Question

What voltage must be applied to an 6 nF capacitor to store 0.14 mC of charge? Give answer in terms of kV.

Answer 1

`V=23.3kV`

Step-by-step explanation:

Definition of the capacitance C, where a voltage V is applied and a charge Q is stored:

`Q=C*V`

We solve to find V:

`V=Q/C=0.14*10^(-3)C/6*10^(-9)F)=2.33*10^(4)V=23.3kV`