A dielectric is inserted into a capacitor while the charge on it is kept constan

Question

A dielectric is inserted into a capacitor while the charge on it is kept constant. What happens to the potential difference and the stored energy? a.) the potential difference decreases and the stored energy increases. b.) both the potential difference and the stored energy increase. c.) the potential difference increases and the stored energy decreases d.) both potential difference and the stored energy decrease e.) both the potential difference and the stored energy remain the same the answer is D. I understand why the stored energy decreases because U of dielectric < U but wouldn't that mean that the potential difference increases? I don't understand why the potential difference decreases...if someone can explain, I'd greatly appreciate it!

Explanation / Answer

This question can be interpreted in 2 different ways according to which characteristics of the system are considered to be constant. Case 1 - The capacitor has fixed charge (Q) - i.e. it has been charged to voltage V, and then disconnected from the source of charging current. We need an expression for energy (E) in terms of relative permittivity (e). Starting with a well-known equation for stored energy - E = 0.5 C V^2, and using an equation for V, C and Q V = Q/C, we get E = 0.5 Q^2/C The capacitance C is proportional to relative permittivity of the dielectric, so we can write E = 0.5 Q^2/Ca*e where Ca is the capacitance with air as a dielectric. It is clear that inserting a dielectric of permittivity e will reduced the stored energy in inverse proportion to e. Where has the energy gone ? The answer to this is that as the dielectric material is inserted, the capacitor will exert an electrostatic force on it, and actually pull it in between the plates. So the capacitor works like a kind of electrostatic motor and does work in the external environment. Some of the stored electrical energy is converted into mechanical work. Case 2. The voltage is maintained constant while the dielectric is inserted. This can be achieved by leaving the capacitor connected to a constant voltage source while the dielectric is inserted. Using E = 0.5 C V^2 as before, it is clear that as C increases in proportion to e, inserting the dielectric will increase the stored energy. As before, the dielectric will be attracted by the capacitor, so the energy drawn from the supply will be larger than that needed to increase the stored energy, and the difference will be converted to mechanical work. In both cases, if the dielectric is withdrawn again, mechanical work will be done, and energy transferred back into the electrical system. In the first case the voltage of the capacitor will increase back to its original value. This is a bit like the electrophorus electrical generator which converts mechanical work to electrical energy stored in the form of increased voltage on a capacitor.

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