# Flux of Electric Field

The electric flux passing through an area $\vec{A}$ in an electric field $\vec{E}$ is given by
\begin{align}
\phi=\vec{E}\cdot\vec{A}.
\end{align}

## 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,

- $2E_{0}a^2$
- $\sqrt{2}E_{0}a^2$
- $E_{0}a^2$
- $\frac{E_{0}a^2}{2}$

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

## Exercise

An uncharged spherical conductor of radius $R$ is placed in a non-uniform electric field (see figure). The electric flux through the sphere is (i) less than zero (ii) zero or (iii) greater than zero?

## Related

- Electric field
- Electric field lines
- Gauss's law and its application