Name: Equipotential Surfaces and Its Properties
Uploaded: 2019-07-19
Duration: 7 min 22 s
Description: The surface over which the electric potential is the same everywhere is called an equipotential surface. Let's more about it and its properties in this video.

Question

The graphical variation of electric potential due to point charge $q_{1}$ and $q_{2}$ lies on the $x$ -axis at some separation $d$ , which is shown in the figure. If the origin is the point between the charges where potential is zero. Distance of $q_{2}$ from origin is $\frac{d}{4}$ . Find the distance of point $P$ (marked in the figure) from charge $q_{2}$ .

Accepted Answer

Given,

The electric potential at the origin due to charges $q_{1}$ and $q_{2}$ is zero.

Position of the charge $q_{2}$ from the origin is $\frac{d}{4}$ .

The separation between these two charges is $d$ .

Thus, the position of the charge $q_{1}$ from the origin is, $(d - \frac{d}{4}) = \frac{3 d}{4}$ .

Since the potential at the origin is zero,

$k \frac{q_{1}}{(\frac{3 d}{4})} + k \frac{q_{2}}{(\frac{d}{4})} = 0$

$\Rightarrow k \frac{q_{1}}{(\frac{3 d}{4})} = - k \frac{q_{2}}{(\frac{d}{4})}$

$\Rightarrow q_{2} = - \frac{q_{1}}{3}$ .

From the graph, the slope of the electric potential $V$ versus $x$ at point $P$ is zero.

That is, $slope = \frac{d V}{d x} = |E| = 0$ .

Where, $E$ is the electric field which is zero at point $P$ .

If $x$ is the distance of point $P$ from the charge $q_{2}$ then the electric field at $P$ is given by,

$E_{P} = \frac{k q_{1}}{{(d + x)}^{2}} + \frac{k q_{2}}{x^{2}} = 0$

$\Rightarrow \frac{q_{1}}{{(d + x)}^{2}} = - \frac{q_{2}}{x^{2}}$

$\Rightarrow \frac{q_{1}}{{(d + x)}^{2}} = - \frac{(- \frac{q_{1}}{3})}{x^{2}}$

$\Rightarrow 3 x^{2} = {(d + x)}^{2}$

$∴ x = \frac{d}{(\sqrt{3} - 1)}$ .

Important Questions on Electric Potential

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