A spherical shell has uniform charge density $8.8 \times 10^{- 11} C / m^{2}$ . If a pin hole is made in the surface of the shell then electric field in the hole in N/C is :
$ϵ_{0} = 8.8 \times 10^{- 12}$ S.I units

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Solution

The correct option is A 5
The electric field in the hole is electric field of complete shell - electric field of material (which occupied the hole) .
$E_{o t h e r} = (E_{o u t s i d e} + E_{i n s i d e}) / 2$
Therefore, $E_{o t h e r} = σ / (2 ϵ_{0}) = 5$

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1 \times 10^{14} Hz

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A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q.

(a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?

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1 \times 10^{14}

ϵ_{0} = 8.8 \times 10^{- 12} C^{2} / N - m^{2}

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