Question

If 4 × 10-3 J of work is done in moving a particles carrying a charge of 16 × 10-6 from infinity to point p. What will be the potential at a point?

Answer 1

w = qV
4 × 10^-3 / 16 × 10^-6 = 250V

Answer 2

$Work \ done (W) = 4\times 10^{-3} J\\Charge (Q) = 16\times 10^{-6}\\Potential \ Difference (V) = ?\\\\ We \ know,\\V = \frac{W}{Q}\\V = \frac{4\times 10^{-3}}{15\times 10^{-6}} \\V = \frac{1000}{4}\\V = 250 \\\\Hence, }{Potential \ Difference = 250 V}$
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$\uwave{@DuragPal Singh}$