Question

State Gauss's theorem of electrostatics. What is the meaning of equipotential surface? Two insulator plates having equal charge density are placed as shown in the figure. Find the electric field intensity at points P and Q.

Hint:

The electric field due to a uniformly charged infinite plane sheet is constant and does not depend on the distance from the sheet.

Answer 1

Step 1: Gauss's Theorem of Electrostatics.
Gauss's theorem states that the electric flux \( \Phi_E \) through a closed surface is proportional to the charge \( Q_{\text{enc}} \) enclosed within the surface. Mathematically, it is expressed as: \[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \] where: - \( \vec{E} \) is the electric field, - \( d\vec{A} \) is the differential area vector on the closed surface, - \( \epsilon_0 \) is the permittivity of free space.
Step 2: Equipotential Surface.
An equipotential surface is a surface on which the potential is constant at every point. No work is required to move a charge along an equipotential surface because the potential difference between any two points on the surface is zero. The electric field is always perpendicular to the equipotential surfaces.
Step 3: Electric Field Intensity at Points P and Q.
Given two insulator plates with equal charge density, the electric field between them is uniform. For an infinite plane sheet with charge density \( \sigma \), the electric field \( E \) at a distance \( r \) from the plate is given by: \[ E = \frac{\sigma}{2\epsilon_0} \] where \( \sigma \) is the surface charge density and \( \epsilon_0 \) is the permittivity of free space. For points \( P \) and \( Q \), if they lie between the two plates, the electric field at both points will be the same in magnitude and direction, assuming the plates are infinitely large. The field will be directed from the positive plate towards the negative plate.