Physics · Electrostatic Potential And Capacitance · NEET
F = QE would use the TOTAL field between the plates, E = sigma/epsilon0. But a plate cannot push itself. Each plate only feels the field made by the OTHER plate, which is half the total: E_other = sigma/(2 epsilon0). So F = Q x E_other = Q x sigma/(2 epsilon0) = Q^2/(2 epsilon0 A). The 1/2 comes from removing the plate's own field.
The total field between the plates (E = sigma/epsilon0) is the SUM of two equal fields: half from the positive plate and half from the negative plate. When you find the force on one plate, you must drop that plate's own contribution and keep only the other plate's half. That surviving half is what gives the 1/2 in F = Q^2/(2 epsilon0 A).
Always attractive. One plate holds +Q and the other holds -Q, and opposite charges attract. The plates pull toward each other. This is why energy is released and work is done BY the field if the plates are allowed to move closer.
For an ISOLATED charged capacitor (Q fixed), no. F = Q^2/(2 epsilon0 A) has no d in it, so moving the plates apart or together does not change the force. This is the exact idea NEET 2018 tested. If instead a BATTERY stays connected (V fixed), then F = epsilon0 A V^2/(2 d^2), which DOES depend on d.
Force is the rate at which energy changes with separation: F = -dU/dx. For a fixed-charge capacitor U = Q^2 x / (2 epsilon0 A) (since C = epsilon0 A/x), so dU/dx = Q^2/(2 epsilon0 A). The magnitude of that slope is exactly the attractive force between the plates.
The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
F = Q^2/(2 epsilon0 A) for a fixed charge Q on plate area A. It can also be written F = (1/2) epsilon0 E^2 A, where E is the field in the gap, or F = QE/2.
Because each plate experiences only the field created by the other plate, which is exactly half the total field in the gap. So the force is Q times E/2, giving QE/2, not QE.
Yes. With a dielectric of constant K filling the gap at fixed charge, the field and the force both drop by a factor K: F = Q^2/(2 K epsilon0 A). The dielectric weakens the attraction.
Dividing F by area A gives the electrostatic pressure P = sigma^2/(2 epsilon0) = (1/2) epsilon0 E^2. This equals the energy density stored in the field, which is a useful cross-check.
