrying a charge of $16\times {10}^{-6}C$ from infinity to a point. What will be the potential at that point?

Answer 1

Given,

Work done(W) = $4\times {10}^{-3}$

Charge(q) = $16\times {10}^{-6}$

We know,

$W=q\Delta V$

Putting all the values

$W=16\times {10}^{-6}(V-0)$

$Or,4\times {10}^{-3}=16\times {10}^{-6}\times V$

$So,V=\frac{1000}{4}=250V$

Hence, potential at that point is 250 V

WD=16×10−6(V−0)WD=16 \times 10^{-6}(V-