   Chapter 16, Problem 45P

Chapter
Section
Textbook Problem

A 25.0-μF capacitor and a 40.0-μF capacitor are charged by being connected across separate 50.0-V batteries. (a) Determine the resulting charge on each capacitor. (b) The capacitors are then disconnected from their batteries and connected to each other, with each negative plate connected to the other positive plate. What is the final charge of each capacitor? (c) What is the final potential difference across the 40.0-μF capacitor?

(a)

To determine
The charge on each capacitor.

Explanation

Formula to calculate the charge on 25.0μF is,

Q1=C25.0μF(ΔV)

• ΔV is the potential difference.

Substitute 25.0μF for C25.0μF and 50.0 V for ΔV

Q1=(25.0μF)(50.0V)=1.25mC

Formula to calculate the charge on 40.0μF is,

Q2=C40

(b)

To determine
The new charge on each capacitor.

(d)

To determine
The final potential difference on 40.0μF

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