Consider a uniform spherical charge distribution of radius R _ 1 centred at the origin O. In this distribution, a spherical cavity of radius R _ 2 , centred at P with distance OP = a = R _ 1 - R _ 2 (see figure) is

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Correct answer: D. → E is uniform and both its magnitude and direction depend on → a

Consider a uniform spherical charge distribution of radius R _ 1 centred at the origin O. In this distribution, a

Considering any point A inside the cavity electric field can be calculated at A by using super position principle.

{\overset{\to}{E}}_{A} = {\left({\overset{\to}{E}}_{A}\right)}_{Duetospherewithoutcavity} - {\left({\overset{\to}{E}}_{A}\right)}_{Duetosphereofradius{R}_{2} & centreatP}

{\overset{\to}{E}}_{A} = \dfrac{ρ}{3{ϵ}_{0}}O\overset{\to}{A} = \dfrac{ρ}{3{ϵ}_{0}}\left(P\overset{\to}{A}\right)

But

O\overset{\to}{P} + P\overset{\to}{A} = O\overset{\to}{P}

∴ O\overset{\to}{A} - P\overset{\to}{A} = O\overset{\to}{P}

∴ {\overset{\to}{E}}_{A} = \dfrac{ρ}{3{ϵ}_{0}}\left(O\overset{\to}{P}\right) = \dfrac{ρ}{3{ϵ}_{0}}\left(\overset{\to}{a}\right)

So E.F. is independent of location of A inside the cavity.

Consider a uniform spherical charge distribution of radius R _ 1...

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