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Given,
Radius of the charged sphere $(R) = 12 \; cm$
Charge $(q) = 6 * 10^{-6}\; C$
Electric field intensity $(E) = ?$
i) On the surface
$E = $$\frac{1}{4 \pi \epsilon_0} \frac{q}{R^2}$$ \;=\; 9 \;*\; 10^9 \;*\; $$\frac{6 \;*\; 10^{-6}}{(0.12)^2}$$ = 3.75 * 10^{6}\;N/C$
ii) Inside the sphere,
$E = 0$; Because charge enclosed by Gaussian surface is zero.
iii) Outside the sphere at distance $15 \;cm$
i.e. $r = 12 + 15 = 27\;cm = 0.27\;m$
$E = $$\frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}$$ = 9 * 10^9 \;*\; $$\frac{6 \;*\; 10^6}{(0.27)^2}$$ = 7.41 \;*\; 10^5\; N/C$
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