JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Electrostatics

An infinitely long solid cylinder of radius

R

has a uniform volume charge density

ρ

. It has a spherical cavity of radius

R /2

with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point

P

, which is at a distance

2 R

from the axis of the cylinder, is given by the expression

\frac{23 ρR}{16 k ε _{0}}

. The value of

k

Answer: 6

Solution

Using superposition, the field at

P

is the sum of the field due to the solid cylinder and the negative field due to the spherical cavity. For the cylinder at distance

2 R

E_{cy l} = \frac{λ}{2 π ε _{0} ( 2 R )} = \frac{ρ π R ^{2}}{4 π ε _{0} R} = \frac{ρR}{4 ε _{0}}

. For the cavity (treating it as a negative sphere):

E_{c a v i t y} = \frac{K ρ ( 4 π R ^{3} /8 )}{3 \times 4 R ^{2}} = \frac{K ρ π R}{24}

. The net field magnitude is

\frac{ρR}{4 ε _{0}} - \frac{ρR}{96 ε _{0}} = \frac{23 ρR}{96 ε _{0}}

. Comparing with

\frac{23 ρR}{16 k ε _{0}}

, we get

k = 6

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Integer answer