# Force on a dielectric inside a capacitor and conservation of energy

• Termotanque
In summary, the problem is to determine the height of the oil (h) between two coaxial cylindrical metal tubes maintained at different potentials in a tank of dielectric oil. The solution involves calculating the capacitance of the system and using conservation of energy to find the height of the oil. However, the provided solution is off by a factor of 2, possibly due to energy not being conserved in an ideal capacitor with a voltage source. Another possible approach is using force balance and virtual work to find the force exerted by the electric field.
Termotanque
The problem is as follows: Two long coaxial cylindrical metal tubes (inner radius a, outer radius b) stand vertically in a tank of dielectric oil (susceptibility χe, density ρ). The inner one is maintained a potential V, and the outer one is grounded. To what height (h) does the oil rise in the space between the tubes? (Griffiths Introduction to Electrodynamics, problem 4.28).

First I calculated the capacitance of the system, as a function of the height of the oil, as C(h) = 2πε0eh + l)/ln(b/a), where l is the total height of the capacitor.

Then I say: let the initial potential energy stored in the electric field be U0 = 1/2 C(h=0)V2, and the final energy be Uf = 1/2 C(h)V2.

The difference of energy has to be the potential energy gained by the system as gravitational, namely by the rising of the oil: ΔU = ρhπ(b2-a2)g(h/2).

This gives, after some rearrangement, h = 2V2ε0χe / ln(b/a)ρg(b2-a2).

This is the solution given by Griffiths, except for the factor 2 I'm writing. He solves it by balance of force, saying that the electrostatic force has to be equal and opposite to the gravitational force, while I solve it by conservation of energy. What is wrong with my solution? Is energy not conserved?

I'm aware that the energy is not conserved in the ideal charge of an ideal capacitor with a voltage source, that is, the energy provided by the source is VQ, while the energy stored in the electric field is VQ/2. Does this also have to do with my solution being off by the same factor?

In a word - yes.

I would have done it by force balance also - using virtual work to find the force exerted by the electric field.

## 1. What is the force on a dielectric inside a capacitor?

The force on a dielectric inside a capacitor is calculated using the formula F = εA(V²/d²), where ε is the permittivity of the dielectric, A is the area of the capacitor plates, V is the potential difference between the plates, and d is the distance between the plates.

## 2. How does the force on a dielectric inside a capacitor affect the energy stored in the capacitor?

The force on a dielectric inside a capacitor does not directly affect the energy stored in the capacitor. However, it does affect the capacitance of the capacitor, which in turn affects the energy stored. The greater the force, the higher the capacitance and therefore the greater the energy stored.

## 3. Why is energy conserved in a capacitor with a dielectric?

Energy is conserved in a capacitor with a dielectric because the force applied on the dielectric by the electric field is equal and opposite to the force applied on the capacitor plates. This results in the energy being stored in the electric field of the capacitor, rather than being lost as heat or other forms of energy.

## 4. Can the force on a dielectric inside a capacitor be negative?

Yes, the force on a dielectric inside a capacitor can be negative. This occurs when the dielectric has a lower permittivity than the surrounding medium, causing it to be attracted towards the opposing capacitor plate. In this case, the force is still calculated using the same formula, but the direction is opposite to that of a positive force.

## 5. How does the force on a dielectric inside a capacitor change when the distance between the plates is increased?

The force on a dielectric inside a capacitor decreases as the distance between the plates is increased. This is because the force is inversely proportional to the square of the distance (F ∝ 1/d²). Therefore, as the distance increases, the force decreases. However, this also results in a decrease in capacitance and therefore a decrease in the energy stored in the capacitor.

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