Two charges +q and -q are placed at points A | vedclass

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(A) The work done in moving a charge $Q$ in an electrostatic field is path-independent and depends only on the initial and final positions. Work done

Vedclass · Accepted Answer

$\frac{-Qq}{6 \pi \epsilon_0 L}, \frac{-Qq}{6 \pi \epsilon_0 L}$

Vedclass · Answer

(A) The work done in moving a charge $Q$ in an electrostatic field is path-independent and depends only on the initial and final positions. Work done $W = Q(V_{final} - V_{initial})$. Here, the initial point is $C$ and the final point is $D$. Distance $AC = L$, $CB = L$, $BD = L$. Potential at $C$ $(V_C)$: $V_C = \frac{1}{4 \pi \epsilon_0} [\frac{q}{AC} + \frac{-q}{CB}] = \frac{1}{4 \pi \epsilon_0} [\frac{q}{L} - \frac{q}{L}] = 0$. Potential at $D$ $(V_D)$: $V_D = \frac{1}{4 \pi \epsilon_0} [\frac{q}{AD} + \frac{-q}{BD}] = \frac{1}{4 \pi \epsilon_0} [\frac{q}{3L} - \frac{q}{L}] = \frac{1}{4 \pi \epsilon_0} [\frac{q - 3q}{3L}] = \frac{-2q}{12 \pi \epsilon_0 L} = \frac{-q}{6 \pi \epsilon_0 L}$. Since the electrostatic force is conservative, the work done along any path from $C$ to $D$ is the same: $W_1 = W_2 = Q(V_D - V_C) = Q(\frac{-q}{6 \pi \epsilon_0 L} - 0) = \frac{-Qq}{6 \pi \epsilon_0 L}$. Thus, $W_1 = W_2 = \frac{-Qq}{6 \pi \epsilon_0 L}$.

Two charges $+q$ and $-q$ are placed at points $A$ and $B$ respectively, which are at a distance $2L$ apart. $C$ is the midpoint of $A$ and $B$. The work done in moving a charge $+Q$ along the semicircle $CSD$ $(W_1)$ and along the line $CBD$ $(W_2)$ are:

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