The electric field in a region is radially ou | vedclass

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(C) According to Gauss's Law,the total electric flux $\Phi_E$ through a closed surface is given by $\Phi_E = \oint E \cdot dA = \frac{Q_{enclosed}}{\v

Vedclass · Accepted Answer

$2.22 	imes 10^{-10}$

Vedclass · Answer

(C) According to Gauss's Law,the total electric flux $\Phi_E$ through a closed surface is given by $\Phi_E = \oint E \cdot dA = \frac{Q_{enclosed}}{\varepsilon_0}$.Given the electric field $E = 250 r$ is radially outward,for a sphere of radius $r = 20 \, cm = 0.2 \, m$,the surface area is $A = 4 \pi r^2$.The flux is $\Phi_E = E \cdot A = (250 r) \cdot (4 \pi r^2) = 1000 \pi r^3$.Substituting $r = 0.2 \, m$:$\Phi_E = 1000 \pi (0.2)^3 = 1000 \pi (0.008) = 8 \pi \, V \cdot m$.Now,using $\Phi_E = \frac{Q}{\varepsilon_0}$,we have $Q = \Phi_E \cdot \varepsilon_0$.$Q = 8 \pi \cdot (8.854 	imes 10^{-12}) \approx 25.13 	imes 8.854 	imes 10^{-12} \approx 2.22 	imes 10^{-10} \, C$.

The electric field in a region is radially outward and at a point is given by $E = 250 r \, V/m$ (where $r$ is the distance of the point from the origin). Calculate the charge contained in a sphere of radius $20 \, cm$ centered at the origin in Coulombs $(C)$.

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