Two parallel plate air capacitors, each of capacitance X Farad are in series with a battery of 12 V. If a dielectric with κ = 3 is inserted between the plates of one of the capacitors, then calculate the change in electrical charge in Coulombs) that occurs on one of its plates.

Question

Quizplus · Accepted Answer

When a dielectric with κ is inserted between the plates of a capacitor, the capacitance of the capacitor increases by a factor of κ. In this case, the capacitance of one of the capacitors becomes 3X.

Since the two capacitors are in series, the equivalent capacitance is given by:

1/C = 1/X + 1/3X

Simplifying, we get:

C = 3X/4

The charge on the plates of the capacitors must be equal, since they are in series and connected to the same battery. Let Q be the charge on each plate of the capacitors.

The voltage across each capacitor is the same and is equal to the voltage of the battery, which is 12 V. Using the capacitance of 3X for one of the capacitors, we can find the charge on one of its plates:

Q = C * V = (3X/4) * 12 V = 9X Coulombs

Therefore, the change in electrical charge on one of the plates is 9X Coulombs.

The correct answer is (C) 3X.

Two Parallel Plate Air Capacitors, Each of Capacitance X Farad