JEE Main · 2009 · 12 Apr 2009 · Offline · Electrostatics

Let

P (r) = \frac{Q}{π R ^{4}} r

be the charge density distribution for a solid sphere of radius

R

and total charge

Q

. for a point '

p

' inside the sphere at distance

r_{1}

from the centre of the sphere, the magnitude of electric field is

Answer: (C)

Solution

The charge enclosed within a radius

r_{1}

comes from integrating the given density,

q = \int_{0}^{r_{1}} \frac{Q}{π R ^{4}} r \cdot 4 π r^{2} d r = \frac{4 Q}{R ^{4}} \cdot \frac{r _{1}^{4}}{4} = \frac{Q r _{1}^{4}}{R ^{4}}

. Applying Gauss's law to a spherical surface of radius

r_{1}

E \cdot 4 π r_{1}^{2} = \frac{q}{ε _{0}}

, and substituting the enclosed charge gives

E = \frac{Q r _{1}^{2}}{4 π ε _{0} R ^{4}}

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option