A 10.0 - MuF capacitor is charged so that the potential difference between its p

Question

A 10.0 - MuF capacitor is charged so that the potential difference between its plates is 10 0 V A 5 0-MuF capacitor is similarly charged so that the potential difference between its plates is 5.0 V The two chafed capacitors are then connected to each other in parallel with positive plate connected to positive plate and negative plate connected to negative plate. How much charge flows from one capacitor to the other when the capacitors a 67 MuC 17 MuC 83 MuC 33 MuC Zero What is the final potential difference across the plates of the capacitors when they are connected in parallel? 7.5 V 10 V 6.7 V 8.3 V 5.0 V

Explanation / Answer

here,

capacitor of capcitor 1, C1 = 10 uF = 10^-5 F
potential Difference acorss C1, V1 = 10 V
Charge across plates, Q1 = C1/v1
Q1 = 10 * 10^-5
Q1 = 10^-4 C

capacitor of capcitor 2, C2 = 5 uF = 0.5 * 10^-5 F
potential Difference acorss C2, V2 = 5 V
Charge across plates, Q2 = C1/v1
Q2 = 5* 0.5* 10^-5
Q2 = 2.5*10^-5 C

When they caonnected in parallel, Eqvilanet capcitance will be, C
C = C1 + C2
C = 10 + 5
C = 15 uF

When they are connected in parallel net charge,

Q = Q1 + Q2
Q = 10^-4 + 0.25*10^-4
Q = 1.251*10^-4

Part b:
Bet Voltage, V = Q/C
V' = 1.251*10-^4 /(15*10^-6)
V' = 8.34 V

Part A:
WHen they are connected, Neq charge on Q1 on c1 ,
Q1' = C1*V'
Q1' = 10^-5 * 8.34
Q1' = 0.834*10^-4 C

net loss of charge on capacitor 1
Q1'' = 0.834*10^-4 - 10^-4
Q1'' = 1.66 * 10^-5 C

Similarly,
Q2' = C2*V
Q2' = 0.5 * 10^-5 * 8.34
Q2' = 4.17*10^-5 C

net loss of charge on capacitor 1
Q2'' = Q2 - Q2'
Q2'' = 2.5*10^-5 - 4.17*10^-5
Q2'' = 1.67 *10^-5

Net charge will be :
Q = Q1'' + Q2''
Q = 1.66 * 10^-5 + 1.67 * 10^-5
Q = 3.33 * 10^-5
Q = 33 uC

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