A charge +q is placed at the origin O of X-Y | vedclass

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(A) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by $V = \frac{q}{4 \pi \varepsilon_{0} r}$.Since the electric field

Vedclass · Accepted Answer

$\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{a-b}{a b}\right)$

Vedclass · Answer

(A) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by $V = \frac{q}{4 \pi \varepsilon_{0} r}$.Since the electric field is conservative,the work done in moving a charge $Q$ from point $A$ to point $B$ is independent of the path and is given by $W = Q(V_B - V_A)$.The potential at point $A(a, 0)$ is $V_A = \frac{q}{4 \pi \varepsilon_{0} a}$.The potential at point $B(0, b)$ is $V_B = \frac{q}{4 \pi \varepsilon_{0} b}$.Therefore,the work done is $W = Q \left( \frac{q}{4 \pi \varepsilon_{0} b} - \frac{q}{4 \pi \varepsilon_{0} a} \right)$.$W = \frac{q Q}{4 \pi \varepsilon_{0}} \left( \frac{1}{b} - \frac{1}{a} \right) = \frac{q Q}{4 \pi \varepsilon_{0}} \left( \frac{a - b}{a b} \right)$.

$A$ charge $+q$ is placed at the origin $O$ of $X-Y$ axes as shown in the figure. The work done in taking a charge $Q$ from $A$ to $B$ along the straight line $AB$ is

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