Why does adding an electron to a 2.1 aF capacitor give zero energy change?

[member 422733]

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:
$U = Q^2 \div 2C$

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:
$\Delta U = U_f - U_i \\<br /> = \frac{(Q_f)^2 - (Q_i)^2}{2C} \\<br /> = \frac{(-0.5e)^2 - (0.5e)^2}{2C}<br /> = 0$

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

 

[NascentOxygen]

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?

 

[member 422733]

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.

 

[member 422733]

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.

 

[rude man]

What is the voltage on the capacitor before and after adding the charge e? What then is the difference in energy?