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Change in energy of a capacitor

  • #1

Homework Statement


A 2.1 aF capacitor has a net charge of 0.5e (a positive charge, the symbol e is taken as a positive number 1.6 x 10-19 coulomb). What is the energy needed to add one electron (charge -e) to this capacitor?


Homework Equations


Energy in a Capacitor:
[itex]U = Q^2 \div 2C[/itex]

Where U is the energy, Q is the charge and C is the capacitance

The Attempt at a Solution


The initial charge is +0.5e and after adding a charge of -e, the final charge would be -0.5e.
To find the energy needed:
[itex]\Delta U = U_f - U_i \\
= \frac{(Q_f)^2 - (Q_i)^2}{2C} \\
= \frac{(-0.5e)^2 - (0.5e)^2}{2C}
= 0[/itex]

Squaring the initial and final charge results in positive e/4 and subtracting these values of equal magnitude gives 0. But having 0 energy change doesn't make sense to me. What am I doing wrong here?

Thanks in advance for any help
 

Answers and Replies

  • #2
NascentOxygen
Staff Emeritus
Science Advisor
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A 2.1 aF capacitor has a net charge of 0.5e (a positive charge, the symbol e is taken as a positive number 1.6 x 10-19 coulomb).
So on its plates this capacitor is storing a charge difference of half an electron?
 
  • #3
So on its plates this capacitor is storing a charge difference of half an electron?
This is indeed the problem as my Professor wrote it.
 
  • #4
After asking him about it, my Professor responded:

"The non integer charge can come from polarization of the metal piece by an electric field, that slightly moves the position of many charges."

If that helps.
 
  • #5
rude man
Homework Helper
Insights Author
Gold Member
7,688
746
What is the voltage on the capacitor before and after adding the charge e? What then is the difference in energy?
 

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