Chapter 6, Problem 6.1P

A 50-µC/kg (approximately 200 mR) pocket dosimeter with air-equivalent walls has a sensitive volume with the dimensions in. (diameter) and 2.5 in. (length); the volume is filled with air at atmospheric pressure. The capacitance of the dosimeter is 10 pF. If 200 V are required to charge the chamber, what is the voltage across the chamber when it reads $\text{50-µC/kg }\left(\sim \text{200 mR}\right)$ ?

To determine

The voltage across the chamber when it reads 50 µC/kg

Answer to Problem 6.1P

The voltage across the chamber when it reads 50 µC/kg is 150.755 V

Explanation of Solution

Given info:

  $\frac{\Delta Q}{m}=50\frac{\mu C}{kg}$

Diameter, D = 0.5 in

Length, L = 2.5 in

Capacitance, C = 10 pF

Initial Voltage, $v_i=200V$

Formula used:

To calculate volume of the chamber, use the formula as

  $V=\pi r^{2}L$ ...... (1)

Calculation:

First convert the unit from ‘in’ to ‘cm’.

We know, 1 in = 2.54 cm

So, L = 2.5 in = 2.5 × 2.54 cm = 6.35 cm

Also, D = 0.5 in, $r=\frac{0.5}{2}in=0.25in=0.25\times 2.54cm=0.635cm$

Substitute the values in equation (1), we get

  $V=\pi {\left(0.635cm\right)}^2\left(6.35cm\right)=8.04c{m}^3$

Also given that, $\frac{\Delta Q}{m}=50\frac{\mu C}{kg}$

Or, $\Delta Q=50\times {10}^{-6}\frac{C}{kg}\times m$ ……......(2)

To calculate mass (m), use the relation

  $Density(\rho )=\frac{mass(m)}{Volume(V)}$

Or, $m=\rho \times V$

Since, volume is filled with air, ${\rho }_{air}=1.225\times {10}^{-3}\frac{g}{c{m}^3}=1.225\times {10}^{-6}\frac{kg}{c{m}^3}$

Substitute the values in equation (2), we get

  $\begin{array}{l}\Delta Q=50\times {10}^{-6}\frac{C}{kg}\times {\rho }_{air}\times V\\ \Delta Q=50\times {10}^{-6}\frac{C}{kg}\times 1.225\times {10}^{-6}\frac{kg}{c{m}^3}\times 8.04c{m}^3\\ \Delta Q=4.9245\times {10}^{-10}C\end{array}$

Now, to calculate final voltage use the relation

  $\begin{array}{l}\Delta Q=C\times \Delta v\\ \Delta Q=C\times \left(v_i-v_f\right)\\ \frac{\Delta Q}{C}=v_i-v_f\\ v_f=v_i-\frac{\Delta Q}{C}\end{array}$

Substitute the values, we get

  $\begin{array}{l}{v}_f=200V-\frac{4.9245\times {10}^{-10}C}{10\times {10}^{-12}pF}\\ v_f=150.755V\end{array}$

Conclusion:

Thus, the voltage across the chamber when it reads 50µC/kg is 150.755V