Question

A(3.000×10²)μF capacitor is fully-charged when it has (9.6×10⁻³)C on its plates. What is the potential difference across the plates of the capacitor?

Answer 1

The potential difference across the plates of a (3.000×10²)μF capacitor with a (9.6×10⁻³)C charge is calculated using the formula V = Q/C and is found to be 32 volts (V).

Step-by-step explanation:

To find the potential difference across the plates of a capacitor, you can use the relationship V = Q/C, where V is the potential difference, Q is the charge, and C is the capacitance. Given a capacitor with a charge of (9.6×10⁻³)C and a capacitance of (3.000×10²)μF, we can calculate the potential difference as follows:

V = Q / C
= (9.6×10⁻³ C) / (3.000×10² μF)

Before we can divide, we need to convert the capacitance from microfarads to farads:

1 μF = 10⁻¶ F
So, (3.000×10²) μF = (3.000×10⁻⁴) F

Now, let's substitute and calculate the potential difference:

V = (9.6×10⁻³ C) / (3.000×10⁻⁴ F)
= 32 V

The potential difference across the plates of the capacitor is 32 volts (V).