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Asked in GSEB Board March 2023 · Shell and uniform solid sphere
Given: R=25 cm=0.25 m; σ=3/π C m⁻².
Idea: on a shell the charge spreads over the whole outer surface, of area 4π R², so Q=σ×4π R².
Q=3/π×4π(0.25)²
The π cancels: Q=3×4×0.0625
Q=0.75 C
So the shell needs 0.75 C.
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