Name: Dielectrics in a Capacitor
Uploaded: 2019-07-19
Duration: 6 min 3 s
Description: Let us use our knowledge about dielectrics and capacitors to find out how the use of dielectrics can increase the capacitance of a system. Learn through this...

Question

A system consists of a ball of radius

R

carrying a spherically symmetric charge, and the surrounding space filled with a charge of volume density

ρ = \frac{α}{r}

, where

α

is a constant and

r

is the distance from the centre of the ball. Find the ball's charge at which the magnitude of the electric field strength vector is independent of

r

, outside the ball. How high is this strength? The permittivity of the ball and the surrounding space are assumed to be equal to unity.

Accepted Answer

In the above figure, a ball with radius $R$ and surrounding space filled with volume charge density $ρ = \frac{α}{r}$ , is shown.

Let the charge carried by the sphere be $q$ .

Then, using Gauss' theorem for a spherical surface having radius $r > R$ ,

$E \times 4 π r^{2} = \frac{q_{enclosed}}{ε_{0}} . . . (1)$

Here, $q_{enclosed} =$ Total charge of the ball $+$ Charge enclosed by the surrounding sphere having radius $r$ .

$= q + \int_{R}^{r} ρ 4 π r^{2} d r$

$= q + \int_{R}^{r} \frac{α}{r} 4 π r^{2} d r$

From equation $(1)$ , we get,

$E \times 4 π r^{2} = \frac{q}{ε_{0}} + \frac{1}{ε_{0}} \int_{R}^{r} \frac{α}{r} 4 π r^{2} d r$

$\Rightarrow E \times 4 π r^{2} = \frac{(q - 2 π α R^{2})}{ε_{0}} + \frac{4 π α r^{2}}{2 ε_{0}}$

$\Rightarrow E = \frac{(q - 2 π α R^{2})}{ε_{0} 4 π r^{2}} + \frac{α}{2 ε_{0}} . . . (2)$

From the above equation, we can clearly analyse that the intensity $E$ does not depend on $r$ , when the expression in the parentheses is equal to zero.

Hence, $q - 2 π α R^{2} = 0$

$\Rightarrow q = 2 π α R^{2}$

Putting this value of $q$ in equation $(2)$ , we get,

$\Rightarrow E = \frac{α}{2 ε_{0}}$

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