Calculate the capacitance of a capacitor when each plate has an area of 1.2 cm2 and are 2.5 mm apart. 2. Calculate the maximum charge that each plate above can hold if a voltage difference of 12.0 V is applied to the capacitor. 3. Calculate the magnitude of the uniform electric field produced by this capacitor if the voltage supplied is 12.0 V

Answer: a) 0.43 pF (10~-12F); 5.16 pC (10^-12C); 4.85 * 10^3 N/C

Explanation: In order to solve this problem we have to consider the expresion for the capacitor of parallel plates, which is

ΔV=Q/C in this case C=A*εo/d where A is the area of the plates and d its separation,

The we have;

C=0,00012 m^2 * 8,85*10~ - 12/0.0025 m = 0.43 pF

Then we have that

Q = ΔV*C = 12V * 0.43 pF=5.16 pC

Finally, the electric field inside the plates is given by:

E=Q / (A*εo) = 5.16 pC / (0.00012 m^2*8.85 * ^10-12 C^2/N*m^2) = 4.85 * 10^3 N/C