The magnitude of charge on each plate of the capacitor is related to the voltage

Question

The magnitude of charge on each plate of the capacitor is related to the voltage difference across the plate and the capacitance, C, of the capacitor: Q = C?V.

If we connect the negative plate of one capacitor of capacitance C1 to the positive plate of a second capacitor of capacitance C2, we say we are connecting them in series. If we connect the positive plate of one capacitor of capacitance C1 to the positive plate of a second capacitor of capacitance C2, we say we are connecting them in parallel.

A. In the figure above, two capacitors connected are in series. Assume that the capacitors were first connected to each other and then to the battery.

1. If the charge on the positive plate of capacitor C1 is Q1 (a positive charge), what is the charge on the negative plate of capacitor C1? On the positive plate of capacitorC2? On the negative plate of capacitor C2? Explain how you know.

2. If the voltage rise in the battery is ?V, what is the voltage drop across each of the individual capacitors? Explain how you know.

3. Using your result above, find the effective capacitance of the combined pair of capacitors. (That is, what you would measure their capacitance as if they were contained in a box and you didn't know they were two separate capacitors.)

B. In the figure above two capacitors are connected in parallel. Assume that the capacitors were first connected to each other and then to the battery.

1. If the charge on the positive plate of capacitor C1 is Q1 (a positive charge), what is the charge on the negative plate of capacitor C1? On the positive plate of capacitor C2? On the negative plate of capacitor C2? Explain how you know.

2. If the voltage rise in the battery is ?V, what is the voltage drop across each of the individual capacitors? Explain how you know.

3. Using your result above, find the effective capacitance of the combined pair of capacitors. (That is, what you would measure their capacitance as if they were contained in a box and you didn't know they were two separate capacitors.)

Explanation / Answer

A1) charge on the negative plate of capacitor C1 = -Q1

Charge on the positive plate of capacitorC2= Q1

charge On the negative plate of capacitor C2 = -Q1

the charges will flow from battery to positive plate of first capacitor , negative plate of first to positive plate os second, negative plate of second to the battery

A2) C1V1 = C2V2 = (C1*C2/(C1+C2))V

V1=C2/(C1+C2))V

V2=C1/(C1+C2))V

because charge is equal on both and voltage is inversally propotional to capacitance for constant charge

A3) Since charge moved is equal, so for equivalent capacitance,

C1V1 = Ceq V

C1*C2/(C1+C2))V =  Ceq V

Ceq=C1*C2/(C1+C2))

B1)

charge on the negative plate of capacitor C1 = -Q1

Charge on the positive plate of capacitorC2= C2V = C2 *Q1/C1= Q1C2/C1

charge On the negative plate of capacitor C2 = - Q1C2/C1

B2) It will be   V in both capasitors because battery is accross them

b3)charge is conserved,

Ceq  V= C1 V+   VC2

Ceq= C1 +C2

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