# Consider the parallel plate capacitor in the figure below.?

Consider the parallel plate capacitor in the figure below. The capacitor is made out of square plates. The sides of the square plates are L = 7 cm . The plates are separated by distance d. The space between the pates is filled with air so that the capacitance is C0 = 8×10^-11 F . The left plate is positively charged and the right most plate is negatively charged. The magnitude of the charge that appears on each plate is q = 1×10^-6 C.

Questions:

Determine the plate separation.

Determine The magnitude of the electric field E0 between the plates.

Determine the energy stored U0 in the capacitor. Relevance

C = εoA/d = 8.854e-12 *0.07²/d = 8e-11

8.854e-12 *0.07²/8e-11 = d = 5.423e-4

Plate separation d = 0.5423mm <<<<<

Q = CV so V = Q/C = Q*d/(εoA) = E*d

E = Q/(εoA) = 1e-6/(8.854e-12*0.07²) = 2.3e7

E0 = 2.3e7 N/C <<<<<<

Energy = ½CV² = ½QV = ½*Q*E0*d

Energy = 6.25e-3 J = 6.25mJ <<<<

• a) BY C = ɛA/d

=>8 x 10^-11  = [8.85 x 10^-12 (0.07)^2]/d

=>d = 5.42 x 10^-4 m OR 0.542 cm

b) BY C = Q/V

=>V = Q/C = 10^-6/(8 x 10^-11)

=>V = 12500 V OR 12.5 kV

c)BY E = 1/2CV^2 = 1/2 x 8x 10^-11 x (12500)^2 = 6.25 x 10^-3 J OR 6.25 mJ