Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0×10−22C/m2. What is E: (a) in the outer region of the first plate. (b) in the outer region of the second plate, and (c) between the plates?
The situation is represented in the following figure.
A and B are two parallel plates close to each other. Outer region of plate A is labelled as I, outer region of plate B is labelled as III, and the region between the plates, A and B, is labelled as II.
Charge density of plate A, σ=17.0×10−22C/m2
Charge density of plate B, σ=−17.0×10−22C/m2
Electric field in regions can be found with the help of Gauss Law. In the regions, I and III, electric field E is zero. This is because charge is not enclosed by the Gaussian surfaces of the plates.
Electric field E in region II is given by the relation,
E=σ∈0
Where,
∈0= Permittivity of free space =8.854×10−12N−1C2m−2
∴E=17.0×10−228.854×10−12
=1.92×10−10N/C
Therefore, electric field between the plates is 1.92×10−10N/C.
Surface charge density is defined as the charge per unit surface area of surface charge distribution. i.e., σ=qS
Two large thin metal plates are parallel and close to each other, on their inner faces, the plates have surface charge densities of opposite sign having magnitude of 1.70×10−22 Cm−2 as shown in figure.
[Use ε0=8.85×10−12 Fm−1]
(i) The electric field E in the region in between the plates A and B is
