Answer to Question #296833 in Physics for Ding

Question

Lesson: Capacitance & Capacitors

Solve the problem:

The two metal objects in the figure below have net charges of +70 pC and -70 pC, which result in a 20 V potential difference between them. (a) What is the capacitance of the system? (b) if the charges are changed to +200 pC and -200pC, what does capacitance become? (c) What does the potential difference become?

Answer 1

Given:

$q_1=70\:\rm pC$

$V_1=20\:\rm V$

(a) What is the capacitance of the system? 

$$C=\frac{q_1}{V_1}=\frac{70*10^{-12}\:\rm C}{20\:\rm V}=3.5*10^{-12}\:\rm F$$

(b) if the charges are changed to +200 pC and -200pC, what does capacitance become?

$$C=\rm const=3.5*10^{-12}\:\rm F$$

(c) What does the potential difference become?

$$V_2=\frac{q_2}{C}=\frac{200*10^{-12}\:\rm C}{3.5*10^{-12}\:\rm F}=57\:\rm V$$