Two points P and Q are maintained at the pote | vedclass

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(D) The work done $W$ in moving a charge $q$ from point $P$ to point $Q$ is given by $W = q(V_Q - V_P)$.Here,the charge $q$ is the total charge of $10

Vedclass · Accepted Answer

$2.24 	imes 10^{-16} \, J$

Vedclass · Answer

(D) The work done $W$ in moving a charge $q$ from point $P$ to point $Q$ is given by $W = q(V_Q - V_P)$.Here,the charge $q$ is the total charge of $100$ electrons,so $q = 100 	imes (-e) = 100 	imes (-1.6 	imes 10^{-19} \, C) = -1.6 	imes 10^{-17} \, C$.The potential difference is $V_Q - V_P = -4 \, V - 10 \, V = -14 \, V$.Therefore,the work done is $W = (-1.6 	imes 10^{-17} \, C) 	imes (-14 \, V) = 22.4 	imes 10^{-17} \, J = 2.24 	imes 10^{-16} \, J$.

Two points $P$ and $Q$ are maintained at the potentials of $10\, V$ and $-4\, V$,respectively. The work done in moving $100$ electrons from $P$ to $Q$ is:

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