An electric dipole is as shown in the figure. | vedclass

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(D) The electric potential due to a point charge $q$ at a distance $r$ is given by $V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r}$.For the given dipole,t

Vedclass · Accepted Answer

$\frac{aq}{2 \pi \epsilon_0(x^2-a^2)}$

Vedclass · Answer

(D) The electric potential due to a point charge $q$ at a distance $r$ is given by $V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r}$.For the given dipole,the distance of point $P$ from charge $-q$ is $(x+a)$ and from charge $+q$ is $(x-a)$.The potential due to $-q$ is $V_{-q} = \frac{1}{4 \pi \epsilon_0} \frac{(-q)}{(x+a)}$.The potential due to $+q$ is $V_{+q} = \frac{1}{4 \pi \epsilon_0} \frac{(+q)}{(x-a)}$.The total potential $V_P$ at point $P$ is the algebraic sum of potentials due to individual charges:$V_P = V_{-q} + V_{+q} = \frac{q}{4 \pi \epsilon_0} \left[ \frac{1}{x-a} - \frac{1}{x+a} \right]$.$V_P = \frac{q}{4 \pi \epsilon_0} \left[ \frac{(x+a) - (x-a)}{(x-a)(x+a)} \right] = \frac{q}{4 \pi \epsilon_0} \left[ \frac{2a}{x^2-a^2} \right]$.$V_P = \frac{2aq}{4 \pi \epsilon_0(x^2-a^2)} = \frac{aq}{2 \pi \epsilon_0(x^2-a^2)}$.

An electric dipole is as shown in the figure. The electric potential at point $P$ due to the dipole is $[\epsilon_0 = \text{permittivity of free space}]$.

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