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1. Electric Charges and Fields
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If $\vec E = \frac{{{E_0}x}}{a}\hat i\,\left( {x - mt} \right)$ then flux through the shaded area of a cube is

A
$E_0a^2$
B
Zero
C
$E_0a^3$
D
$-E_0a^3$
Solution
$\overrightarrow{\mathrm{E}}=\frac{\mathrm{E}_{0}(\mathrm{a})}{\mathrm{a}} \hat{\mathrm{i}}=\mathrm{E}_{0} \hat{\mathrm{i}}$
Flux $=\mathrm{E}_{0}(\hat{\mathrm{i}}) \cdot \mathrm{a}^{2}(\hat{\mathrm{i}}) $
$=\mathrm{E}_{0} \mathrm{a}^{2} $
Standard 12
Physics
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