Question

A parallel-plate air capacitor with a capacitance of 244 pF has a charge of magnitude 0.137 mu C on each plate. The plates have a separation of 0.265 mm. What is the potential difference between the plates? What is the area of each plate? Use 8.85 times 10^-12 F/m for the permittivity of free space. What is the electric field magnitude between the plates? What is the surface-charge density on each plate?

Answer 1

Here:

(a) potential diff = Q / C =   0.137 x 10^-6 / 244 x10^-12 =    561.48 volts

(b)   C = εA /d     

so        

A = d C / ε = 0.000265 * 244 x 10^-12 / 8.85 x10^-12 =   0.007303m²

(c) E = V / d = 561.48 /0.000265 = 2118792.5 V/m

(d) σ = Q / A = 0.137 x10^-6 / 0.007303 =   1.876 x 10^-5 C/m²