Answer to Question #296833 in Physics for Ding

Question #296833

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?

1
Expert's answer
2022-02-13T12:15:08-0500

Given:

q1=70pCq_1=70\:\rm pC

V1=20VV_1=20\:\rm V


(a) What is the capacitance of the system? 

C=q1V1=701012C20V=3.51012FC=\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=const=3.51012FC=\rm const=3.5*10^{-12}\:\rm F

(c) What does the potential difference become?

V2=q2C=2001012C3.51012F=57VV_2=\frac{q_2}{C}=\frac{200*10^{-12}\:\rm C}{3.5*10^{-12}\:\rm F}=57\:\rm V


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