Chapter 16, Problem 67GP

To determine

To find: The Magnitude and sign of the point charge.

Answer to Problem 67GP

  $9.64\times {10}^{-7}\text{C}$ , negative

Explanation of Solution

Given:

The given figure is shown below.

  

Formula Used:

  

Here, mg is the force of gravity, $\mathrm{T}\cos \mathrm{\theta }$ is the horizontal component of the tension, $\mathrm{T}\sin \mathrm{\theta }$ is the vertical component of tension, and $\text{H}$ is the height.

The positive charge is away from the electric field. Therefore, it must be a positive charge.

In vertical direction,

  $\begin{array}{l}\mathrm{F}\mathrm{y}={\cos }^{}\mathrm{\theta }-\mathrm{m}\mathrm{g}=0\\ \Rightarrow \mathrm{T}=\frac{\mathrm{m}\mathrm{g}}{\cos \mathrm{\theta }}\end{array}$

In Horizontal Direction:

  $\mathrm{E}\mathrm{x}=\mathrm{F}\mathrm{E}-\mathrm{T}\sin \mathrm{\theta }=0$

  $\Rightarrow \mathrm{F}\mathrm{E}-\mathrm{T}\sin \mathrm{\theta }$

  $\Rightarrow \mathrm{F}\mathrm{E}=\frac{\mathrm{m}\mathrm{g}}{\cos \mathrm{\theta }}\cos \mathrm{\theta }$

  $\Rightarrow \mathrm{Q}\mathrm{F}=\tan \mathrm{\theta }$

  $\mathrm{Q}=\mathrm{m}\mathrm{g}\tan \mathrm{\theta }/\mathrm{E}$

Calculation:

Solve for $\mathrm{q}$ by substituting 1 g for m , 12,000 N/C for E , $9.8\text{m}/{\text{s}}^2$ for g and ${38.57}^{\circ }$ for $\mathrm{\theta }$ in the equation

  $\begin{array}{l}\mathrm{q}=\frac{1}{\mathrm{E}}\mathrm{m}\mathrm{g}\tan \mathrm{\theta }\\ \begin{array}{c}\mathrm{q}=\frac{1}{12,000\text{N}/\text{C}}\left(1.0\text{g}\right)\left(9.8\text{m}/{\text{s}}^2\right)\tan {38.57}^{\circ }\\ =\frac{1}{12,000\text{N}/\text{C}}\left(1.0\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1.00\text{g}}\right)\left(9.8\text{m}/{\text{s}}^2\right)\tan {38.57}^{\circ }\\ =9.64\times {10}^{-7}\text{C}\end{array}\end{array}$

Conclusion:

Hence, the charge is equal to $9.64\times {10}^{-7}\text{C}$ and is positive in sign.