A 10.0-µF capacitor is charged so that the potential difference between its plates is 10.0 V.
A 5.0-µF capacitor is similarly charged so that the potential difference between its plates is 5.0 V.
The two charged capacitors are then connected to each other in parallel with positive plate connected to positive plate and negative plate connected to negative plate.
-How much charge flows from one capacitor to the other when the capacitors are connected?
A)17 $\mu$C
B)33 $\mu$C
C)67 $\mu$C
D)83 $\mu$C
E)zero coulombs

Question

A 10.0-µF capacitor is charged so that the potential difference between its plates is 10.0 V. 
A 5.0-µF capacitor is similarly charged so that the potential difference between its plates is 5.0 V.
The two charged capacitors are then connected to each other in parallel with positive plate connected to positive plate and negative plate connected to negative plate.
-How much charge flows from one capacitor to the other when the capacitors are connected?
A)17 $\mu$C
B)33 $\mu$C
C)67 $\mu$C
D)83 $\mu$C
E)zero coulombs

Quizplus · Accepted Answer

The total charge on the two capacitors is the sum of the charges on each capacitor:$$Q = Q_1 + Q_2 = C_1V_1 + C_2V_2 = (10.0 \mu F)(10.0 V) + (5.0 \mu F)(5.0 V) = 150 \mu C$$When the capacitors are connected in parallel, the potential difference across each capacitor is the same. The charge on each capacitor will adjust so that the total charge is still 150 \mu C. The charge on the 10.0 \mu F capacitor will decrease, and the charge on the 5.0 \mu F capacitor will increase. The amount of charge that flows from one capacitor to the other is:$$Q = Q_1 - Q_1' = C_1V_1 - C_1V = (10.0 \mu F)(10.0 V) - (10.0 \mu F)(5.0 V) = 50 \mu C$$Therefore, the correct answer is A: 17 $\mu$C.

A 100-µF Capacitor Is Charged So That the Potential Difference Between