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Electrostatics Part 1

Section: Physics

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Question : 80 of 100

Marks: +1, -0

Let

P(r)=\frac{Q}{\pi R^4} r

be the change density distribution for a solid sphere of radius

R

and total charge

Q

p'

inside the sphere at distance

r_1

from the center of the sphere, the magnitude of electric field is :

$\;\frac{Q}{4\pi E_0 r_1^2}$
$\;\frac{Q r_1^2}{4\pi \epsilon_0 R^4}$
$\;\frac{Q r_1^2}{3\pi \epsilon_0 R^4}$
$0$

Solution:

Let us consider a spherical shell of thickness

d x

and radius

x

. The volume of this spherical shell

=4\pi r^2 d r

.

The charge enclosed within shell

=\;\frac{Q_r}{\pi R^4} [4\pi r^2 d r]

The charge enclosed in a sphere of radius

r_1

is

\;\frac{4Q}{R^4} \int\limits_{0}^{r_1} r^3 d r

=\;\frac{4Q}{R^4} \left[\;\frac{r^4}{4}\right]_{0}^{r_1}

=\;\frac{Q}{R^4} r_1^4

\therefore

The electric field at point

p

inside the sphere at a distance

r_1

from the center of the sphere is

\;E=\;\frac{1}{4\pi E_0} \; \frac{ \left[ \frac{Q}{R^4} r_1^4 \right] }{r_1^2}

\;=\;\frac{1}{4\pi E_0} \; \frac{Q}{R^4} r_1^2

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