A 2.75 F capacitor is charged to 495 V and a 3.70 F capacitor is charged to 505

Question

A 2.75 F capacitor is charged to 495 V and a 3.70 F capacitor is charged to 505 V . These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each capacitor? What will be the charge on each capacitor? What is the voltage for each capacitor if plates of opposite sign are connected? What is the charge on each capacitor if plates of opposite sign are connected?

Explanation / Answer

1. when the the positive plates are now connected to each other and the negative plates are connected to each other

the voltages equalize, charge is conserved, and energy is NOT conserved.

Charge on the two caps is Q = CV
Q = 2.75 x 495 + 3.7 x 505 = 1361.25 + 1868.50 = 3229.75 µC

after they are connected, C = 2.75 + 3.7 = 6.45µF, so V = Q/C
V = 3229.75 µC / 6.45µF = 500.73 volts. (same voltage on both)...........Ans.

Q(2.75) = CV = 500.73 x 2.75 = 1377.025 µC........Ans.
Q(3.7) = CV = 500.73 x 3.7 = 1852.70 µC........Ans.

2. if plates of opposite sign are connected

Now the charges subtract, Q = 1361.25 - 1868.50 = 507.25 µC

after they are connected in series , C = 2.75*3.7/ 6.45 = 1.577 µF,

so V = Q/C
V = 507.25 µC / 1.577 µF = 321.54 volts. (same voltage on both)...........Ans.

Q(2.75) = CV = 321.54 x 2.75 = 884.26 µC........Ans.
Q(3.7) = CV = 321.54 x 3.7 = 1189.698 µC........Ans.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.