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Q19.5-52PE

Expert-verifiedFound in: Page 698

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 \(667{\rm{ }}pF\).

**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.**

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

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

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 \(667{\rm{ }}pF\).

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