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Question

The electric field for r>R2 is given by E=1Xπε0Qr2. Find X?
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Solution

Let us imagine a surface of radius r>R2 inside the cavity. By symmetry the electric field (if any) will be radial and only a function of radius r.
therefore, by Gauss's law,
E.4πr2=QinϵE=Q4πϵr2forr>R2

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