SUMMARY
The forum discussion centers on calculating the work required to increase the separation of an isolated parallel-plate capacitor from 1.2 mm to 4.5 mm while maintaining a constant charge of Q=1.4x10^-5 C. The capacitance at the initial separation is given as C1=3.1x10^-11 F. Participants discuss the formulas for capacitance and potential energy, ultimately determining that the work done is W=192.76 J after calculating the potential energies at both separations. The correct capacitance at 4.5 mm is found to be C2=8.26x10^-12 F.
PREREQUISITES
- Understanding of parallel-plate capacitor theory
- Familiarity with capacitance formulas, specifically C=ε₀A/d
- Knowledge of potential energy in capacitors, U=(1/2)(Q²)/C
- Ability to perform unit conversions, particularly between millimeters and meters
NEXT STEPS
- Study the relationship between capacitance and plate separation in parallel-plate capacitors
- Learn how to derive the work done in electrostatic systems
- Explore the implications of constant charge on capacitor behavior
- Investigate the role of electric field strength in capacitor calculations
USEFUL FOR
Students studying electromagnetism, electrical engineers, and anyone involved in capacitor design or analysis.