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JEE Advanced 2013Paper 1
Exam
JEE Advanced
Year
2013
Paper
Paper 1
Subject
Physics
Chapter
Electrostatics

JEE Advanced 2013 · Paper 1 · Physics · Electrostatics

Two non-conducting solid spheres of radii R and 2R , having uniform...

Two non-conducting solid spheres of radii RR and 2R2R, having uniform volume charge densities ρ1{ρ}_{1} and ρ2{ρ}_{2}, respectively, touch each other. The net electric field at a distance 2R2R from the centre of the smaller sphere, along the line joining the centres of the spheres, is zero. The ratio ρ1ρ2\dfrac{{ρ}_{1}}{{ρ}_{2}} can be

Two non-conducting solid spheres of radii R and 2R , having uniform volume charge densities ρ _ 1 and ρ _ 2 ,Two non-conducting solid spheres of radii R and 2R , having uniform volume charge densities ρ _ 1 and ρ _ 2 ,

Options

  1. A−4- 4
  2. B−3225- \dfrac{3 2}{2 5}
  3. C3225\dfrac{3 2}{2 5}
  4. D44

Correct answer

B. −3225- \dfrac{3 2}{2 5}

Step-by-step solution

Two non-conducting solid spheres of radii R and 2R , having uniform volume charge densities ρ _ 1 and ρ _ 2 ,Field due to small sphere
=Kρ143πR3(2R)2= \dfrac{\text{K} {ρ}_{1} \dfrac{4}{3} \pi {\text{R}}^{3}}{{\left(2 \text{R}\right)}^{2}}
Field due to bigger sphere
=Kρ243π(2R)3×R(2R)3= \dfrac{\text{K} {ρ}_{2} \dfrac{4}{3} \pi {\left(2 \text{R}\right)}^{3} \times \text{R}}{{\left(2 \text{R}\right)}^{3}}
The net electric field at point PP will be

E→net=0\overset{\to}{E}net = 0

Equating above two, we get ρ1ρ2=4\dfrac{{ρ}_{1}}{{ρ}_{2}} = 4
[Charges cannot be of an opposite sign otherwise field will not zero]
Two non-conducting solid spheres of radii R and 2R , having uniform volume charge densities ρ _ 1 and ρ _ 2 ,

Similarly, the net electric field at point QQ becomes zero. Then ρ1ρ2=−3225\dfrac{{ρ}_{1}}{{ρ}_{2}} = - \dfrac{3 2}{2 5} here both spheres should have opposite polarity.

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