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(D) The electric field is given by $\overrightarrow{E} = E_0 x \hat{i}$.The electric flux through the cube is only through the faces perpendicular to

Vedclass · Accepted Answer

$288$

Vedclass · Answer

(D) The electric field is given by $\overrightarrow{E} = E_0 x \hat{i}$.The electric flux through the cube is only through the faces perpendicular to the $x$-axis.At $x = 0$,the flux $\phi_1 = \overrightarrow{E} \cdot \overrightarrow{A} = (E_0 \cdot 0) \cdot (a^2 \hat{i}) = 0$.At $x = a$,the flux $\phi_2 = \overrightarrow{E} \cdot \overrightarrow{A} = (E_0 a \hat{i}) \cdot (a^2 \hat{i}) = E_0 a^3$.The net flux $\phi_{	ext{net}} = \phi_2 - \phi_1 = E_0 a^3$.According to Gauss's Law,$\phi_{	ext{net}} = \frac{q_{	ext{en}}}{\varepsilon_0}$,so $q_{	ext{en}} = \varepsilon_0 E_0 a^3$.Given $E_0 = 4 	imes 10^4 	ext{ N C}^{-1} 	ext{m}^{-1}$,$a = 2 	ext{ cm} = 0.02 	ext{ m} = 2 	imes 10^{-2} 	ext{ m}$,and $\varepsilon_0 = 9 	imes 10^{-12} 	ext{ C}^2 	ext{ N}^{-1} 	ext{m}^{-2}$.$q_{	ext{en}} = (9 	imes 10^{-12}) 	imes (4 	imes 10^4) 	imes (2 	imes 10^{-2})^3$.$q_{	ext{en}} = 36 	imes 10^{-8} 	imes 8 	imes 10^{-6} = 288 	imes 10^{-14} 	ext{ C}$.Comparing with $Q 	imes 10^{-14} 	ext{ C}$,we get $Q = 288$.

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