JEE Main 2015 · 10 Apr · Shift 1 · Physics Q14

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PhysicsSingle correct+4 / -1Question 14 of 90

A thin disc of radius b = 2a has a concentric hole of radius 'a' in it (see figure). It carries uniform surface charge 'σ' on it. If the electric field on its axis at height 'h' (h<<a) from its centre is given as 'Ch' then value of 'C' is :

A shaded annular disc (outer radius 2a, inner hole radius a) seen in perspective, with its axis dashed.

\dfrac\sigma{a\epsilon_0}

\dfrac\sigma{2a\epsilon_0}

\dfrac\sigma{4a\epsilon_0}

\dfrac\sigma{8a\epsilon_0}

Answer and worked solution

Official answer: C

For a disc of radius R at height $h\ll R$ : $E=\dfrac\sigma{2\epsilon_0}\left(1-\dfrac h{\sqrt{h^2+R^2}}\right)$ . The annulus is the difference of two discs:
$E=\frac{\sigma h}{2\epsilon_0}\left(\frac1a-\frac1{2a}\right)=\frac{\sigma h}{4a\epsilon_0},\qquad C=\frac\sigma{4a\epsilon_0}.$

Topics:Electric field

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JEE Main 2015 · 10 Apr · Shift 1, Physics question 14