Find electric field due to infinitely  long cylindrical charge distribution of radius R having linear charge density \lambda at a distance half of the radius from its axis .

  • Option 1)

    \frac{\lambda}{4 \pi \epsilon _{o}R}

  • Option 2)

    \frac{\lambda}{ \pi \epsilon _{o}R}

  • Option 3)

    0

  • Option 4)

    \frac{2K\lambda}{R^{2}}

 

Answers (1)
A Avinash

As we learnt ,

 

Electric Field at a point P lies inside the Cylinder -

E_{in}=0 for conducting cylinder

E_{in}=\frac{\lambda r}{2\pi \varepsilon _{0}R^{2}} for non-conducting cylinder

-

 

 

E =\frac{\lambda}{2\pi \epsilon _{o}R}       but r =\frac{R}{2}

E = 0 inside if r = R    \frac{\lambda}{2\pi \epsilon _{o}R}

 


Option 1)

\frac{\lambda}{4 \pi \epsilon _{o}R}

Option 2)

\frac{\lambda}{ \pi \epsilon _{o}R}

Option 3)

0

Option 4)

\frac{2K\lambda}{R^{2}}

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