Circuit and electromagnetism basics
A charged parallel-plate capacitor is disconnected from its source, and the gap between the plates (air, relative permittivity taken as 1) is completely filled with a dielectric of relative permittivity εr. How do the voltage between the plates and the stored energy change?
- AThe voltage becomes 1/εr times and the energy 1/εr times
- BThe voltage becomes εr times and the energy εr times
- CThe voltage is unchanged and the energy becomes εr times
Show answer and explanation
Answer: A. The voltage becomes 1/εr times and the energy 1/εr times
Because the capacitor is disconnected, its charge Q stays constant. Filling it with dielectric multiplies the capacitance by εr, so the voltage V = Q/C becomes 1/εr times and the energy W = Q²/(2C) also becomes 1/εr times. The lost energy goes into the work done by the force pulling the dielectric into the gap. If the source stays connected (constant voltage), the charge and energy both become εr times larger.
- A. Correct. With constant charge, V = Q/C and W = Q²/(2C) both become 1/εr times.
- B. The capacitance increases, so with constant charge the voltage falls.
- C. An unchanged voltage with energy εr times larger applies when the source remains connected.
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