If the electric field is given by $(6 \hat{i}+3 \hat{j}+4 \hat{k}) V/m$. The electric flux through a surface of area $20 m^2$ lying in Y-Z plane would be (in V-m).

120

The correct answer is Option (1) → 120

The electric flux $(\phi_E)$ through a surface is given by -

$\phi_E=\vec E.\vec A$

where,

$\vec E$ (Electric field) = $(6\hat i+3\hat j+4\hat k)V/m$

$\vec A$ (Area vector perpendicular to y-z direction) = $20\hat im^2$

$∴\phi_E=\vec E(6\hat i+3\hat j+4\hat k).(20\hat i+0\hat j+0\hat k)$

$=6×20+0×3+0×4=120$