Question

Hello guys..


Answer the attachment.​

Answer 1

Answer:

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

−22

C/m

2

Charge density of plate B, σ=−17.0×10

−22

C/m

2

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=

∈

0

σ

Where,

∈

0

= Permittivity of free space =8.854×10

−12

N

−1

C

2

m

−2

∴E=

8.854×10

−12

17.0×10

−22

=1.92×10

−10

N/C

Therefore, electric field between the plates is 1.92×10

−10

N/C.

solution

Explanation:

I hope this is helpful