Question

acitor A was measured to have a capacitance of 80nF. if the charge it could hold is 30mC and the plates have distance of 0.5cm a) find the voltage that thr capacitor was run through and b) find the electric field between the plates. if the circuit diagram belowonly used a capacitor with the same capacitance of the capacitor A. c) find the total capacitance of the circuit

Answer 1

a) The voltage in the capacitor is

$$V=Q/C=0.03/(80\cdot10^{-9})=375000\text{ V}.$$

b) The electric field:

$$E=V/d=375000/0.005=75 000 000\text{ V/m}.$$