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=Qinϵ⇒E=Q4πϵr2forr>R2