Capacitance Homework Word Problem

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SUMMARY

The problem involves two capacitors: a 2.50µF capacitor charged to 857V and a 6.80µF capacitor charged to 652V. After disconnecting them from their batteries and connecting the positive plates together and the negative plates together, the conservation of charge principle is applied. The charge on each capacitor is calculated using the formula Q=CV, resulting in Q1 = 0.0021C and Q2 = 0.0044C. To find the potential difference across each capacitor, the capacitance of the combined system in parallel is computed, followed by the application of Q=CV to determine the new potential difference.

PREREQUISITES
  • Understanding of capacitor fundamentals, including capacitance and charge.
  • Familiarity with the formulas Q=CV and Vb-Va for potential difference.
  • Knowledge of energy concepts in capacitors, specifically PE=V/Q.
  • Basic principles of charge conservation in electrical circuits.
NEXT STEPS
  • Study the principles of capacitors in parallel configurations.
  • Learn about charge conservation in electrical circuits.
  • Explore the derivation and applications of the formulas Q=CV and V=Ed.
  • Investigate potential energy calculations in capacitive systems.
USEFUL FOR

Students studying electrical engineering, physics enthusiasts, and anyone tackling capacitor-related problems in circuit analysis.

mer584
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The problem.
A 2.50uF capacitor is charged to 857V and a 6.80uF capacitor is charged to 652V. These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potential difference across each and the charge on each? (note that charge is conserved)


Homework Equations


PE= V/Q ; V=Ed; Q=CV; C= Eo (A/D); Vb-Va


Attempt:
What I started problem I used the formula Q=CV to find the charge that is occurring in each case finding Q1 = .0021C and Q2= .0044 C. I really wasnt sure if this was even helpful or where to go from there. It appears to be a uniform field so I know I can use V=Ed but we don't have a distance or an area in this problem.

In order to get the potential difference I know you need to work with Vb-Va and possibly the potential energy. Should I use PE= V/Q then subtract the potential energies to find the work and then the poential difference?
 
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I would compute the capacitance of the two capacitors in parallel, then use Q=CV for this capacitor to get the potential difference across both, and then use Q=CV for the 2 original capacitors to get the charge on each
 

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