Two capacitors (C 1 = 5.1 uF, C 2 = 19.6 uF) are charged individually to (V 1 =

Question

Two capacitors (C1 = 5.1 uF, C2 = 19.6 uF) are charged individually to (V1 = 15.0 V, V2 = 3.4 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: 5.795E+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.

Correct, computer gets: 4.69E-05 C

Hint: Pick one capacitor, you know the charge on this capacitor before they were connected. Now that you know the potential difference after they are connected - remember the potential drop is the same for both of them - you can calculate the charge: Q=CV. You have to calculate the difference of Q before and after.

9. [1pt]
By how much (absolute value) is the total stored energy reduced when the two capacitors are connected?

need help with the last question

Explanation / Answer

1)Final Potential difference

charge equals capacitance times voltage

Q1 = C1 * V1 = 5.1 * 10^-6 * 15 = 7.6 * 10^-5

Q2 = C2 * V2 = 19.6 *10^-6 *3.4 = 6.66* 10^-5

The total charge on the two capacitors is 14.26 * 10^-5

Since the combined parallel caapacitance is 24.7 * 10^-6 or 2.47 * 10^-5 , that corresponds to a voltage of 14.26 *10^-5 / 2.47*10^-5 = 5.77 volts.

2) Charge flow

The charge on the smaller cap is its caapacitance times the final voltage

Q1 = C1 * V1 = 5.1 * 10^-6 * 5.77 = 2.94 * 10^-5

since its charge started out at 7.63*10^-5 , it has lost 4.7 *10^-5

Q2 = C2 * V2 = 19.6 *10^-5 * 5.77 = 11.30 *10^-5 , which is 4.7 *10^-5 greater than its initial value .

3) Total stored energy

J = ( c* v ^2) / 2

J1 = (5.1 * 10^ -6 * 15^2 ) / 2 = 5.73 * 10^-4 J

J2 = (19.6 *10^-6 * 3.4^2) / 2 = 11.3*10^-4 J

When paralleled

J = ( 24.7 *10^-6 * 5.77^2)/ 2 = 41.1*10^-4 J

reduction in stored energy = 29.8 *10^-4 J

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