Capacitor Tasks: Flat Vacuum Capacitor Changes & Work Needed to Increase Plate Distance

1. The flat vacuum capacitor is charged and disconnected from the voltage source. How will Q, E, U, C, Ep change if we move the capacitor plates apart? Justify your answer. Q-cond charge, E- field strength between covers, U-voltage between okla, C-capacity, Ep- cond energy.
2. The vacuum flat capacitor is connected to the voltage source U, How will Q, E, U, C, Ep change if we move the capacitor plates apart.
3. The flat air condenser consists of two plates with the area of s = 200 cm2 each, located at a distance of l1 = 0.3 cm from each other. What work needs to be done to increase the distance between the covers to l2 = 0.5 cm? Solve the problem for two cases:
a) Charge the capacitor to U0 = 600V and disconnect the batteries
b) The capacitor remains connected to the battery at all times, maintaining a constant 600V potential difference.

This is their content. Help me.

1 - Q will be constant because the cargo has nowhere to escape and has no place to appear. C will be small (according to the formula for capacity). U will increase (U = Q / C - Q constantly, C decreases, so U increases)

As for the energy and field strength, I don't remember the patterns at the moment ;)

2 - U is constant - comes from a voltage source. We are removing the covers - C is decreasing. Q = UC, U constant, C decreasing, Q must also decrease. E and Ep as in the previous quest

3 - Eh, these patterns You have to use the principle of conservation of energy here

thanks and for that

Quote:

1. The flat vacuum capacitor is charged and disconnected from the voltage source. How will Q, E, U, C, Ep change if we move the capacitor plates apart? Justify your answer. Q-cond charge, E- field strength between covers, U-voltage between okla, C-capacity, Ep- cond energy.

Charge on the capacitor

Q = U * C

By charging a capacitor, we provide it with a charge Q, which will remain constant in an isolated system.

Vacuum Flat Capacitor Capacity:

$$C=\frac{\eps_0*S}{d}$$

where

?0 - electric permittivity of the vacuum
S - plate surface
d - plate distance

Increasing the distance of the plates causes:

Q - charge does not change
C - the capacity decreases because the denominator in the formula for C increases
U = Q / C - the voltage on the capacitor increases because C decreases and Q is constant.

E, the field strength = U / d will remain constant

Q = eps0 * S / d * U -> U = Q * d / (eps0 * S)

E = U / d = Q / (eps0 * S)

Energy in the capacitor:

Ep = 0.5 * C * U ^ 2 = (Q ^ 2 * d) / (2 * eps0 * S)

that is, it grows with increasing d.

So the first is:
Q = const (no charge added, no charge decreased)
U = increases (since Q / C and Q const, C decreases)
C = decreasing (d increasing)
E = U / d (i.e. constants because i U and id increases)
Ep = (Q * U) / 2 So Ep increases as Q cons and U increases
And the second is:
U = const
C = decreases because d increases
E = U / d decreases because d increases
Q = U * C so it decreases (Uconst C decreases)
Ep = (Q * U) / 2 so it decreases because Q decreases