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JEE Main 2017 · Physics · Electrostatics · hard

A uniformly charged non-conducting sphere of radius R and volume charge density ρ has a spherical cavity of radius R/2 whose centre is at a distance…

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A uniformly charged non-conducting sphere of radius RR and volume charge density ρ\rho has a spherical cavity of radius R/2R/2 whose centre is at a distance R/2R/2 from the centre of the sphere. The electric field at any point inside the cavity has magnitude
  1. A. ρR6ε0\dfrac{\rho R}{6\varepsilon_0}
  2. B. ρR3ε0\dfrac{\rho R}{3\varepsilon_0}
  3. C. ρR12ε0\dfrac{\rho R}{12\varepsilon_0}
  4. D. 00
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Answer: (A) ρR6ε0\dfrac{\rho R}{6\varepsilon_0}

Use superposition: a full sphere of density +ρ+\rho plus a sphere of density −ρ-\rho filling the cavity. Inside a uniformly charged sphere, E⃗=ρ r⃗3ε0\vec E = \dfrac{\rho\,\vec r}{3\varepsilon_0} measured from its centre. For a point PP in the cavity, E⃗=ρ3ε0(r⃗1−r⃗2)\vec E = \dfrac{\rho}{3\varepsilon_0}\left(\vec r_1 - \vec r_2\right), where r⃗1,r⃗2\vec r_1,\vec r_2 are position vectors from the two centres, and r⃗1−r⃗2=a⃗\vec r_1 - \vec r_2 = \vec a is the vector joining the centres. So the field is uniform, with magnitude ρa3ε0=ρR6ε0\dfrac{\rho a}{3\varepsilon_0} = \dfrac{\rho R}{6\varepsilon_0} since a=R/2a = R/2.
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