[tex]\Delta [/tex]Uand W, increasing the separation of capacitor, with V=const.?

In summary: The work exerted on the system by moving the plates is also equal to 0.5[1/x - 1/(x+\Deltax)], and this work is exerted by the electrostatic pressure. In summary, the work done by the battery and the work exerted by the electrostatic pressure are both equal to the energy stored in the system as potential energy.
  • #1
Monnn
2
0
I expected it to be quite simple as it was without the battery, in which case [tex]\Delta[/tex]U = W, but now I am confused.

with the separation increased from x to x+[tex]\Delta[/tex]x,

It is obvious that [tex]\Delta[/tex]U changes with 0.5[1/x - 1/(x+[tex]\Delta[/tex]x)] since C changes with 1/r. (Without the common coefficient, Q^2 over A epsilon0 or something)

And the battery which maintains the voltage difference between the plates gains energy from the system with V[tex]\Delta[/tex]Q = [1/x - 1/(x+[tex]\Delta[/tex]x)] (Note that it is without 0.5 which sticks with the energy difference)

And the work exerted on the system by moving the plates is (pressure)*(area)*(distance) an integrand over 1/r^2, which gives 0.5[1/x - 1/(x+[tex]\Delta[/tex]x)] as well (the factor 0.5 originates from that of the electrostatic pressure).

So, 0.5 is exerted on the system and it seems that it is stored in the system, as [tex]\Delta[/tex]U. Then how should I explain the work with the battery?
 
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  • #2
Is it the same 0.5[1/x - 1/(x+\Deltax)] or is it something else?The work done by the battery is equal to the energy stored in the system, which is \DeltaU = 0.5[1/x - 1/(x+\Deltax)]. The battery provides energy to the system to maintain the voltage difference between the plates, and this energy is stored in the system as potential energy.
 

Related to [tex]\Delta [/tex]Uand W, increasing the separation of capacitor, with V=const.?

1. What is the relationship between [tex]\Delta[/tex]U and W in a capacitor with constant voltage?

The change in potential energy, [tex]\Delta[/tex]U, and the work, W, done on a capacitor with constant voltage are directly proportional. This means that as the separation of the capacitor plates increases, both [tex]\Delta[/tex]U and W will also increase by the same factor.

2. How does increasing the separation of a capacitor affect its stored energy?

Increasing the separation of a capacitor results in an increase in its stored energy. This is because the potential energy of a capacitor is directly proportional to the square of its electric field, which in turn is directly proportional to the inverse of the separation between the plates.

3. Can [tex]\Delta[/tex]U and W ever be negative in a capacitor with constant voltage?

No, [tex]\Delta[/tex]U and W cannot be negative in a capacitor with constant voltage. This is because the potential energy and work done on a capacitor are always positive values, as they represent the amount of energy stored and expended in the system.

4. What happens to the electric field and charge in a capacitor when the separation between the plates is increased?

When the separation between the plates of a capacitor is increased, the electric field between the plates decreases. This also results in a decrease in the amount of charge stored on the plates, as the electric field is directly proportional to the charge on the plates.

5. How does the separation of a capacitor affect its capacitance?

The capacitance of a capacitor is inversely proportional to the separation between its plates. This means that as the separation between the plates increases, the capacitance decreases. This relationship is described by the equation C = [tex]\frac{\epsilon_0 A}{d}[/tex], where [tex]\epsilon_0[/tex] is the permittivity of free space, A is the area of the plates, and d is the separation between them.

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