Charges +q and -q are placed at points A and | vedclass

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(A) The work done $W$ in moving a charge $Q$ from point $C$ to point $D$ is given by $W = Q(V_D - V_C)$.At point $C$ (midpoint of $AB$): The distance

Vedclass · Accepted Answer

$-\frac{qQ}{6\pi \epsilon_0 L}$

Vedclass · Answer

(A) The work done $W$ in moving a charge $Q$ from point $C$ to point $D$ is given by $W = Q(V_D - V_C)$.At point $C$ (midpoint of $AB$): The distance from $A$ is $L$ and from $B$ is $L$. $V_C = \frac{kq}{L} + \frac{k(-q)}{L} = 0$.At point $D$ (which is at a distance $L$ from $B$ and $2L$ from $C$,thus $3L$ from $A$): The distance from $A$ is $3L$ and from $B$ is $L$. $V_D = \frac{kq}{3L} + \frac{k(-q)}{L} = \frac{kq}{L} (\frac{1}{3} - 1) = -\frac{2kq}{3L}$.Substituting these values into the work formula:$W = Q(V_D - V_C) = Q(-\frac{2kq}{3L} - 0) = -\frac{2kqQ}{3L}$.Since $k = \frac{1}{4\pi \epsilon_0}$,we have:$W = -\frac{2qQ}{3L(4\pi \epsilon_0)} = -\frac{qQ}{6\pi \epsilon_0 L}$.

Charges $+q$ and $-q$ are placed at points $A$ and $B$ separated by a distance $2L$. $C$ is the midpoint between $A$ and $B$. The work done in moving a charge $+Q$ along the semi-circular path $CRD$ is .......

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