A 5.0\text{ }\mu\text{F} parallel-plate capacitor is connected across a potential differen… — JAMB UTME

A $5.0\text{ }\mu\text{F}$ parallel-plate capacitor is connected across a potential difference of $20.0\text{ V}$ . What is the magnitude of the electric charge stored on either plate of the capacitor, in microcoulombs ( $\mu\text{C}$ )?

Answer: 100 \mu C

Answer

The magnitude of the electric charge stored on either plate of the capacitor is

100.0\text{ }\mu\text{C}

Using the capacitor charge equation

Q = C \times V

, substituting

C = 5.0\text{ }\mu\text{F}

and

V = 20.0\text{ V}

yields

Q = 5.0 \times 20.0 = 100.0\text{ }\mu\text{C}

Step-by-Step Solution

Identify the given physical quantities and the formula for electric charge on a capacitor.

Capacitance

C = 5.0\text{ }\mu\text{F}

, voltage

V = 20.0\text{ V}

. Formula:

Q = C V

The charge stored by a capacitor is directly proportional to the potential difference across its terminals.

Perform the multiplication to determine the charge magnitude in microcoulombs.

Q = 5.0 \times 20.0 = 100.0\text{ }\mu\text{C}

Multiplying capacitance in microfarads by potential difference in volts yields charge directly in microcoulombs.

Key Concept

Fundamental relationship between capacitance, charge, and potential difference (

Q = C V

)

Estimated Time:45s

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