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(D) The total electric flux $\phi$ through a closed surface is given by Gauss's Law: $\phi = \frac{q}{\varepsilon_0}$.Since the charge is uniformly di

Vedclass · Accepted Answer

$280 \ kV \cdot m$

Vedclass · Answer

(D) The total electric flux $\phi$ through a closed surface is given by Gauss's Law: $\phi = \frac{q}{\varepsilon_0}$.Since the charge is uniformly distributed on the surface,$q = \sigma 	imes A$,where $A = 4 \pi r^2$ is the surface area of the sphere.Given: diameter $d = 14 \ cm$,so radius $r = 7 \ cm = 7 	imes 10^{-2} \ m$.Surface charge density $\sigma = 40 \ \mu C/m^2 = 40 	imes 10^{-6} \ C/m^2$.Substituting these values:$\phi = \frac{4 \pi r^2 \sigma}{\varepsilon_0}$$\phi = \frac{4 	imes 3.14 	imes (7 	imes 10^{-2})^2 	imes 40 	imes 10^{-6}}{8.85 	imes 10^{-12}}$$\phi = \frac{4 	imes 3.14 	imes 49 	imes 10^{-4} 	imes 40 	imes 10^{-6}}{8.85 	imes 10^{-12}}$$\phi = \frac{2461.76 	imes 10^{-10}}{8.85 	imes 10^{-12}}$$\phi \approx 278.16 	imes 10^2 = 2.78 	imes 10^4 \ V \cdot m$ (or $280 \ kV \cdot m$ approximately).Thus,the correct option is $D$.

$A$ uniformly charged conducting sphere of diameter $14 \ cm$ has a surface charge density of $40 \ \mu C/m^2$. The total electric flux leaving the surface of the sphere is nearly (Permittivity of free space $\varepsilon_0 = 8.85 \times 10^{-12} \ C^2/N \cdot m^2$)

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