IIT-JEE 2012 · Paper 1 · Physics Q3

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PhysicsSingle correct+3 / -1Question 3 of 60

Consider a thin spherical shell of radius $R$ with its centre at the origin, carrying uniform positive surface charge density. The variation of the magnitude of the electric field $|\vec E(r)|$ and the electric potential $V(r)$ with the distance $r$ from the centre, is best represented by which graph?

E (solid) and V (dashed) against r: E zero inside, jumps at R then falls as 1/r²; V rises linearly from 0 to a peak at R then falls as 1/r.

E (solid) constant inside up to R then falling; V (dashed) constant inside up to R then falling as 1/r.

E (solid) constant inside up to R then falling; V (dashed) rising linearly from 0 to a peak at R then falling.

E (solid) zero inside, jumping at R then falling as 1/r²; V (dashed) constant inside up to R then falling as 1/r.

Answer and worked solution

Official answer: D

Field. By Gauss's law $E=0$ inside the shell ( $r<R$ ) and $E=\dfrac{kQ}{r^2}$ outside, with a jump at $r=R$ .

Potential. $V=\dfrac{kQ}R$ , constant, inside; $V=\dfrac{kQ}r$ outside, continuous at $R$ .

Only (D) shows $E$ zero inside and $V$ flat inside.

(D)

Topics:Electric flux and Gauss's law Electric potential

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IIT-JEE 2012 · Paper 1, Physics question 3