Question

Two large thin plane sheets, each having surface charge density \(\sigma\), are held close and parallel to each other in air. What is the net electric field at a point (i) inside and (ii) outside the sheets ?

Hint:

For two large parallel sheets carrying charges \(+\sigma\) and \(-\sigma\): \[ \boxed{ E_{\text{inside}} = \frac{\sigma}{\varepsilon_0} } \] and \[ \boxed{ E_{\text{outside}}=0 } \] This arrangement forms a parallel-plate capacitor, where the electric field is uniform between the plates and zero outside.

Answer 1

Concept: The electric field due to a single infinite plane sheet having uniform surface charge density \(\sigma\) is \[ E=\frac{\sigma}{2\varepsilon_0}. \] The field is independent of the distance from the sheet and is directed:
• Away from the sheet if the sheet is positively charged.
• Towards the sheet if the sheet is negatively charged. The net electric field due to two charged sheets is obtained by applying the principle of superposition of electric fields, according to which the resultant electric field at any point is the vector sum of the individual electric fields produced by each sheet. Since the question asks for the electric field inside and outside the sheets, we consider two large parallel sheets carrying equal and opposite surface charge densities \(+\sigma\) and \(-\sigma\).

Step 1: Determine the electric field due to each sheet.
The magnitude of the electric field produced by each sheet is \[ E=\frac{\sigma}{2\varepsilon_0}. \] The direction of the field due to the positively charged sheet is away from it, whereas the direction of the field due to the negatively charged sheet is towards it.

Step 2: Calculate the electric field at a point inside the sheets.
At a point between the two sheets, the electric fields due to both sheets are in the same direction. Hence, the resultant electric field is \[ E_{\text{inside}} = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0}. \] Therefore, \[ \boxed{ E_{\text{inside}} = \frac{\sigma}{\varepsilon_0} } \] The direction of this electric field is from the positively charged sheet towards the negatively charged sheet.

Step 3: Calculate the electric field at a point outside the sheets.
At any point outside the two sheets, the electric fields due to the two sheets are equal in magnitude but opposite in direction. Therefore, \[ E_{\text{outside}} = \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = 0. \] Hence, \[ \boxed{ E_{\text{outside}}=0 } \] Therefore, \[ \boxed{ E_{\text{inside}} = \frac{\sigma}{\varepsilon_0} } \] and \[ \boxed{ E_{\text{outside}} = 0. } \]