# Heat generated is zero even when current is passed through wire

A parallel-plate capacitor having plate area $$20\,\text{cm}^2$$ and separation between the plates $$1.00\,\text{mm}$$ is connected to a battery of $$12.0\,\text{V}$$. The plates are pulled apart to increase the separation to $$2.0 \, \text{mm}$$. (a) Calculate the charge flown through the circuit during the process. (b) How much energy is absorbed by the battery during the process? Calculate the stored energy in the electric field before and after the process. (d) Using the expression for the force between the plates, find the word done by the person pulling the plates apart. (e) Show and justify that no heat is produced during this transfer of charge as the separation is increased.

1. When a parallel plate capacitor is charged by connecting it with a battery, only half the work done by the battery is stored as the internal energy of the capacitor and the other half is lost as heat due to the flow of charge in the wire.

2. Now, if the separation between the capacitor plates is changed with the battery still connected, charge flows through the connecting wires but the heat generated is zero.

I’m unable to understand how both can be true simultaneously. Kindly help me understand.

My doubt arises due to the non zero answer of the first question (a) combined with the last question (e).

• If the wires have a resistance and a current passes through them, heat is generated. However it might be possible for the energy to come from work done by whatever changes the separation of the plates, rather than from the battery. Commented Feb 3 at 11:52
• “charge flows through the connecting wires but the heat generated is zero.” What makes you think that? Commented Feb 3 at 12:45
• @BobD A parallel-plate capacitor having plate area 20 cm 2 and separation between the plates 1.00 mm is connected to a battery of 12.0 V. The plates are pulled apart to increase the separation to 2.0 mm. (a) Calculate the charge flown through the circuit during the process. (b) How much energy is absorbed by the battery during the process ? (c) Calculate the stored energy in the electric field before and after the process. (d) Using the expression for the force between the plates, find the work done by the person pulling the plates apart.
– Tom
Commented Feb 3 at 16:31
• @BobD e) Show and justify that no heat is produced during this transfer of charge as the separation is increased. This is a question from my physics text book. The non zero answer of the first question (a) combined with question (e) is the cause of my question
– Tom
Commented Feb 3 at 16:32
• Hello, welcome to Physics Stack Exchange. Please always type your question. I have two suggestions. First, do not use pictures of text. Text can be edited and searched, whereas images cannot. Second, any time you ask a circuit question, include a circuit diagram. The resolution to your doubt lies in whether or not you include a resistor in the circuit. If you do not include a resistor when charging the capacitor from a battery, then the situation is not so simple. I would be happy to explain this but only after the question is clarified. Commented Feb 4 at 0:31

You just need to write the governing equations of the system, for both the electromagnetic and the mechanical part of the system.

## Flat capacitor

Before moving on, let's recall some relations about a capacitor:

• relation between total charge on a plate $$Q$$ and the incoming electric current $$i$$ $$\frac{d Q}{dt} = i \ ,$$
• relation between surface charge distribution $$\sigma$$, total charge $$Q = \sigma A$$ and electric field $$e$$ inside the capacitor $$e = \frac{\sigma}{\varepsilon} = \frac{Q}{\varepsilon A}$$
• relation between the voltage difference across the capacitor and the charge $$v_c = x e = x \frac{Q}{\varepsilon A} \qquad , \qquad Q = \varepsilon A \frac{v_c}{x} \qquad , \qquad i = \dfrac{dQ}{dt} = \dfrac{d}{dt}\left( \frac{\varepsilon A}{x}v_c \right) =: \dfrac{d}{dt} \left( C(x) v_c\right)$$
• attraction force between the plates of a flat capacitor, $$F = \frac{1}{2} Q \, e = \frac{Q^2}{2 \varepsilon A}$$

## Electro-mechanical system: governing equations

Now, let's write the governing equations of an electro-mechanical system consisting in a RC circuit with a tension generator with prescribed voltage $$e$$, with a flat-plate capacitor with a plate of mass $$m$$ that is free to move, under the action of an external force $$F^e$$ and the attraction by the other constrained plate of the capacitor:

• electric circuit $$v_c + R \, i = e \tag{1}$$
• mechanical system $$m \ddot{x} = F^e - \frac{Q^2}{2 \varepsilon A} \ \tag{2}$$

## Electro-mechanical system: energy balance

It's possible to derive an energy balance equation, multiplying the first equation by $$i$$, and the latter one by $$\dot{x}$$, and summing,

$$\dot{x} m \ddot{x} + i \, v_c + R \, i^2 = i \,e + \dot{x} \, F^e - \dot{x} \, \frac{Q^2}{2 \varepsilon A} \ ,$$

and get a more explicit equation, after manipulation of some of these terms:

• kinetic energy of the moving plate $$\dot{x} m \ddot{x} = \dfrac{d}{dt} \left( \dfrac{1}{2} m \dot{x}^2 \right)$$
• energy of the capacitor \begin{aligned} i \, v_c & = v_c \, \frac{d}{dt} \left( C(x) v_c \right) = && \text{(integration by parts)} \\ & = \dfrac{d}{dt}\left( \dfrac{1}{2} C(x) v_c^2 \right) + \dfrac{1}{2} \dot{C} v_c^2 \end{aligned} with the last term for flat-plate capacitors equal to $$\dfrac{1}{2} \dot{C} v_c^2 = - \frac{1}{2} \dfrac{\varepsilon A}{x^2} \dot{x} \left( \frac{x \, Q}{\varepsilon A} \right)^2 = - \dot{x} \, \frac{Q^2}{2 \varepsilon A} \ ,$$ i.e. a term that simplifies with the force of the power of the force of attraction between the plates of the capacitor.

Putting everything together, the energy balance reads $$\dfrac{d}{dt} \underbrace{\left( \frac{1}{2} m \dot{x}^2 \right)}_{\text{kin. en.}} + \dfrac{d}{dt} \underbrace{\left( \dfrac{1}{2} C v_c^2\right)}_{\text{EM pot. en. stored in the capacitor}} + \underbrace{R \, i^2}_{\text{dissipation in el. resistance}} = \underbrace{F^e \, \dot{x} + i \, e}_{\text{external (mech+EM) power into the system}} \ .$$

From the last equation, it should be easy to realize that the energy balance equation for the model used in this answer has no energy dissipation if the electric circuit has zero resistance, $$R=0$$, and thus producing no heat, and the resulting energy balance equation can be written as $$\dfrac{d}{dt} E^{tot} = P^{ext} \ ,$$ that is nothing more than the first principle of thermodynamics (or total energy balance) for a closed system, $$\dot{E}^{tot} = P^{ext} + \dot{Q}^{ext}$$, with no heat exchange from the system to the environment.