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Q19.5-52PE

Expert-verified
Found in: Page 698

### College Physics (Urone)

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

# What is the capacitance of a large Van de Graaff generator's terminal, given that it stores $$8.00{\rm{ }}mC$$of charge at a voltage of $$12.0{\rm{ }}MV$$?

The capacitance of a large Van de Graaff generator's terminal is $$667{\rm{ }}pF$$.

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 Provided

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 Van de Graaff has a charge stored in it.

$$\begin{array}{c}Q = (8.00{\rm{ }}mC)\left( {\frac{{1{\rm{ }}C}}{{{{10}^3}{\rm{ }}\mu C}}} \right)\\ = 8.00 \times {10^{ - 3}}{\rm{ }}C{\rm{ }}\end{array}$$

The Van de Graaff generator's potential difference is

$$\begin{array}{c}V = (12.0\;V)\left( {\frac{{{{10}^6}\;V}}{{1{\rm{ }}mV}}} \right)\\ = 12.0 \times {10^6}\;V\end{array}$$

## Step 3: Value of the capacitor

Equation gives the capacitance of the Van de Graaff generator's terminal.

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

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

$$\begin{array}{c}C = \frac{{8.00 \times {{10}^{ - 3}}{\rm{ }}C}}{{12.0 \times {{10}^6}\;V}}\\ = 6.67 \times {10^{ - 10}}\;F\\ = \left( {667 \times {{10}^{ - 12}}\;F} \right)\left( {\frac{{{{10}^{12}}{\rm{ }}pF}}{{1\;F}}} \right)\\ = 667{\rm{ }}pF\end{array}$$

Therefore, the capacitance needed is $$667{\rm{ }}pF$$.