Question

A thin metallic spherical shell of radius 40.6 cm

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Answer #1

Draw the Gaussian surface of 16.4cm from center it enclosed only 3.83\mu C charge

By Gause law, we have

\oint E.ds=\frac{Q}{\epsilon _0}\textup{, where Q is charge enclosed within the Gaussian surfface}   

E=\frac{Q}{A\epsilon _0}

E=1.61\times 10^6N/C

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