Answer :

Given:

Volume charge density inside cylinder=ρ

We need to find the electric field at point P which is at a distance x from the axis of the cylinder

Consider a cylindrical gaussian surface of radius x and height h

By Gauss’s law, flux of net electric field (E⃗ ) through a closed surface S equals the net charge enclosed by the surface (q_{in}) divided by ϵ_{0}

..(i)

Volume of this gaussian surface =πx^{2}h

Curved surface area of this surface=2πxh

∴ charge enclosed by this surface=

→ ….(ii)

Now,

All the points on the curved part of this gaussian surface are at the same perpendicular distance from the line charge. Therefore all these points are equivalent. The electric field at all these points will have the same magnitude E and direction of field at any point on the curved surface is normal to the line and hence normal to the cylindrical surface.

Therefore flux through this surface can be written as,

……(iii)

From gauss law(i) and using eqns (ii) and (iii)

**Therefore electric field at a point P inside cylindrical volume at a distance x from its axis is given by ρx/2****ϵ**_{0}

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