PART A These capacitors are then disconnected from their batteries, and the posi

Question

PART A

These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each capacitor?  

Enter your answers  numerically separated by a comma.  

PART B

What will be the charge on each capacitor?

Enter your answers numerically separated by a comma.

PART C

What is the voltage for each capacitor if plates of opposite sign are connected?   

Express your answers using two significant figures. Enter your answers numerically separated by a comma.

PART D

What is the charge on each capacitor if plates of opposite sign are connected?

Express your answers using two significant figures. Enter your answers numerically separated by a comma.

Explanation / Answer

Part (A)

Capacitors are connected in parallel combination.

Q1 = C1V1 = 2.9*500 = 1450 uC

Q2 = C2V2 = 3.5*505 = 1767.5 uC

Q net = Q1+Q2 = 3217.5 uC

C net = C1 + C2 ( parallel combination) = 2.9+3 5=6.4uC

Potential difference across each capacitors would be same .

V net = Qnet/Cnet = 3217.5/6.4=502.73 V

Ans { 502.73V , 502.73V}

Part (B)

Q1 = 2.9*502.73=1457.93 uC

Q2 = 3.5*502.73=1759.57 uC

Ans {1457.93uC, 1759.57uc} or { 1.4 mC , 1.7 mC}

Part (C)

Now they are connected in series combination

Cnet = C1C2/(C1+C2) = 2.9*3.5/(2.9+3.5)= 1.58 uC

Vnet = V1 + V2 = 500+505 = 1005 V

Qnet = Can't*Vnet = 1.58*1005= 1593.87 uC

V1new = Qnet/C1 = 1593.87/2.9 = 549.6V

V2new = Qnet/C2 = 1593.87/3.5 = 455.39 V

Ans {549.60V , 456.39V}

Part (D)

Calculated in part C . Charge will be same on both as they are in series combination.

Ans{ 1593.87uC , 1593.87 uC}

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