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JEE Advanced 2013Paper 2
Exam
JEE Advanced
Year
2013
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics

JEE Advanced 2013 · Paper 2 · Physics · Electrostatics

Two non-conducting spheres of radii R _ 1 and R _ 2 and carrying...

Two non-conducting spheres of radii R1{R}_{1} and R2{R}_{2} and carrying uniform volume charge densities +ρ+ ρ and −ρ- ρ, respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region,
Two non-conducting spheres of radii R _ 1 and R _ 2 and carrying uniform volume charge densities + ρ and - ρ ,

Options

  1. Athe electrostatic field is zero.
  2. Bthe electrostatic potential is constant.
  3. Cthe electrostatic field is constant in magnitude.
  4. Dthe electrostatic field has same direction.

Correct answer

C. the electrostatic field is constant in magnitude.

Step-by-step solution

In triangle PC1C2P{C}_{1}{C}_{2},
r→2=d→+r→1\overset{\to}{r}2 = \overset{\to}{d} + \overset{\to}{r}1


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

The electrostatic field at point PP is,
E→=K(ρ43πR13)(r→2)R13+K(ρ43πR23)(−r→1)R23\overset{\to}{E} = K\left(ρ\dfrac{4}{3}\pi{R}_{1}^{3}\right)\left(\overset{\to}{r}2\right){R}_{1}^{3} + K\left(ρ\dfrac{4}{3}\pi{R}_{2}^{3}\right)\left(- \overset{\to}{r}1\right){R}_{2}^{3}
E→=Kρ43π(r→2−r→1)\overset{\to}{E} = Kρ\dfrac{4}{3}\pi\left(\overset{\to}{r}2 - \overset{\to}{r}1\right)
E→=ρ3ε0d→\overset{\to}{E} = \dfrac{ρ}{3{ε}_{0}}\overset{\to}{d}

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