The diagrams below show regions of equipotent | vedclass

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(A) The work done in moving a charge $q$ from point $A$ to point $B$ is given by the formula $W = q(V_B - V_A) = q \Delta V$.In each of the four diagr

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In all the four cases the work done is the same.

Vedclass · Answer

(A) The work done in moving a charge $q$ from point $A$ to point $B$ is given by the formula $W = q(V_B - V_A) = q \Delta V$.In each of the four diagrams provided:- Figure $(I)$: $V_A = 10 	ext{ V}$ and $V_B = 40 	ext{ V}$,so $\Delta V = 40 - 10 = 30 	ext{ V}$.- Figure $(II)$: $V_A = 10 	ext{ V}$ and $V_B = 40 	ext{ V}$,so $\Delta V = 40 - 10 = 30 	ext{ V}$.- Figure $(III)$: $V_A = 10 	ext{ V}$ and $V_B = 40 	ext{ V}$,so $\Delta V = 40 - 10 = 30 	ext{ V}$.- Figure $(IV)$: $V_A = 10 	ext{ V}$ and $V_B = 40 	ext{ V}$,so $\Delta V = 40 - 10 = 30 	ext{ V}$.Since the potential difference $\Delta V$ is $30 	ext{ V}$ in all four cases,the work done $W = q(30 	ext{ V})$ is the same for all diagrams.

The diagrams below show regions of equipotentials. $A$ positive charge $q$ is moved from $A$ to $B$ in each diagram.

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