Chapter 16, Problem 68GP

To determine

To Find: Value of electric field due to sheet of charge along with direction.

Answer to Problem 68GP

  $1.06\times {10}^7\text{N/C}$

The direction of electric field is downward.

Explanation of Solution

Given:

The given figure is shown below.

  

Formula used:

Using free body diagram of forces acting on point charge

  ${\displaystyle \sum _{}^{}\mathrm{F}\text{}\text{}=\mathrm{T}-\mathrm{m}\mathrm{g}-\mathrm{F}\mathrm{E}=0}$

  $\Rightarrow {\mathrm{F}}_{\mathrm{E}}=\mathrm{T}-\mathrm{m}\mathrm{g}$

  $\Rightarrow \mathrm{E}\mathrm{a}=\mathrm{T}-\mathrm{m}\mathrm{g}$

  $\Rightarrow \mathrm{E}=\frac{\mathrm{T}-\mathrm{m}\mathrm{g}}{\mathrm{a}}$

Calculation:

  

Tension in the string $\mathrm{T}=5.67\text{N}$

The weight of the mass is

  $\begin{array}{cc}\mathrm{W}& =\mathrm{m}\mathrm{g}\\ & =0.210\text{kg}\left(9.8\text{m}/{\text{s}}^2\right)\\ & =2.058\text{N}\end{array}$

But the tension in the cord is $5.67\text{N}$ , which is more than $2.05\text{N}$ .

There must be a downward electric force on the positive charge, which means that the electric field must be directed downwards.

  

From the free body diagram the magnitude of electric field is

  $\begin{array}{cc}\sum \mathrm{F}& ={\mathrm{F}}_{\mathrm{r}}-\mathrm{m}\mathrm{g}-{\mathrm{F}}_{\mathrm{E}}=0\\ {\mathrm{F}}_{\mathrm{E}}& ={\mathrm{q}}_{\mathrm{E}}={\mathrm{F}}_{\mathrm{r}}-\mathrm{m}\mathrm{g}\\ \mathrm{E}& =\frac{{\mathrm{F}}_{\mathrm{T}}-\mathrm{m}\mathrm{g}}{\mathrm{q}}\\ & =\frac{(5.67\text{N})-(2.058\text{N})}{0.34\times {10}^{-6}\text{C}}\\ \mathrm{E}=& 1.06\times {10}^7\text{N}/\text{C}\end{array}$

The direction of electric field is downward.