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Q19.5-51E

Expert-verified
Found in: Page 698

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# What capacitance is needed to store $$3.00{\bf{ }}\mu C$$ of charge at a voltage of $$120\;V$$?

$$2.50 \times {10^{ - 8}}\;F$$ capacitance is needed to store $$3.00{\rm{ }}\mu C$$ of charge at a voltage of $$120\;V$$.

See the step by step solution

## Step 1: Defining capacitor

A capacitor is a device used to store electrical energy and works in an electric field. It is a passive electrical component with two terminals. Capacitance is the term used to describe the effect of a capacitor.

## Step 2: Work of Capacitor and Information Given

Any pair of conductors separated by an insulating substance is referred to as a capacitor. When the capacitor is charged, the two conductors have charges of equal magnitude $$Q$$ and opposite sign, and the positively charged conductor's potential $$\Delta V$$ with respect to the negatively charged conductor is proportional to $$Q$$ The ratio of $$Q$$ to $$\Delta V$$ determines the capacitance $$C$$

$$C = \frac{Q}{{\Delta V}}$$

The farad is the SI unit of capacitance ($$F$$): $$F = 1C/V$$

The capacitor holds the following charge:

$$\begin{array}{c}Q = (3.00{\rm{ }}\mu C)\left( {\frac{{1{\rm{ }}C}}{{{{10}^6}{\rm{ }}\mu C}}} \right)\\ = 3.00 \times {10^{ - 6}}{\rm{ }}C\end{array}$$

Across the capacitor, the potential difference is $$\Delta V = 120\;V$$.

## Step 3: Value of the capacitor

The capacitor's capacitance is:

$$C = \frac{Q}{{\Delta V}}$$

Substitute the values of $$Q$$ and $$\Delta V$$:

$$\begin{array}{c}C = \frac{{3.00 \times {{10}^{ - 6}}{\rm{ }}C}}{{120\;V}}\\ = 2.50 \times {10^{ - 8}}\;F\end{array}$$

Therefore, the capacitance needed is $$2.50 \times {10^{ - 8}}\;F$$.