JEE Advanced 2013 · Paper 1 · Physics Q14

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PhysicsMultiple correct+4 / -1Question 14 of 60

Two non-conducting solid spheres of radii $R$ and $2R$ , having uniform volume charge densities $\rho_1$ and $\rho_2$ respectively, touch each other. The net electric field at a distance $2R$ from the centre of the smaller sphere, along the line joining the centres of the spheres, is zero. The ratio $\dfrac{\rho_1}{\rho_2}$ can be

-4

-\dfrac{32}{25}

\dfrac{32}{25}

Answer and worked solution

Official answer: B, D

Two possible points at distance $2R$ from the small sphere's centre on the line of centres (the large centre is $3R$ away).

Between the centres (inside the large sphere, $R$ from its centre): $E_\text{small}=\dfrac{\rho_1R^3}{3\varepsilon_0(2R)^2}=\dfrac{\rho_1R}{12\varepsilon_0}$ , $E_\text{large}=\dfrac{\rho_2R}{3\varepsilon_0}$ , pointing in opposite directions for like charges. Zero when $\dfrac{\rho_1}{\rho_2}=4$ .
On the far side of the small sphere ( $5R$ from the large centre): both fields point the same way for like charges, so the charges must be opposite: $\dfrac{\rho_1R^3}{4R^2}=\dfrac{\rho_2(8R^3)}{25R^2}$ , $\dfrac{\rho_1}{\rho_2}=-\dfrac{32}{25}$ .

(B), (D)

Topics:Electric flux and Gauss's law Electric field

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JEE Advanced 2013 · Paper 1, Physics question 14