Name: Electric Field of a Charged Plane Sheet
Uploaded: 2019-07-19
Duration: 6 min 58 s
Description: Another distribution of charge. This time on an infinite plane sheet. Let’s see once again how Gauss’s law comes to the rescue in electric field calculations.

Question

A hollow dielectric sphere, as shown in figure, has inner and outer radii of $R_{1}$ and $R_{2}$ , respectively. The total charge carried by the sphere is $+ Q$ , this charge is uniformly distributed $R_{1}$ and $R_{2}$

(a) Then, the electric field for $r < R_{1}$ is zero. (Yes/No)

(b) Then, the electric field for $R_{1} < r < R_{2}$ is given by ______.

(c) Then, the electric field for $r > R_{2}$ is given by _____.

Accepted Answer

Given:

The inner radius of the hollow dielectric sphere, $R_{1}$

The outer radius of the hollow dielectric sphere, $+ Q$

The charge carried by the sphere, $+ Q$

As, at $r < R_{1}$ , there is no charge enclosed by the sphere, then,

According to Gauss law

$ϕ = \oint \vec{E} \cdot \vec{A} = \frac{Q}{ε_{0}} = 0$

Here, $E$ is electric field, $A$ is area and $ε_{0}$ is electric constant.

Yes, there will be no electric field at this range.

$E = 0$

The charge enclosed by the sphere at $R_{1} < r < R_{2}$ is given by $q = ρ \frac{4}{3} π (r^{3} - R_{1}^{3})$ ... $(1)$

where $ρ$ is the volume charge density given by

$ρ = \frac{Q}{\frac{4}{3} π ({R_{2}}^{3} - {R_{1}}^{3})}$

Putting this value in equation $(1)$ , we get

$q = \frac{Q (r^{3} - R_{1}^{3})}{(R_{2}^{3} - R_{1}^{3})}$

Using Coulomb's law,

$E = \frac{k q}{r^{2}}$ , here $k$ is Coulomb's constant.

Electric field is $E = \frac{Q}{4 π ε_{0} r^{2}} (\frac{r^{3} - R_{1}^{3}}{R_{2}^{3} - R_{1}^{3}})$

The charge enclosed by the sphere at $R_{2} < r$ is $Q$ .

Applying Coulomb's law, electric field is given by

$E = \frac{k Q}{r^{2}}$ , (Here the electric field outside the sphere will be equal to the electric field due to point charge.)

Important Questions on Electric Flux and Gauss's Law

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