# Why is heat generated when dielectric slab is inserted between the plates of a capacitor?

Why is heat generated when we insert a dielectric slab of dielectric constant $k$ between the plates of a parallel plate capacitor?

Like if the capacitor is uncharged if we insert a slab then $\dfrac{1}{2}CV^2$ amount of heat will be generated because battery force does $CV^2$ work but capacitor gets only $\dfrac 12 CV^2$ energy so the rest must go in the form of heat. But why exactly does this heat get produced?

Because of resistance in the wire? Then shouldn't it depend on the resistance of the wires but it is independent of that.

What if I connect the battery to the capacitor through a wire of 0 resistance?

## 3 Answers

Say we have some initial voltage difference, $\Delta V$, between a battery and a capacitor.

This means that first electron arriving from the negative terminal of the battery to the negative terminal of the capacitor and another electron electron simultaneously arriving from the positive plate of the capacitor to the positive terminal of the battery have to lose the energy corresponding to that voltage difference, i.e., $e\Delta V$.

As the voltage on the capacitor increases, the voltage difference decreases and, therefore the losses per unit charge decrease as well, averaging at $e\Delta V/2$.

If, as you are suggesting the wires had no resistance, the energy would be burned on the internal resistance of the battery.

If the battery had no internal resistance, then the electrons would speed up in the wires and have an abrupt stop at their destinations with the excessive kinetic energy converted to heat anyway.

We can potentially avoid these losses, if we insert an inductor in series with the capacitor.

In this case, the initial current will be zero and the current will be ramping up gradually. When the voltage on the capacitor equalizes the voltage on the battery, the current will reach its peak and will start declining, while still charging the capacitor. When the voltage on the capacitor reaches 2x of the battery voltage, the current drops back to zero. If, at that very moment, we break the circuit, we'll end up with the charged capacitor and no losses (except minor losses in the wires).

• I prefer this answer to my own for its explanation of how the factor of 2 arises. Commented Jul 18, 2018 at 1:40

Well there is no other form of this extra energy to escape and so heat is produced due to collisions of electrons inside the conductor

Note that entropy is generated every time energy moves down a gradient; this entropy increase is often (but not always) associated with heating.

In this case, the gradient is the constant applied voltage $V$ from the battery versus the initial zero voltage in the dielectric before it is inserted. Think of the separation of charge carriers within the dielectric as analogous to a damped spring with a load suddenly hung on it; in the process of reaching a new positional equilibrium of the charge carriers (analogous to oscillation of the spring), internal "friction" converts some of the electrical energy into thermal energy, corresponding to the required entropy generation.