Electric Current: Finding Energy lost in a capacitor

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SUMMARY

The discussion centers on a physics problem involving a 100-pF capacitor charged to 100 volts, which is then connected in parallel to another capacitor, resulting in a final voltage of 30 volts. The key equations used include the energy stored in a capacitor, U = 1/2 C V^2, and the relationship between charge, capacitance, and voltage, Q = C V. The energy lost during the process can be calculated by determining the initial and final energy states of the capacitors. The problem emphasizes the conservation of charge and the need to calculate the capacitance of the second capacitor based on the voltage drop.

PREREQUISITES
  • Understanding of capacitor fundamentals, including capacitance and energy storage
  • Familiarity with the equations governing electric charge and voltage relationships
  • Knowledge of parallel capacitor configurations and their implications on voltage and charge
  • Basic skills in algebra for manipulating equations to solve for unknowns
NEXT STEPS
  • Calculate the capacitance of the second capacitor using the formula C = Q/V
  • Learn about energy conservation in electrical circuits, specifically in capacitor systems
  • Explore the implications of connecting capacitors in parallel versus series
  • Study the concept of energy loss in electrical systems and how it relates to circuit components
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in understanding capacitor behavior in electrical circuits.

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Homework Statement



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?

Homework 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)



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 into 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!
 
Physics news on Phys.org
phi is a voltage? Ok

Q = 70
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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