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 ball of radius $R$ carries a positive charge whose volume density depends only on a separation $r$ from the ball's centre, as $ρ = ρ_{0} (1 - \frac{r}{R}),$ where $ρ_{0}$ is a constant. Assuming the permittivities of the ball and the environment to be equal to unity, find:

$(a)$ the magnitude of the electric field strength as a function of the distance $r$ , both inside and outside the ball;
$(b)$ the maximum intensity $E_{\max}$ and the corresponding distance $r_{m}$ .

Accepted Answer

(a) Let us consider a sphere of radius $R$ and volume charge density $ρ = ρ_{0} (1 - \frac{r}{R})$ , where $ρ_{0}$ is a constant.

For a point inside the sphere, $r \leq R$ .

Charge enclosed by the considered sphere is,

$q_{enclosed} = \int_{0}^{r} 4 π r^{2} d r ρ = \int_{0}^{r} 4 π r^{2} ρ_{0} (1 - \frac{r}{R}) d r . . . (1)$

Now, applying Gauss' theorem,

$E_{r} 4 π r^{2} = \frac{q_{enclosed}}{ε_{0}}$

(where, $E_{r}$ is the projection of electric field along the radial line)

$= \frac{ρ_{0}}{ε_{0}} \int_{0}^{r} 4 π r^{2} (1 - \frac{r}{R}) d r$

or, $E_{r} = \frac{ρ_{0} r}{3 ε_{0}} [1 - \frac{3 r}{4 R}] . . . (2)$

For a point outside the sphere, $r > R$ .

$q_{enclosed} = \int_{0}^{R} 4 π r^{2} d r ρ_{0} (1 - \frac{r}{R})$

(As there is no charge outside the ball, only the charge of the ball will be considered)

Again, from Gauss' theorem,

$E_{r} 4 π r^{2} = \int_{0}^{R} \frac{4 π r^{2} d r ρ_{0} (1 - \frac{r}{R})}{ε_{0}}$

$E_{r} = \frac{ρ_{0}}{r^{2} ε_{0}} [\frac{R^{3}}{3} - \frac{R^{4}}{4 R}] = \frac{ρ_{0} R^{3}}{12 r^{2} ε_{0}}$

(b) As the magnitude of electric field decreases with increasing $r$ , for $r > R$ , the electric field will be maximum for $r \leq R .$

Now, for $E_{r}$ to be maximum,

$\frac{d E_{r}}{d r} = 0$

From equation $(2)$ ,

$\frac{d}{d r} (r - \frac{3 r^{2}}{4 R}) = 0$

$\Rightarrow 1 - \frac{3 r}{2 R} = 0$

$\Rightarrow r = r_{m} = \frac{2 R}{3}$

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

$E_{\max} = \frac{ρ_{0} R}{9 ε_{0}}$

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