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Electric Current: Finding Energy lost in a capacitor

1. The problem statement, all variables and given/known data

E& M
Purcell, 3.11

A 100-pF capacitor is charged to 100 volts. After the charging battery is disconnected, the capacitor is connected in parallel to another capacitor. If the final voltage is 30 volts, what is the capacitance of the second capacitor? How much energy was lost, and what happened to it?

2. Relevant equations

Q = 1/4pi times integral of E dA or the integral of σ dA

C = Q/phi not = a

σ = E/4pi = δ1 - δ2 / s4pi

capacitance of conductor: Q = C(δ1 - δ2)



3. The attempt at a solution

Using the known equations and values and keeping in mind that the plate is charged in parallel, I know that the energy stored in a capacitor is

U = 1/2 A/4pi(s) (Es)^2 or E^2/8pi * volume

The voltage difference is 100-30 = 70 volts, so we start to plug in to find capacitance by denoting Q = 70, C = 70/phi. I don't really know how to find phi, or the energy afterwards, but I do know the energy equation. Any help would be greatly appreciated!
 
32,566
8,457
phi is a voltage? Ok

That is wrong.
You can calculate Q with the capacity and known initial voltage.
This Q stays the same when the other capacitor is connected, this allows to find its capacity.

Energy loss can be calculated afterwards.

You don't need details about the capacitors, like volume, electric field and so on - they are irrelevant here.
 

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