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. 

 

 
 
 
 
 

Answers (1)
S safeer

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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