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1. Electric Charges and Fields
hard
An electric field $\overrightarrow{\mathrm{E}}=(2 \mathrm{xi}) \mathrm{NC}^{-1}$ exists in space. $\mathrm{A}$ cube of side $2 \mathrm{~m}$ is placed in the space as per figure given below. The electric flux through the cube is .................. $\mathrm{Nm}^2 / \mathrm{C}$
A
$13$
B
$14$
C
$15$
D
$16$
(JEE MAIN-2024)
Solution
$\overrightarrow{\mathrm{E}}=2 x \hat{\mathrm{i}}$
$\phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}$
$\phi_{\text {in }}=-4 \times 4=-16 \mathrm{Nm}^2 / \mathrm{c}$
$\phi_{\text {out }}=8 \times 4=32 \mathrm{Nm}^2 / \mathrm{c}$
$d_{\text {net }}=\phi_{\text {in }} \phi_{\text {out }}=-16+32=16 \mathrm{Nm}^2$
Standard 12
Physics
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