• question_answer The electric field in a region is given by E$E=\frac{3}{5}{{E}_{0}}i+\frac{4}{5}{{E}_{0}}j$with${{E}_{0}}=2.0\times {{10}^{3}}N/C.$ The flux of this field through a rectangular surface area of $0.2\,{{m}^{2}}$parallel to the Y-Z plane is A) $60\,N{{m}^{2}}/C$                     B) $120\,N{{m}^{2}}/C$C) $180\,N{{m}^{2}}/C$                   D) $240\,N{{m}^{2}}/C$

The area vector $\overrightarrow{dA}=0.2\,\hat{i}$ $\therefore$    flux $=\vec{E}.\overrightarrow{dA}=\left( \frac{3}{2}{{E}_{0}}\hat{i}+\frac{4}{5}{{E}_{0}}\hat{j} \right).0.2\hat{i}$ $=0.2\times \frac{3}{5}\times 2\times {{10}^{3}}$ $=240\,N{{m}^{2}}/C$ Hence, the correction option is [d].