{"id":746,"date":"2023-04-28T09:15:00","date_gmt":"2023-04-28T03:45:00","guid":{"rendered":"http:\/\/physicscatalyst.com\/article\/?p=746"},"modified":"2024-01-15T17:42:58","modified_gmt":"2024-01-15T12:12:58","slug":"electric-field-due-charged-conductor","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/electric-field-due-charged-conductor\/","title":{"rendered":"Electric field due to charged conductor"},"content":{"rendered":"\n

We know that\u00a0electric field inside a conductor is zero and any charge the conductor may carry shall\u00a0be distributed on the surface of the conductor. The distribution of these surface charges depends on the shape of the conductor. For example, on a spherical conductor, the charge distribution is uniform, but it may vary on conductors of different shapes. we will check out now how to Electric field due to charged conductor<\/p>\n\n\n\n

For our discussion consider a conductor carrying charge on its surface again consider a small surface element ds over which we can consider surface charge density $\\sigma$ to be approximately constant.
For positive charge distributed over the surface of the conductor , electric field\u00a0E\u00a0<\/strong>would be directed at right angels to the surface pointing in outwards direction. Now\u00a0E\u00a0<\/strong>due to charge carrying conductor can be calculated using Gauss’s law. For this draw a Gaussin cylendrical surface as shown below in the figure<\/p>\n\n\n\n

\"Electric<\/a><\/figure>\n\n\n\n

Now S is the area of cross-section of the surface. The flux due to cylendrical surface is zero because electric field and the normal to the surface are perpandicular to each other. Since electric field inside the conductor is zero hence only contribution to the flux is due to the chare on area S lying outside the  surface of the conductor. So total flux through the surface would be<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

From Gauss’s law,<\/p>\n\n\n\n

Gauss’s Law is crucial for understanding electric fields due to conductors. It states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space.<\/p>\n\n\n\n

$E\\cdot S=\\frac {q}{\\epsilon_{0}}=\\frac{\\sigma S}{\\epsilon_{0}}$
or,
$E=\\frac{\\sigma}{\\epsilon_{0}}$
and this is the required relation for the field of charged conductor<\/p>\n\n\n\n

Important points<\/strong><\/p>\n\n\n\n