Q19.4-49PE - College Physics (Urone)

Question

Calculate the voltage applied to a  \({\bf{2}}{\bf{.00}}\;{\bf{\mu F}}\) capacitor when it holds \({\bf{3}}{\bf{.10}}\;{\bf{\mu C}}\) of charge.

Step-by-Step Solution

Answer

The voltage applied between the plates of the capacitor is \(1.55\;V\).

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 capacitor's capacitance is: \(C = 2.00{\rm{ }}\mu F\)

The capacitor holds the following charge: \(Q = 3.10{\rm{ }}\mu C\)

Step 3: Value of the capacitor

Equation is used to calculate the potential difference across the capacitor.

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

Substitute the values of \(Q\) and \(C\):

\(\begin{array}{c}V = \frac{{3.10{\rm{ }}\mu C}}{{2.00{\rm{ }}\mu F}}\\ = 1.55\;V\end{array}\)

Therefore, the voltage applied is \(1.55\;V\).