Q19.4-47PE - College Physics (Urone)

Question

Find the charge stored when \(5.50{\rm{ }}V\) is applied to an \(8.00{\rm{ }}\mu F\) capacitor.

Step-by-Step Solution

Answer

The amount of charge stored in a \(8.00{\rm{ }}\mu F\) capacitor when \(5.50{\rm{ }}V\) is applied to it, is \(Q = 44.0{\rm{ }}\mu C\).

Step 1: Given Information

• Voltage applied to the Capacitor – \(5.50{\rm{ }}V\)
• Capacitance value for the Capacitor – \(8.00{\rm{ }}\mu F\)

Step 2: Concept Introduction

Capacitors: Any pair of conductors divided by an insulating substance is referred to as a capacitor. The potential \(\Delta V\) of the positively charged conductor with respect to the negatively charged conductor is proportional to \(Q\) when the capacitor is charged, and there are charges on the two conductors of equal magnitude and opposite sign. The ratio of \(Q\) to \(\Delta V\) is what determines the capacitance \(C\).

\(C = \frac{Q}{{\Delta V}}...(1)\)

Step 3: Calculation for Charge Stored

The charge stored in the capacitor is found by solving Equation \((1)\) for \(Q\) –

\(Q = C\Delta V\)

Entering the values for \(C\) and \(\Delta V\), it is obtained –

\(\begin{array}{c}Q = (8{\rm{ }}\mu F)(5.50\;V)\\ = 44.0{\rm{ }}\mu C\end{array}\)

 

Therefore, the charge stored is  \(Q = 44.0{\rm{ }}\mu C\).