A parallel-plate capacitor has a capacitance of 10 mF and is charged with a 20-V power supply.The power supply is then removed and a dielectric material of dielectric constant 4.0 is used to fill the space between the plates.What is the voltage now across the capacitor?
A) 80 V
B) 20 V
C) 10 V
D) 5.0 V
E) 2.5 V

Question

Quizplus · Accepted Answer

The capacitance of the parallel-plate capacitor increases when a dielectric material is inserted between the plates. The capacitance is given by:
C' = κC
where κ is the dielectric constant and C is the original capacitance.
In this case, the new capacitance is:
C' = 4.0 x 10 mF = 40 mF
The charge on the capacitor remains the same after the dielectric is inserted, since charge is conserved. The voltage across the capacitor is inversely proportional to the capacitance, and since the charge remains the same, we have:
V' = V(C/C')
where V is the original voltage and V' is the new voltage.
Substituting the given values, we get:
V' = 20 V (10 mF/40 mF) = 5.0 V
Therefore, the voltage across the capacitor is now 5.0 V. Answer is D).

A Parallel-Plate Capacitor Has a Capacitance of 10 MF and Is