Capacitors: charge, potential difference and stored energy

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SUMMARY

This discussion focuses on calculating the charge, potential difference, and stored energy for three capacitors (C1 = 4.0 µF, C2 = 8.0 µF, C3 = 9.0 µF) connected to a 50V source. The equation CV = Q is essential for determining charge, where C is capacitance and V is voltage. The discussion emphasizes the difference in charge storage between capacitors in series and parallel configurations, highlighting that the equivalent capacitance (Ceq) dictates the charge distribution. Understanding electric potential and energy is crucial for solving these types of problems.

PREREQUISITES
  • Understanding of capacitor configurations (series and parallel)
  • Familiarity with the equation CV = Q
  • Knowledge of electric potential and energy concepts
  • Basic grasp of circuit theory and charge flow
NEXT STEPS
  • Learn how to calculate equivalent capacitance for series and parallel capacitors
  • Study the relationship between voltage, charge, and energy in capacitors
  • Explore practical applications of capacitors in electronic circuits
  • Investigate the effects of different dielectric materials on capacitance
USEFUL FOR

Students studying electrical engineering, physics enthusiasts, and anyone looking to deepen their understanding of capacitor behavior in circuits.

CIERAcyanide
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Homework Statement



The 50Volts is going through 3 capacitors.
C1 = 4.0 µF, C2 = 8.0 µF, and C3 = 9.0 µF,

The URL below is the image of how the capacitors are connected:
http://www.webassign.net/hrw/26_27.gif

For each capacitor i need to find the charge, the potential difference and the stored energy.

Homework Equations



CV = Q

The Attempt at a Solution


I really don't understand what potential difference is, and I'm not sure what the problem asks for with the stored energy.
I tried CV = Q for the charge of C1 and got 4E-6 * 50 = 2E-4, which is wrong.
 
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CIERAcyanide said:

Homework Statement



The 50Volts is going through 3 capacitors.
C1 = 4.0 µF, C2 = 8.0 µF, and C3 = 9.0 µF,

The URL below is the image of how the capacitors are connected:
http://www.webassign.net/hrw/26_27.gif

For each capacitor i need to find the charge, the potential difference and the stored energy.

Homework Equations



CV = Q

The Attempt at a Solution


I really don't understand what potential difference is, and I'm not sure what the problem asks for with the stored energy.
I tried CV = Q for the charge of C1 and got 4E-6 * 50 = 2E-4, which is wrong.

Remember that a capacitor is a break in the circuit. Capacitors in parallel can store a lot more charge than two in series, because the parallel combination allows each plate to fill up with its "rated" charge. In the series combination, the break in the circuit prevents this simple process. Instead, charge is stored on the first plate the voltage source comes in contact with and then charge departs the bottom plate toward the second capacitor's first plate. The second capacitor's second plate then departs charge toward to voltage source allowing for an apparent current (though no charge actually flows THROUGH a capacitor). Specific to this problem, all of the charge depleted from the voltage source due to the equivalent capacitance, Ceq, must be stored on the first two capacitors that are in parallel. The third capacitor holds the same charge as the two in parallel do (because it only receives charge out of the departure of charge from the first two capacitors which is equal to the stored charge on the first capacitor's plates)

Voltage is the energy per unit charge needed to move a charge from point p(1) to point p(2). A person must expend energy per unit charge in moving some charge q only if there is an electric field between the points. (Remember, f = Eq. If E = 0, f = 0 and w = 0).

The mechanical analog to voltage usually helps students visualize the concept:
gravitational potential energy due to a gravitational field is mgh where m is the measurement of the material's sensitivity to the field's tendency, g is the field, and h is the distance over which the constant field acts. If both sides were divided by mass, we would arrive to the never used but now dubbed gravitational potential (work per mass) -- gh.

Similarly, electric potential energy is qEd where q is the measurement of the material's sensitivity to the field's tendency, E is the field, and d is the distance over which the constant field acts. If both sides are divided by charge, we would arrive to the always used electric potential (i.e. voltage, work per charge) -- Ed.
 
Last edited:
Thank you so much! I really appreciate your in depth response; it cleared up a lot.
 

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