Answer to Question #226689 in Electricity and Magnetism for Paulinus

Question

An electric field is generated in between two parallel hundred-square-centimeter electrodes that are separated by a distance of two centimeters. We have the one plate charged to 5.0 nano- Coulombs and the other one exactly equal but opposite. An additional "square-shaped" surface ( with side length of a centimeter) is placed in between the plates and rotated such that it makes an angle of pie over four radians with the electric field. Determine the flux of the electric field through the square surface.

Answer 1

Given,

Area of the plate of the capacitors $(A) =100cm^2 = 0.01m^2$

Separation between the plates $(d)=2cm$

Charge on the plates $(q)=5nC = 5\times 10^{-9}C$

Angle $(\theta)= \pi$

Flux $(\phi) =E S \cos\theta$

$\Rightarrow E=\frac{q}{cd}$

$\Rightarrow E = \frac{qd}{\epsilon_o A . d}$

$\Rightarrow E = \frac{q}{\epsilon_o A}$

Now, substituting the values,

$\Rightarrow \phi = \frac{q}{\epsilon_o A} S \cos\theta$

$\Rightarrow$ $\phi = \frac{5\times 10^{-9}\times 0.01 \times \cos(\pi)}{8.85\times 10^{-12}\times 0.01}$

$\Rightarrow \phi = 5.6 \times 10^3 N/C.m^2$