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

A spherical drop of water carrying a charge of

30 pC

has a potential of

500 V

at its surface (with

V = 0

at infinity). (a) What is the radius of the drop? (b) If two such drops of same charge and radius combine to form a single spherical drop, what is the potential at the surface of the new drop? (c) What is the ratio of the surface charge density on the new drop to that on the original drop?

Accepted Answer

Given,

charge on water drop, $q = 30 pC$

$\Rightarrow q = 30 \times 10^{- 12} C$ .

Potential on its surface, $V = 500 V$ .

Potential at infinity, $V_{\infty} = 0$ .

(a) Potential of a sphere of radius $R$ ,

$V = \frac{k q}{R} . . . (1)$

$\Rightarrow 500 = \frac{9 \times 10^{9} \times 30 \times 10^{- 12}}{R}$

$\Rightarrow R = 5.4 \times 10^{- 4} m$ .

(b) $Volume of two small drops = Volume of large drop made by two small drops$

$\Rightarrow \frac{4}{3} π R^{3} + \frac{4}{3} π R^{3} = \frac{4}{3} π R_{b}^{3}$ .

Here, $R =$ radius of small drop, $R_{b} =$ radius of bigger drop, $Q_{b} =$ charge on bigger drop.

$\Rightarrow 2 R^{3} = R_{b}^{3}$

$\Rightarrow R_{b} = 2^{\frac{1}{3}} R . . . (2)$

Potential of larger drop, $(V_{b})$

$V_{b} = \frac{k Q_{b}}{R_{b}} . . . (3)$

$Charge on bigger drop = Total charge of both drops$

$\Rightarrow Q_{b} = 2 q . . . (4)$

From equations $(1)$ , $(2)$ , $(3)$ and $(4)$ , we get,

$V_{b} = \frac{k (2 q)}{(2^{\frac{1}{3}} R)} = {(2)}^{\frac{2}{3}} \times V$

$\Rightarrow V_{b} = {(2)}^{\frac{2}{3}} \times 500 = 793.7 V$ .

(c) Surface charge density $(σ_{1})$ of original drop,

$σ_{1} = \frac{q}{4 π R^{2}} . . . (5)$

Surface charge density $(σ_{2})$ of new drop,

$σ_{2} = \frac{Q}{4 π R_{b}^{2}} . . . (6)$

From equation $(5)$ and $(6)$ , we get,

$\frac{σ_{2}}{σ_{1}} = \frac{Q}{q} \times \frac{R^{2}}{R_{b}^{2}}$

$\Rightarrow \frac{σ_{2}}{σ_{1}} = \frac{2 q}{q} \times \frac{R^{2}}{{(2^{\frac{1}{3}} R)}^{2}}$ , from equation $(2)$

$\Rightarrow \frac{σ_{2}}{σ_{1}} = 2^{\frac{1}{3}} \approx 1.3$ .

Important Questions on Electric Potential

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