The parallel plate capacitor is connected with the power supply after charging, and then the electrode plates are filled with isotropic uniform dielectric with relative permittivity of ε. How many times are the electric quantity on the two plates? How many times are the electric field strength? How many times are the electric field energy?

The parallel plate capacitor is connected with the power supply after charging, and then the electrode plates are filled with isotropic uniform dielectric with relative permittivity of ε. How many times are the electric quantity on the two plates? How many times are the electric field strength? How many times are the electric field energy?

The original capacitance of the capacitor is C. after adding the dielectric with permittivity ε, the capacitance C1 will be increased
C1=εC
So the new electric quantity is ε times of the original one. Because the voltage is constant and the distance between the capacitors is constant, the electric field is strong
E = u / D unchanged
What is the energy of the electric field
C1U^2=εCU^2
It's an epsilon

The parallel plate capacitor after physical charging is placed horizontally, and the distance between the two plates is 5cm
The charged parallel plate capacitor is placed horizontally with a distance of 5cm between the two plates. At a point 2cm away from the lower plate, there is an uncharged ball with a mass of 2mg, which starts to fall from rest. When the ball collides with the lower plate, it can get a negative charge of 2 * 10 ^ - 8C and rebound to a height 4cm away from the lower plate, It is known that the charge of the upper plate is + 1.0 * 10 ^ - 6C. Take g = 10m / s. calculate the electric field strength e between the two plates and the capacitance C of the capacitor! (the charge obtained by the ball does not affect the change of the electric field between the plates)
MG is mg

I will, you wait for me to teach you
Use the acceleration to calculate

The voltage of power supply a is 6V, and the voltage of power supply B is 8V. When the switch is switched from a to B (the capacitance of capacitor is 2 micro method), the amount of charge passing through the galvanometer is 0
I'm too lazy to make a picture of myself

If it's divided, it's the answer
Together with you is 4 * 10 ^ - 6
Which is it·

As shown in the figure: the capacitance of the capacitor is 2 μ F, the voltage of power supply a is 6V, and the voltage of power supply B is 8V, when the switch s turns from a to B
What is the amount of charge passing through the galvanometer?

From C = q / Δ u
q=c*Δu=2x10^-6x2=4x10^-6