ElectrostaticsEasy
Question
A thin disc of radius b = 2a has 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 :
Options
A.
B.
C.
D.
Solution
Electric field due to complete disc (R = 2a)
E1 =
E1 =
Electric field due to disc (R = a)
E2 =
Electric field due to given disc.
E = E1 - E2
=
option (2)
E1 =
E1 =
Electric field due to disc (R = a)
E2 =
Electric field due to given disc.
E = E1 - E2
=
option (2)
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