A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho=\kappa r^{a}$, where $\kappa$ and $a$ are constants and $r$ is the distance from its centre. If the electric field at $r=\frac{R}{2}$ is $\frac{1}{8}$ times that at $r=R$, find the value of $a$.

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Electrostatics

A solid sphere of radius

R

has a charge

Q

distributed in its volume with a charge density

ρ = κ r^{a}

, where

κ

and

a

are constants and

r

is the distance from its centre. If the electric field at

r = \frac{R}{2}

\frac{1}{8}

times that at

r = R

, find the value of

a

Answer: 2

Solution

The electric field at radius

r

inside the sphere is

E (r) = \frac{q _{e n c l ose d}}{4 π ε _{0} r ^{2}}

, where

q_{e n c l ose d} = \int_{0}^{r} κ r^{' a} \cdot 4 π r^{'2} d r^{'} = \frac{4 π κ}{a + 3} r^{a + 3}

. Thus

E (r) = \frac{κ r ^{a + 1}}{ε _{0} ( a + 3 )}

. Given

E (R /2) = \frac{1}{8} E (R)

\frac{κ ( R /2 ) ^{a + 1}}{ε _{0} ( a + 3 )} = \frac{1}{8} \times \frac{κ R ^{a + 1}}{ε _{0} ( a + 3 )}

. This gives

(1/2)^{a + 1} = 1/8 = (1/2)^{3}

, so

a + 1 = 3

and

a = 2

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Integer answer

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