Capacitance Calculation: Understanding Charge and Voltage

A capacitor is an electronic component that stores electrical energy in an electric field. The ability of a capacitor to store this electrical charge is measured by its capacitance. This property is fundamental to understanding how capacitors function in various electrical circuits.

Capacitance Definition and Formula

Capacitance (\(C\)) quantifies how much electric charge (\(Q\)) a capacitor can store for a given potential difference or voltage (\(V\)) across its plates. The relationship between these three quantities is defined by the formula:

\(Q = C \times V\)

Where:

• \(Q\) represents the electric charge stored, measured in coulombs (C).
• \(C\) represents the capacitance, measured in farads (F).
• \(V\) represents the voltage or potential difference across the capacitor, measured in volts (V).

To calculate the capacitance when the charge and voltage are known, we can rearrange the formula as follows:

\(C = \frac{Q}{V}\)

Given Values for Capacitor Problem

From the problem statement, we are provided with the following information about the capacitor:

• The amount of charge (\(Q\)) that the capacitor can store = 100 μC (microcoulombs).
• The voltage (\(V\)) across the plates of the capacitor = 10 V (volts).

Step-by-Step Capacitance Calculation

To determine the capacitance value, we will use the formula \(C = \frac{Q}{V}\). It's important to be mindful of the units. Since the options are in microfarads (μF), it's convenient to work with microcoulombs and volts, which will directly yield microfarads.

Let's substitute the given values into the formula:

• Charge, \(Q = 100 \text{ } \mu C\)
• Voltage, \(V = 10 \text{ V}\)

Applying the formula:

\(C = \frac{Q}{V} = \frac{100 \text{ } \mu C}{10 \text{ V}}\)

Performing the division:

\(C = 10 \text{ } \mu F\)

The units work out directly: when charge is in microcoulombs and voltage in volts, capacitance is in microfarads.

Final Capacitance Value

Based on our calculation, the capacitance value for the capacitor that stores 100 μC of charge with 10 V across its plates is 10.0 μF. This result demonstrates a direct application of the fundamental capacitance formula.