Answer :
Answer:
Φ = 361872 N.m^2 / C
Explanation:
Given:-
- The area of the two plates, [tex]A_p = 180 cm^2[/tex]
- The charge on each plate, [tex]q = 17 * 10^-^6 C[/tex]
- Permittivity of free space, [tex]e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}[/tex]
- The radius for the flux region, [tex]r = 3.3 cm[/tex]
- 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:
[tex]A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2[/tex]
- 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 = [tex]\frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\[/tex]
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.
[tex]E+ = E- = \frac{sigma}{2*e_o} \\\\[/tex]
- 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:
[tex]E_n_e_t = (E+) + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C} \\[/tex]
- 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