IIT-JEE 2011 · Paper 1 · Physics Q31

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PhysicsMultiple correct+4Question 31 of 69

A spherical metal shell A of radius $R_A$ and a solid metal sphere B of radius $R_B$ ( $<R_A$ ) are kept far apart and each is given charge ' $+Q$ '. Now they are connected by a thin metal wire. Then

E_A^\text{inside}=0

Q_A>Q_B

\dfrac{\sigma_A}{\sigma_B}=\dfrac{R_B}{R_A}

E_A^\text{on surface}<E_B^\text{on surface}

Answer and worked solution

Official answer: A, B, C, D

Same potential after connection: $\dfrac{Q_A}{R_A}=\dfrac{Q_B}{R_B}$ , so $Q_A>Q_B$ (B ✓).

Surface densities. $\sigma\propto\dfrac Q{R^2}\propto\dfrac1R$ , so $\dfrac{\sigma_A}{\sigma_B}=\dfrac{R_B}{R_A}$ (C ✓), and the surface field $\frac\sigma{\varepsilon_0}$ is smaller on A (D ✓).

Inside the shell there is no field (A ✓).

(A), (B), (C), (D)

Topics:Conductors and charge distribution Electric potential

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IIT-JEE 2011 · Paper 1, Physics question 31