A parallelplate air capacitor of capacitance of 270 pF has a

A parallel-plate air capacitor of capacitance of 270 pF has a charge of magnitude 0.128 pC on each plate. The plates are 0.336 mm apart. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the electric-field magnitude between the plates? (d) What is the surface charge density on each plate?

Solution

Q = CV

=> V = 0.128*10^-6)/(270*10^-12) = 474.074 V

b) C = A*apsalan/d

=> A = 270*10^-12*0.336*10^-3)/(8.85*10^-12) = 0.010250 m^2

c) E = V/d = 474.074)/(0.336*10^-3) = 1.41*10^6 N/C

d) surface charge density = Q/A = 0.128*10^-6)/(0.010250 = 0.0000124 C/m^2