Chapter 16, Problem 64GP

To determine

To Find: The value of charge applied to electroscope.

Answer to Problem 64GP

  $6.1\times {10}^{-6}\text{C}$

Explanation of Solution

Given:

  $m=24\text{g}=(24\text{g})\left(\frac{{10}^{-3}\text{kg}}{\text{g}}\right)=24\times {10}^{-3}\text{kg}$

  $\begin{array}{l}l=78\text{cm}\\ =(78\text{cm})\left(\frac{{10}^{-2}\text{m}}{\text{cm}}\right)\\ =78\times {10}^{-2}\text{m}\end{array}$

  $\theta =30°$

Formula Used:

Consider the given below,

  

Using from free body diagram, the tension can be resolved into components,

  $\begin{array}{l}T\sin \theta =F_E\\ T\cos \theta =mg\end{array}$

Taking the ration of these equations,

  $\begin{array}{l}\tan \theta =\frac{F_E}{mg}\\ F_E=mg\tan \theta \end{array}$

Electric force is given by Coulomb's law, horizontally

  $F_E=\frac{k{(Q/2)}^2}{r^2}$

From the geometry of the figure, the distance between the two charged spheres

  $r=(2l\sin \theta )$

then

  $F_E=\frac{k{Q}^2}{4{(2l\sin \theta )}^2}$

Substitute $F_E=\frac{k{Q}^2}{4{(2l\sin \theta )}^2}$ in the equation $F_E=mg\tan \theta$

  $\frac{k{Q}^2}{4{(2l\sin \theta )}^2}=mg\tan \theta$

Re-arrange this equation forQ $Q=2(2l\sin \theta )\sqrt{\frac{mg\tan \theta }{k}}$

Calculation:

To determine charge, substitute $0.78\text{m}$ for $l,30°$ for $\theta ,9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k$ , $24\times {10}^{-3}\text{kg}$ for $m,9.8\text{m}/{\text{s}}^2$ for g in the equation $Q=2(2l\sin \theta )\sqrt{\frac{mg\tan \theta }{k}}$

  $\begin{array}{l}Q=2(2l\sin \theta )\sqrt{\frac{mg\tan \theta }{k}}\\ =(4\times 0.78\text{m})\sin 30°\sqrt{\frac{\left(24\times {10}^{-3}\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\tan 30°\right)}{9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2}}\\ =6.064\times {10}^{-6}\text{C}\\ =6.1\times {10}^{-6}\text{C}\end{array}$

Conclusion:

Therefore, charge applied is $6.1\times {10}^{-6}\text{C}$ .