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I was wondering what would happen if us were to calculate electric field because of a finite heat charge. Most publications have this because that an limitless line charge. In the given figure if I eliminate the portion of the line beyond the end of the cylinder. I believe the price would stay the same. Additionally if ns imagine the line to be along the \$x\$-axis then would certainly it be correct to to speak that electric field would always be perpendicular come the line and also would never make any other edge (otherwise the currently of pressure would intersect)?

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edited may 3 "20 at 10:50

Vishnu
asked Apr 25 "15 in ~ 18:29

Karan SinghKaran singh
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I have taken that line charge is put vertically and also one test fee is placed.

Now the electrical field knowledgeable by test charge dude to limited line optimistic charge.

\$\$E_x = int dx cos alpha\$\$

\$E_y\$ will be cancel out as they will be the contrary to every other.

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\$\$E_x = int k fracdqx^2+y^2cosalpha\$\$

\$\$E_x = int k fraclambda dyx^2+y^2cosalpha\$\$

Here \$lambda dy\$ is the direct charge density distribution whereby \$dy\$ is small section of that line whereby \$y\$ is perpendicular distance and also \$x\$ is horizontal street to the test fee placed.

\$\$E_x = int k fraclambda xsec^2alpha dalphax^2sec^2alphacosalpha\$\$

Since \$\$ analpha = fracyx\$\$

\$\$dy = xsec^2 alpha dalpha\$\$

Now

\$\$E_x = k fraclambdaxint_alpha^eta cosalpha dalpha\$\$

(In over \$alpha\$ is negative and \$eta\$ is positive)

\$\$E_x = k fraclambdax\$\$

Now because you have actually taken limited line fee you deserve to put the value of edge which can be figured out by placing any kind of test charge between or all over in former of the ling fee or for easy method you have the right to use Gauss theorem to prove the which is much simpler than this.