If vec E = frac{E_0 x}{a} hat i,then find the | vedclass

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(A) The electric field is given by $\vec E = \frac{E_0 x}{a} \hat i$.The shaded face of the cube is located at $x = a$ and lies in the $yz$-plane.At t

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$E_0 a^2$

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(A) The electric field is given by $\vec E = \frac{E_0 x}{a} \hat i$.The shaded face of the cube is located at $x = a$ and lies in the $yz$-plane.At the shaded face $(x = a)$,the electric field is $\vec E = \frac{E_0 (a)}{a} \hat i = E_0 \hat i$.The area vector for this face is $\vec A = a^2 \hat i$ (since the face is a square of side $a$ and the normal points in the $+x$ direction).The electric flux $\phi$ is given by $\phi = \int \vec E \cdot d\vec A$.Since the electric field is uniform over this face,$\phi = \vec E \cdot \vec A = (E_0 \hat i) \cdot (a^2 \hat i) = E_0 a^2$.

If $\vec E = \frac{E_0 x}{a} \hat i$,then find the electric flux through the shaded area of the cube as shown in the figure,where the shaded face is at $x = a$.

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