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An electric field along x-axis is given by \overrightarrow{E}= 100 \widehat{i}N/C for x > 0 and \overrightarrow{E}= -100 \widehat{i}N/C for x <0. A right circular cylinder of length 20cm and radius 5cm lies parallel to the x-axis, with its centre at the origin and one face at x = + 10cm, the other face at x = -10cm. Calculate the net outward flux through the cylinder.

 

 

 

 
 
 
 
 

Answers (1)

x=-10cm                                        x=10cm

From the figure, we can see that on each face the electric field E and small area elements S are parallel. Therefore, outward flux, \phi = E\Delta S

On the left side \phi_{1}= E \Delta S

                               = 100 \widehat{i}\Delta S

                               = +100 \Delta S, sincec\Delta S = - \Delta S

                               = 0.785Nm^{2}C^{-1}

On the side of the cylinder electric field is perpendicular to \Delta S and hence E.\Delta S=0.

Therefore, outward flux on the side of the cylinder 

\phi = 0.785 + 0.785 + 0= 1.57N m^{2}C^{-1}

Posted by

Safeer PP

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