Please show all work. Thank you in advance. A potential difference of 300 V is a

Question

Please show all work. Thank you in advance.

A potential difference of 300 V is applied to a series connection of two capacitors, of capacitance C_1 = 2.0 mu F and capacitance C_2 = 8.0 mu F. What are the charge on and the potential difference across each capacitor? (b) The charged capacitors are disconnected from each other and from the battery. They are then reconnected, positive plate to positive plate and negative plate to negative plate, with no external voltage being applied. What are the charge and the potential difference for each now? (c) Suppose the charged capacitors in (a) were reconnected with plates of opposite sign together. What then would be the steady-state charge and potential difference for each?

Explanation / Answer

For series connection,

Cnet = C1*C2/(C1+C1)

So, Cnet = 2*8/(2+8) = 1.6 uF

a)

So, charge on each capacitor = Cnet*V = 1.6*10^-6*300

Q = 4.8*10^-4 C

Voltage across C1 = Q/C1 = 4.8*10^-4/(2*10^-6) = 240 V

Voltage across C2 = 300 - 240 V = 60 V

b)

The connection is then changed to parallel connection,

So, Cnet' = C1 +C2 = 2+8 = 10 uF

Let the voltage across both be V

So, total charge on both , Qnet = Cnet*V which must be two times as before

So, 10*10^-6*V = 2*4.8*10^-4

So, V = 96 V <------ voltage across both

So, So, charge on C1 = 2*10^-6*96 = 1.92*10^-4 C

and charge on C2 = 8*10^-6*96 = 7.68*10^-4 C

c)

When connected like this the charges cancel each other as they are same.

So, the charge and voltage are 0

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