When the voltage U is applied to the two plates, the edge effect is ignored and the distance between the two plates is f What is the interaction force F The answer is ε 0su * 2 / 2D * 2

When the voltage U is applied to the two plates, the edge effect is ignored and the distance between the two plates is f What is the interaction force F The answer is ε 0su * 2 / 2D * 2

The capacitance of capacitor C = ε 0s / 4 π KD
The electric quantity of capacitor q = Cu = ε 0su / 4 π KD
Electric field strength between plates e = u / D
F=QE=ε0su^2/4πkd^2

The two plates a and B of the parallel plate capacitor are respectively connected to a constant power supply with a voltage of 60V. The distance between the two plates is 3cm, and the charge of the capacitor is 6 * 10-8c,
A plate grounding
(1) Capacitance of parallel plate capacitor;
(2) The electric field strength between two plates of parallel plate capacitor;
(3) The potential at C 2cm away from B plate

(1) C = q / u = 6 * 10-8 / 60 = 1 * 10 ^ - 9C (2) e = u / D = 60 / 0.03 = 2000N / M (3) if a is connected to the positive pole of power supply, UAC = φ a - φ C, φ C = φ a-uac = 0-uac = - ed '= - 2000 * 0.01 = - 20V; if B is connected to the positive pole of power supply, UCA = φ C - φ Ca, φ C = φ a + UCA = 0 + UCA = e * d' = 2000 * 0.01 = 20V

After charging, the parallel plate capacitor is disconnected from the power supply, keeping the charge of the two plates unchanged. But why does the voltage between the two plates change?

Q＝UC
C is the capacitor. If it is disconnected from the power supply, Q does not change
If C is constant, u cannot be changed. As you say that the voltage will change, C must have changed. The factors of C change include the distance between the two plates, the area facing each other, and whether there is any other medium between the two plates