How much energy must a 6.0 V battery expend to fully charge a ccapacitor

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SUMMARY

The discussion focuses on calculating the energy required by a 6.0 V battery to charge capacitors of 0.13 µF and 0.21 µF. Participants clarify that the energy can be calculated using the formula \(E = \frac{1}{2}CV^2\), where \(C\) is the capacitance and \(V\) is the voltage. Additionally, the electric field strength between the plates of an 0.86 µF capacitor with a charge of 74 µC and a separation of 1.9 mm is also discussed. Participants emphasize the importance of precision in calculations, noting that results should be expressed in microjoules for accuracy.

PREREQUISITES
  • Understanding of capacitor charging equations, specifically \(E = \frac{1}{2}CV^2\)
  • Familiarity with capacitance units, such as microfarads (µF)
  • Knowledge of electric field calculations between capacitor plates
  • Basic proficiency in scientific notation and unit conversions
NEXT STEPS
  • Learn about energy storage in capacitors and related formulas
  • Explore electric field calculations for various capacitor configurations
  • Study unit conversions and precision in scientific calculations
  • Investigate the effects of different dielectric materials on capacitance
USEFUL FOR

Students in physics or electrical engineering, educators teaching capacitor concepts, and anyone interested in understanding energy storage in capacitors.

jhcollaz
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Homework Statement



How much energy must a 6.0 V battery expend to fully charge a 0.13 µF and a 0.21 µF capacitor when they are placed in the following configurations?

Another question if anyone can help we get started on this one. I'm not sure where to begin

How strong is the electric field between the plates of an 0.86 µF air-gap capacitor if they are 1.9 mm apart and each has a charge of 74 µC?

Homework Equations



In my attempt at a solution.

The Attempt at a Solution



http://imgur.com/a/ELKdZ#1

Part b series

I keep getting 0.000001 J but it says it not right, can anyone help me out?

Homework Statement


Homework Equations


The Attempt at a Solution

 
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I just did the same calculation and also got 0.000001, however, if you multiply it by 10^6, you get 1.445, which means the calculator is rounding off and not showing all the digits. It would probably be more accurate if you wrote 1.445 micro joules rather than 0.000001 J.
 

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