Question

A power source supplies an electric potential difference of 9.0 V to the plates of a capacitor. The energy stored in the capacitor is 9.8 × 10–4 J. The dielectric constant of the dielectric in the capacitor is 5.1.

What is the capacitance (C) of the capacitor?
A) 1.92×10−5F
B) 1.92×10−4F
C) 1.92×10−3F
D) 1.92×10−2F

Answer 1

The capacitance (C) of the capacitor is 1.92×10−5 F.

Step-by-step explanation:

To find the capacitance (C) of the capacitor, we can use the equation:

C = Q/V

Where Q is the charge stored in the capacitor, and V is the potential difference between the plates. We are given the energy stored in the capacitor (9.8 × 10–4 J), which can be related to the charge and capacitance through the equation:

U = 1/2 * C * V^2

Solving for C, we have:

C = 2 * U / V^2

Substituting the values given in the problem, we get:

C = 2 * (9.8 × 10–4 J) / (9.0 V)^2

After performing the calculations, we find that the capacitance of the capacitor is 1.92×10−5 F.