Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are charged with equal amount of opposite charges, ±17 µC. The charges on the plates face each other. Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates when the normal to the circle makes an angle of 4° with a line perpendicular to the plates. Note that this angle can also be given as 180° + 4°. N · m2/C

Φ = 361872 N.m^2 / C

Explanation:

Given:-

- The area of the two plates,

- The charge on each plate,

- Permittivity of free space,

- The radius for the flux region,

- The angle between normal to region and perpendicular to plates, θ = 4°

Find:-

Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.

Solution:-

- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:

- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):

                           σ =

                           σ = 0.00094 C / m^2

- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.

- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:

- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:

                        Φ = E_net * Ar * cos ( θ )

                        Φ = (106214689.26553) * (0.00342) * cos ( 5 )

                        Φ = 361872 N.m^2 / C

c

explanation: