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College Physics (Urone)
Found in: Page 698

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Short Answer

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 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 \(2.50 \times {10^{ - 8}}\;F\).


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