Problem 4:

A parallel plate capacitor of area 10 cm2 and separation 0.5 cm, with no material between the plates, has its plates connected to the opposite terminals of a battery providing 5 kV potential difference.
  1. What is the magnitude of the charge on each plate?
  2. If an electron is placed just inside the negatively charged plate and allowed to travel (in vacuum) to the positive plate, what is its energy (in eV) upon impact with the positive plate?
  3. Suppose an uncharged fine mesh grid which is so thin as to let almost all the electrons pass through is placed 2/3 of the distance from the negative plate to the positive plate. What is the kinetic energy of the electron as it passes through the grid?
Solution:

This is really a "level I" problem (except, perhaps for part c).

  1. We are given the voltage on a capacitor (5000 V) and asked for the charge; we thus use Q = CV. Here

       C = eoA/d

    There is no `kappa' (dielectric constant) in the above since there is no material between the plates, so kappa = 1. The area = 10 cm2 = 10 x10-4 m2 (please be careful in converting units!) and d = 0.5 cm = 5 x 10-3 m so

       C = (8.85 x 10-12 C2/Nm2)(10-3 m2)/(5 x 10-3 m) = 1.77 x 10-12 F

    thus

       Q = CV = (1.77 x 10-12 F)(5 x 103 V) = 8.85 x 10-9 C = 8.85 nC

  2. The energy gained when a charge moves through a potential difference (i.e. the work done on the charge) is just U = q V. Since we are asked for the energy in electron-volts (eV) this is just U = (one electron)(5000 V) = 5000 eV.

    Many people confused this with the total energy stored in the capacitor (1/2 CV2). This stored energy is a different beast entirely - it is the energy one would get out if all the charges moved from one plate to the other, which is much larger than the energy one poor solitary electron gets when it crosses the potential.

  3. This last one requires a bit of thought. Since the electric field is uniform in a parallel-plate capacitor, the potential difference is just proportional to the distance from one plate to the next (E = V/d). Thus at the grid the electron has gone 2/3 of the way, hence the potential difference is 2/3 of the total potential, thus the electron would have

       2/3 (5000 eV) = 3333 eV

    of kinetic energy.

    Test 2
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    last updated: March 25 1998