E in case of Charge distribution with cylindrical symmetry?

Asked by Aditya Singh Solanki | 24th Aug, 2011, 12:00: AM

Expert Answer:

Field Due to an Infinitely Long Straight Charged Wire

Consider a thin long charged wire. Let the charge unit length of the wire be l. To calculate the field at P we consider a Gaussian surface with wire as axis, radius r and length l as shown in the figure. This Gaussian surface that is, the cylinder is closed at each end by planes normal to the axis.

The electric lines of force are parallel to the end faces of the cylinder and hence the component of the field along the normal to the end faces is zero.

The field is radial everywhere and hence the electric flux crosses only through the curved surface of the cylinder.

If E is the electric field intensity at P, then the electric flux through the Gaussian surface is E x 2prl(2prl is the surface area of the curved part)

The charge enclosed by the Gaussian surface is ll

is directed radially outwards if q is positive and radially inwards if q is negative

Answered by  | 25th Aug, 2011, 03:56: PM

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