Q48PE - College Physics (Urone)

Question

What charge is stored in the capacitor in Example 19.8?

Step-by-Step Solution

Answer

The amount of charge stored in a capacitor when $3.00\times {10}^3 \mathrm{V}$ is applied to it, is $\mathrm{Q}=26.6\mathrm{\mu C}$ .

Step 1: Given data

Area of plates (A) = 1.00 m²

Separation between plates (d) = 1.00 mm 

Voltage between the plates (V) = 3.00 x 10³ V 

Step 2: Concept Introduction

The ability of any capacitor to store energy in the form of electric field is known as Capacitance. The capacitance  C is given as –

 $\mathbf{C}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{\Delta V}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here Q is the charge accumulated at the plates and $\mathbf{∆}\mathbf{V}$ is the potential difference between the plates.

Step 3: Calculation for Charge Stored

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

 $\mathbf{Q}\mathbf{=}\mathbf{C}\mathbf{\times }\mathbf{\Delta V}$

Entering the values for C and $∆\mathrm{V}$ , it is obtained –

  $\begin{array}{rcl}\mathbf{Q}& \mathbf{=}& \left(8.85\times {10}^{-9} F\right)\left(3.00\times {10}^3 V\right)\\ & \mathbf{=}& \mathbf{26}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}\mathbf{ }}\mathbf{C}\\ & \mathbf{=}& \left(26.6\times {10}^{-6} C\right)\left(\frac{{10}^6 \mathrm{\mu C}}{1 C}\right)\\ & \mathbf{=}& \mathbf{26}\mathbf{.}\mathbf{6}\mathbf{ }\mathbf{\mu C}\end{array}$

Therefore, the charge stored is $\mathbf{Q}\mathbf{=}\mathbf{26}\mathbf{.}\mathbf{6}\mathbf{ }\mathbf{\mu C}$ .