Electrostatics

JEE Main level, single correct, medium

208 of 311

Question 208 of 311

A charge QQ is distributed over three concentric spherical shells of radii a,b,c(a<b<c)a,b,c\left(a\lt b\lt c\right) such that their surface charge densities are equal to one another.

The total potential at a point at distance rr from their common centre, where r<a,r\lt a, would be:

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Answer: (D) Q(a+b+c)4πϵ0(a2+b2+c2)\dfrac{Q\left(a+b+c\right)}{4\pi {ϵ}_{0}\left({a}^{2}+{b}^{2}+{c}^{2}\right)}

Surface charge densities of all sphere is given as q1a2=q2b2=q3c2=constant\dfrac{{q}_{1}}{{a}^{2}}=\dfrac{{q}_{2}}{{b}^{2}}=\dfrac{{q}_{3}}{{c}^{2}}=constant, where q1{q}_{1}, q2{q}_{2} and q3{q}_{3} is charge on sphere of radius aa, bb and cc respectively.

q1+q2+q3=Q{q}_{1}+{q}_{2}+{q}_{3}=Q

q1=a2(Q)a2+b2+c2{q}_{1}=\dfrac{{a}^{2}(Q)}{{a}^{2}+{b}^{2}+{c}^{2}}

q2=b2(Q)a2+b2+c2{q}_{2}=\dfrac{{b}^{2}(Q)}{{a}^{2}+{b}^{2}+{c}^{2}}

q3=c2(Q)a2+b2+c2{q}_{3}=\dfrac{{c}^{2}\left(Q\right)}{{a}^{2}+{b}^{2}+{c}^{2}}Potential at a point where r<ar\lt a is given as V=Kq1a+Kq2b+Kq3cV=\dfrac{K{q}_{1}}{a}+\dfrac{K{q}_{2}}{b}+\dfrac{K{q}_{3}}{c}, where KK is a constant.

V=14πϵ0[Q](a+b+ca2+b2+c2)V=\dfrac{1}{4\pi {\epsilon }_{0}}\left[Q\right]\left(\dfrac{a+b+c}{{a}^{2}+{b}^{2}+{c}^{2}}\right)

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