Q54PE - College Physics (Urone)

Question

(a) What is the capacitance of a parallel plate capacitor having plates of area 1.50 m²  that are separated by 0.0200 mm of neoprene rubber? (b) What charge does it hold when 9.00 V is applied to it?

Step-by-Step Solution

Answer

The required solution is:

 

 $$\begin{gathered} \left(\mathrm{a}\right)\mathrm{C}=4.4\mathrm{mF} \\ \mathrm{(}\mathrm{b})\mathrm{Q}{=4.0\times 10}^{\mathrm{ -\; 5}} \mathrm{C} \end{gathered}$$

Step 1: Given Data

The plates of the capacitor have the following area:  A = 1.50 m²

The capacitor plates are separated by the following distance:

  $\begin{array}{rcl}\mathbf{d}& \mathbf{=}& \mathbf{(}\mathbf{0}\mathbf{.}\mathbf{0200}\mathbf{ }\mathbf{mm}\mathbf{)}\left(\frac{1 m}{1000 \mathrm{mm}}\right)\\ & \mathbf{=}& \mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{ }\mathbf{m}\\ & & \end{array}$

Neoprene rubber has a dielectric constant of:  k = 6.7

Across the capacitor, the potential difference is:  $\mathrm{\Delta V}=9.00 \mathrm{V}$

Step 2: Defining capacitor

A device that is capable of storing energy in the form of electric field is called as capacitor and its ability to store energy is called capacitance.

Step 3: Work of 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 $\mathrm{\Delta V}$ with respect to the negatively charged conductor is proportional to Q The ratio of Q to $\mathrm{\Delta V}$  determines the capacitance C.

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

Step 4: Work of parallel plate Capacitor

The parallel plate capacitor's capacitance is  $\mathbf{C}\mathbf{=}\mathbf{k}\frac{{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{A}}{\mathbf{d}}$

where A is the area of each plate,  d is the distance between them, and ${\mathrm{\epsilon }}_{\text{0}}$ is the vacuum permittivity-

 $\begin{array}{rcl}{\mathbf{\epsilon }}_{\mathbf{0}}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi k}}\\ & \mathbf{=}& \mathbf{8.854}\mathbf{\times }{\mathbf{10}}^{\mathbf{ -\; 12}}\mathbf{ }{\mathbf{C}}^{\mathbf{2}}\mathbf{/}\left(N·m^2\right)\\ & & \end{array}$

If the plates are separated by vacuum, k = 1; otherwise, the dielectric constant of the dielectric is  k>1. 

Step 5: Value of the parallel plate capacitor’s capacitance

a)

Equation is used to calculate the parallel plate capacitor's capacitance.

 $\mathbf{C}\mathbf{=}\mathbf{k}\frac{{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{A}}{\mathbf{d}}$

Substitute the values of k, ${\mathrm{\epsilon }}_{\mathrm{o}}$, A and d 

 $\begin{array}{rcl}\mathbf{C}& \mathbf{=}& \mathbf{(}\mathbf{6}\mathbf{.}\mathbf{7}\mathbf{)}\left[8.854\times {10}^{-12} C^2/\left(N·m^2\right)\right]\frac{\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{ }{\mathbf{m}}^{\mathbf{2}}}{\mathbf{0}\mathbf{.}\mathbf{0200}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathbf{m}}\\ & \mathbf{=}& \mathbf{4}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{ }\mathbf{F}\\ & \mathbf{=}& \left(4.4\times {10}^{-6} F\right)\left(\frac{{10}^6 \mathrm{\mu F}}{1 F}\right)\\ & \mathbf{=}& \mathbf{4}\mathbf{.}\mathbf{4}\mathbf{ }\mathbf{\mu F}\end{array}$

Therefore, the parallel plate capacitor's capacitance is   $4.4 \mu F$

Step 6: Value of Capacitor

b)

The capacitor holds the following charge:  $\mathrm{Q}=\mathrm{C}\times \mathrm{\Delta V}$

Substitute the values of C and $\mathrm{\Delta V}$:

 $\begin{array}{rcl}\mathbf{Q}& \mathbf{=}& \left(4.4\times {10}^{-6} F\right)\mathbf{(}\mathbf{9}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{V}\mathbf{)}\\ & \mathbf{=}& \mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{ }\mathbf{C}\end{array}$

Therefore, the capacitor's capacitance is \({\rm{4}}{\rm{.0 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{C}}{\rm{.}}\) .