Two capacitors (C 1 = 6.5 uF, C 2 = 15.5 uF) are charged individually to (V 1 =

Question

Two capacitors (C1 = 6.5 uF, C2 = 15.5 uF) are charged individually to (V1 = 17.8 V, V2 = 5.3 V). The two capacitors are then connected together in parallel with the positive plates together and the negative plates together.
7. [1pt]
Calculate the final potential difference across the plates of the capacitors once they are connected.

Correct, computer gets: 8.993E+00 V

Hint: This problem was discussed in class. Remember capacitance in parallel: C_T = C_1 + C_2 The total charge is the sum of the individual charges. The voltage drop across both capacitors is the same after they are connected.

8. [1pt]
Calculate the amount of charge (absolute value) that flows from one capacitor to the other when the capacitors are connected together.

Explanation / Answer

apply the formula for final volatge V = C1V1 +C2V2/(C1+C2)

here

V1 = 17.8 V

V2 = 5.3 V

C1 = 6.5 uF

C2 = 15.5uF

V = (6.5 * 17.8)   +(15.5 * 5.3)/(6.5 +15.5)

V = 8.99 *10^3 Volts

-------------------------------------------

apply Charge Q = CV

when Connected in parallel, Cnet = C1+C2 = 15.5+6.5

Cnet = 22uF

so

Charge Q = 22e-6* 8000

Q   = 0.176 Coulombs

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