Question

A cylinder of radius $R$ and length $L$ is placed in a uniform electric field of magnitude $E$ parallel to the axis of the cylinder. The outward flux over the surface of the cylinder is

• A. $2\pi R^{2}E$
• B. $\frac{\pi R^{2}E}{2}$
• C. $2\pi RLE$
• D. $\pi R^{2}E$

Answer 1

The correct option is D $\pi R^{2}E$


As shown in the figure, the electric field is parallel to the axis of the cylinder.

We know that from definition of flux

$\varphi =\oint \overrightarrow{E}.\overrightarrow{dA}=\oint E.dA\cos \theta$


At cross sectional area $I$, the area vector $\overrightarrow{A}$ is antiparallel to $\overrightarrow{E}$ which is in inwards direction. So at this surface the flux will be inwards.

For the curved surface of the cylinder the area vector and electric field will be perpendicular at all points. So $\cos \theta =\cos {90}^{\circ }=0$.

Thus, ${\varphi }_{\text{curved surface}}=0$

At cross sectional area $II$, the electric field and area vector are parallel. Due to the outward direction of $\overrightarrow{E}$ flux will be outwards. So, outward flux is through the circular face of the cylinder, that is surface $II$.

${\varphi }_{\text{outwards}}={\varphi }_{II}$

${\varphi }_{\text{outwards}}=\oint \overrightarrow{E}.\overrightarrow{dA}=\oint E.dA\cos \theta$

${\varphi }_{\text{outwards}}=E.A.\cos 0^{\circ }=EA$

$\Rightarrow {\varphi }_{\text{outwards}}=\pi R^{2}E$

So, option $\left(d\right)$ is correct.

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: If the direction of intersecting field lines is}\\ \text{such that it is coming outwards, then flux at that}\\ \text{surface will be outwards.}\end{array}$