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

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(A) The work done in moving a charge $Q$ from point $C$ to point $D$ is given by $W = Q(V_D - V_C)$.Point $C$ is the midpoint of $AB$ (distance $2L$),

Vedclass · Accepted Answer

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

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(A) The work done in moving a charge $Q$ from point $C$ to point $D$ is given by $W = Q(V_D - V_C)$.Point $C$ is the midpoint of $AB$ (distance $2L$),so $AC = L$ and $CB = L$. The potential at $C$ due to charges $+q$ at $A$ and $-q$ at $B$ is $V_C = \frac{1}{4\pi \epsilon_0} (\frac{q}{L} - \frac{q}{L}) = 0$.Point $D$ is on the semicircle with diameter $CD$. Since $C$ is the midpoint of $AB$,$D$ is at a distance $3L$ from $A$ and $L$ from $B$. The potential at $D$ is $V_D = \frac{1}{4\pi \epsilon_0} (\frac{q}{3L} - \frac{q}{L}) = \frac{1}{4\pi \epsilon_0} (\frac{q - 3q}{3L}) = -\frac{2q}{12\pi \epsilon_0 L} = -\frac{q}{6\pi \epsilon_0 L}$.Therefore,$W = Q(V_D - V_C) = Q(-\frac{q}{6\pi \epsilon_0 L} - 0) = -\frac{qQ}{6\pi \epsilon_0 L}$.

Column-$I$	Column-$II$
$A$. National Botanical Research Institute	$I$. Carolus Linnaeus
$B$. Indian Botanical Garden	$II$. Taxon
$C$. Binomial Nomenclature	$III$. Howrah (India)
$D$. $A$ unit of Classification	$IV$. Lucknow

Charges $+q$ and $-q$ are placed at points $A$ and $B$ respectively,which are at a distance $2L$ apart. $C$ is the midpoint between $A$ and $B$. The work done in moving a charge $+Q$ along the semicircle $CRD$ is:

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