Disconnected and Reconnected Capacitors

  • Thread starter nns91
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Homework Statement


A 100pF and a 400pF capacitor are both charged to 2kV. They are then disconnected from the voltage source and are connected together, positive plate to negative plate.

(a). Find the resulting potential difference across each capacitor
(b). Find the energy lost when the connections are made


Homework Equations



Q=C.V

The Attempt at a Solution



So when they are reconnected they are in parallel. However the charge is redistributed because positive plate is hooked to negative plate.

I have calculated the total charge using Q=C.V= (100pF+400pF)*2kV=1[tex]\mu[/tex]C

The I use Q1/V1=Q2/V2 and Q1+Q2=1[tex]\mu[/tex]C to calculate the new charge and get Q1= 0.2[tex]\mu[/tex]C and Q2=0.8[tex]\mu[/tex]C.

However, the new voltage calculated using V=Q/C is 2*10^9 V which is larger than initial voltage. It does not make sense at all. Where did I screw up ?
 

Answers and Replies

  • #2
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Do I have to calculate the initial charge of each capacitor and average them to 0.5microC and then calculate the voltage from there ?
 
  • #3
ehild
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Do I have to calculate the initial charge of each capacitor and average them to 0.5microC and then calculate the voltage from there ?

Calculate the initial charges. As the capacitors are connected positive plate to negative, the total charge will be the difference of the initial charges. The new capacitor has capacitance C1+C2, the charge is Q2-Q1, from that you get the voltage.

ehild
 
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