# An electric field along the x-axis is given by  $\vec{E}=250\hat{i}\: N/C$ for  $x>0$ and $\vec{E}=-250\hat{i}\: N/C$ for $x<0$. A right circular cylinder of length 24 cm and radius 5 cm lies parallel to the x-axis, with its centre at the origin and one face at x = + 12 cm, the other face at x = – 12 cm. Calculate the net outward flux through the cylinder.

From the figure, we can see that on each face the electric field E and small area element  $\Delta S$ are parallel.

Therefore, outward flux : $\phi =E\cdot \Delta S$

On the left side:  $\phi_L =E\cdot \Delta S$

$=-250\hat{i}\cdot \Delta S$

$=250\times \pi \times (0.05)^2$

$=1.96Nm^2C^{-1}$

On the right side:  $\phi_R =E\cdot \Delta S$

$=250\hat{i}\cdot \Delta S$

$=250\cdot \Delta S$

$=1.96Nm^2C^{-1}$

On the circumference of the cylinder, the electric field is perpendicular to  $\Delta S$ and hence $E\cdot \Delta S=0$

Therefore, net outward flux through the cylinder

$\phi =1.96+1.96+0=3.92Nm^2C^{-1}$

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