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(D) The surface charge density $\sigma = 20 \ \mu C \ m^{-2} = 20 	imes 10^{-6} \ C \ m^{-2}$.The diameter of the sphere $d = 3.5 \ cm$,so the radius

Vedclass · Accepted Answer

$8.7 	imes 10^3 \ N \cdot m^2 / C$

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(D) The surface charge density $\sigma = 20 \ \mu C \ m^{-2} = 20 	imes 10^{-6} \ C \ m^{-2}$.The diameter of the sphere $d = 3.5 \ cm$,so the radius $r = 1.75 \ cm = 1.75 	imes 10^{-2} \ m$.The surface area of the sphere $A = 4 \pi r^2 = 4 	imes 3.14 	imes (1.75 	imes 10^{-2})^2 \ m^2$.$A = 12.56 	imes 3.0625 	imes 10^{-4} \approx 3.848 	imes 10^{-3} \ m^2$.The total charge $q = \sigma 	imes A = (20 	imes 10^{-6}) 	imes (3.848 	imes 10^{-3}) \approx 7.696 	imes 10^{-8} \ C$.According to Gauss's Law,the total electric flux $\phi = q / \epsilon_0$.$\phi = (7.696 	imes 10^{-8}) / (8.85 	imes 10^{-12}) \approx 0.8696 	imes 10^4 \approx 8.7 	imes 10^3 \ N \cdot m^2 / C$.

$A$ uniformly charged conducting sphere of diameter $3.5 \ cm$ has a surface charge density of $20 \ \mu C \ m^{-2}$. The total electric flux leaving the surface of the sphere is nearly:
[permittivity of free space,$\epsilon_0 = 8.85 \times 10^{-12} \ SI \ unit$]

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$A$ uniformly charged conducting sphere of diameter $3.5 \ cm$ has a surface charge density of $20 \ \mu C \ m^{-2}$. The total electric flux leaving the surface of the sphere is nearly:[permittivity of free space,$\epsilon_0 = 8.85 \times 10^{-12} \ SI \ unit$]