When a charge (say positive) is placed between an upper positively charged plate and a negatively charged plate, it should experience a repulsive force from the top plate and an attractive force to the bottom plate, should it not? Then if I were to calculate the net electrostatic force on that charge, wouldn’t I have to double the magnitude of the force between the charge and one of the plates, since it is experiencing two forces? Why is this not the case?

  • 3
    $\begingroup$ "Why is this not the case?" But it is the case. The $E$-field between a pair of oppositely charged plates is twice that of a single charged plate. See here. $\endgroup$ – lemon Jul 8 '15 at 16:50
  • $\begingroup$ The field (in V/m) across the gap will be the potential difference divided by the gap. It one plate is at 4MV and the other is at ground, a + charge released at the top plate will acquire 4MeV of energy crossing the gap. If the top plate is at 2MV and the bottom plate is at -2MV (relative to a common ground), the + charge will still only acquire 4MeV of energy crossing the gap. $\endgroup$ – Jon Custer Jul 8 '15 at 18:05

Let me ask you: how are you characterizing the force between the two plates? Typically, the two plates are connected to a voltage source, and the force is found by multiplying the charge by the voltage, and dividing by the plate distance. The voltage already takes into account the charge on both plates, so there's no need to worry. If you're doing a more complicated problem where you look at the charge on each plate explicitly (maybe the charge on each plate is different, or something), then yes, you need to the charge on both plates into account.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.