IIT JEE Physics (1978 to 2018: 41 Years) Topic-wise Complete Solutions

# Flux of Electric Field

## Problems from IIT JEE

Problem (IIT JEE 2011): Consider an electric field $\vec{E}=E_{0}\,\hat\imath$, where $E_0$ is a constant. The flux through the shaded region (as shown in figure) due to this field is,

1. $2E_{0}a^2$
2. $\sqrt{2}E_{0}a^2$
3. $E_{0}a^2$
4. $\frac{E_{0}a^2}{2}$

Solution: The flux through area vector $\vec{S}$ due to an electric field $\vec{E}$ is, \begin{alignat}{2} \label{caa:eqn:1} &\phi=\oint\vec{E}\cdot\mathrm{d}\vec{S}=\vec{E}\cdot\oint\mathrm{d}\vec{S}=\vec{E}\cdot\vec{S}. \quad \text{(since $\vec{E}$ is constant.)} \end{alignat} The area vector of shaded region is cross product of vectors representing two sides i.e., \begin{alignat}{2} \label{caa:eqn:2} &\vec{S}=(a\,\hat\jmath)\times(a\,\hat\imath+a\hat{k})=a^2\,(\hat\imath-\hat{k}). \end{alignat} Use above equations to get, \begin{alignat}{2} &\phi=(E_0\,\hat\imath)\cdot a^2\,(\hat\imath-\hat{k})=E_0 a^2.\nonumber \end{alignat}