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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.

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.

Grade:12

2 Answers

vinnu bhardwaj
8 Points
12 years ago

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

AskiitianExpert Shine
10 Points
12 years ago

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.

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