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        two long,charged,concentric cylinders have radii of 3.0 and 6.0 cm. The charge per unit length is 5.0*10^-6 C/m on the inner cylinder and -7*10^-6 C/m on the outer cylinder. Find the electric field at (a)r=4cm (b) r =8 cm,where r is the radial distance from the common central axis.
8 years ago

vinnu bhardwaj
8 Points
										it is a basic question of the GAUSS LAW. Simply apply the Gauss Law at the two cylindrical gausian surfaces having its axis same as the cylinders and the radius 4 and 8 cms respectively.  In case of second the total charge wud be -2 micro coulumb. See  the length of the hollow cylinde does not matter . Hope u understand why!!!!!!!!1

8 years ago
10 Points
										Hi
Acc to gauss law, electric field wud be only because of the charge enclosed.
Gauss's law, also known as Gauss's flux theorem, is a law relating the distribution of electric charge to the resulting electric field. Gauss's law states that:
The electric flux through any closed surface is proportional to the enclosed electric charge.
In integral form  the equation is: $\oint_S \mathbf{E} \cdot \mathrm{d}\mathbf{A} = \frac{Q_{\mathrm{enclosed}}}{\varepsilon_0}$
In this case , for r=4cm , electric field will be only because of the inner cylinder , bcoz as we draw the gaussian surface outside the inner cylinder but inside the outer cyl, the charge enclosed is just 5x10-6 . So, in the gauss law, put q enclosed equal to 5x10-6 .  .
In second case, draw gaussian surface outside the second cyl, hence net charge enclosed is now the sum of the two charges , hence the net q enclosed is ( -2x 10-6 ), use this in the eq and find E.

8 years ago
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