Chapter 17, Problem 45P

To determine

The capacitance and area of the plate will be determined.

Answer to Problem 45P

The answers are $4.82\times {10}^{-9}F\&0.283m^2$ .

Explanation of Solution

Given data:

Charge on each plate, $Q=0.755\times {10}^{-6}C$

Electric field $E=8.24\times {10}^{4}V/m$

Distance between plates, $d=1.95mm$

Voltage of battery $V=22V$

Formula used:

It is known that: $C=\frac{Q}{Ed}$ and $A=\frac{Cd}{K{\epsilon }_0}$

Where,

  $Q$ is the charge.

  $K$ is the dielectric constant.

  $\text{C}$ is the capacitance of capacitor.

  $A$ is the area.

  $d$ is the distance between plates.

Calculation:

The capacitance will be calculated by the formula $C=\frac{Q}{Ed}$ ,

  $\therefore C=\frac{0.775\times {10}^{-6}C}{\left(8.24\times {10}^{4}V/m\right)\left(1.95\times {10}^{-3}m\right)}$

  $\therefore C=4.82\times {10}^{-9}F$

Now, the plate area will be calculated, $A=\frac{Cd}{K{\epsilon }_0}$

  $\therefore A=\frac{\left(4.82\times {10}^{-9}F\right)\left(1.95\times {10}^{-3}m\right)}{\left(3.75\right)\left(8.85\times {10}^{-12}{C}^2/N\cdot m^2\right)}$

  $\therefore A=0.283m^2$

Conclusion: The capacitance and area of the plate are $4.82\times {10}^{-9}F$ and $0.283m^2$ , respectively.