Find the Final Voltage across a 2.0-microfarad Capacitor

[yxgao]

62 on GRE. Two capacitors of capacitances 1.0 microfarad and 2.0 microfarads are each charged by being connected across a 5.0-volt battery. They are disconnected from the battery and then connected to each other with resistive wires so that plates of opposite charge are connected together. What will be the magnitude of the final voltage across the 2.0-microfarad capacitor?
a. 0 V
b. 0.6 V
c. 1.7 V
d. 3.3 V
e. 5.0 V

The answer is C. Can someone outline the process to arrive at this?

By the way, does anyone know a good website that explains the answers to the practice GRE for physics posted on gre.org?

 

[csmines]

Capacitance adds backwards of how resistance adds; Therefore when the two capacitors are put together in series the capacitance adds as if the two capacitors were resistors in parallel. This yeilds that the series capacitance on either capacitor when the two are connected in series is equal to

1/C1+1/C2=1/Cs or...
Cs=C1*C2/(C1+C2)

plugging in numbers you get that Cs=2/3 microfarads. Comparing this with the initial capacitance it is easy to see that the series capacitance on the 2 uF capacitor is exactly 1/3 of the initial. Following from this if 5V are on the 2uF capacitor initially then 1/3 of 5V will be on the 2uF capacitor when in series with the 1 uF capacitor. 5V/3 =1.6667 or about 1.7V.

Bored CSMPhysicist.

 

[M98Ranger]

To find the final voltage across the 2.0-microfarad capacitor, we can use the formula for capacitance in series:

Ceq = C1*C2 / (C1 + C2)

First, we need to find the equivalent capacitance of the two capacitors when connected in series. Plugging in the values, we get:

Ceq = (1.0 microfarad)*(2.0 microfarads) / (1.0 microfarad + 2.0 microfarads) = 2/3 microfarads

Next, we can use the formula for capacitance in parallel to find the total charge stored in the capacitors when they are connected in parallel:

Q = Ceq * V = (2/3 microfarads) * (5.0 volts) = 10/3 microcoulombs

Since the capacitors were initially charged by a 5.0-volt battery, the total charge on each capacitor is also 10/3 microcoulombs.

When the capacitors are connected in parallel, the plates of opposite charge will be connected together, resulting in the redistribution of charge. This will cause the voltage on each capacitor to equalize until they both have the same charge.

Using the formula for capacitance, we can find the final voltage on the 2.0-microfarad capacitor:

V = Q / C = (10/3 microcoulombs) / (2.0 microfarads) = 5/6 volts = 0.833 volts

Therefore, the magnitude of the final voltage across the 2.0-microfarad capacitor is approximately 0.833 volts, which is closest to option C (1.7 volts).

As for a website that explains the answers to practice GRE for physics, you can try searching for specific topics on websites like Khan Academy or Physics Classroom. You can also check out online forums and communities where students discuss their solutions and explanations for GRE practice questions.