On the axis and on the equator of an electric | vedclass

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Question

(D) The electric potential $V$ due to an electric dipole at a point $(r, 	heta)$ is given by the formula:$V(r, 	heta) = \frac{1}{4 \pi \varepsilon_0

Vedclass · Accepted Answer

on the axis $V 
eq 0$ and on equator $V = 0$

Vedclass · Answer

(D) The electric potential $V$ due to an electric dipole at a point $(r, 	heta)$ is given by the formula:$V(r, 	heta) = \frac{1}{4 \pi \varepsilon_0} \frac{p \cos 	heta}{r^2}$$1$. On the axis of the dipole,the angle $	heta$ is either $0$ or $\pi$ radians. Substituting this into the formula:$V = \frac{1}{4 \pi \varepsilon_0} \frac{p \cos(0)}{r^2} = \frac{p}{4 \pi \varepsilon_0 r^2}$ or $V = \frac{1}{4 \pi \varepsilon_0} \frac{p \cos(\pi)}{r^2} = -\frac{p}{4 \pi \varepsilon_0 r^2}$Thus,on the axis,$V 
eq 0$.$2$. On the equatorial plane of the dipole,the angle $	heta = \frac{\pi}{2}$ radians. Substituting this into the formula:$V = \frac{1}{4 \pi \varepsilon_0} \frac{p \cos(\pi/2)}{r^2} = 0$ (since $\cos(\pi/2) = 0$)Thus,on the equator,$V = 0$.Therefore,the correct option is $D$.

On the axis and on the equator of an electric dipole for all points . . . . . . .

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