Question

Calculate the electric field intensity at a point just near the surface of a charged plane sheet, measured from its mid-point. [ \( \sigma = 8.85 \, \mu C/m^2 \) ]

Hint:

The electric field from an infinite sheet is constant everywhere in space. For a conducting sheet, the formula would be \( E = \sigma / \epsilon_0 \), which is twice as strong.

Answer 1

For a large (ideally infinite) charged plane sheet, the electric field intensity (E) at a point near it is uniform and given by Gauss's law. Assuming it's a non-conducting sheet: \[ E = \frac{\sigma}{2\epsilon_0} \] where \(\sigma\) is the surface charge density and \(\epsilon_0\) is the permittivity of free space (\( 8.85 \times 10^{-12} \, C^2/N \cdot m^2 \)).
Given: \(\sigma = 8.85 \, \mu C/m^2 = 8.85 \times 10^{-6} \, C/m^2 \). \[ E = \frac{8.85 \times 10^{-6} \, C/m^2}{2 \times 8.85 \times 10^{-12} \, C^2/N \cdot m^2} \] \[ E = \frac{10^{-6}}{2 \times 10^{-12}} = 0.5 \times 10^6 \, N/C \] The electric field intensity is \( 5 \times 10^5 \, N/C \). The distance from the mid-point does not affect the result for a large plane sheet.