Chapter 16, Problem 62GP

Solution

To Find: The location of the place(s) along the x axis where the electric field is zero due to the two charges.

From the origin, the points where electric field would be zero are:

  $x_1=3.62\text{m}$ and $x_2=1.38\text{m}$

Given:

Two charges $Q_1$ and $Q_2$ are separated by a distance $2\text{m}$ on the x axis

The origin is at $Q_1$

Positive charge, $Q_1=2.5\times {10}^{-5}\text{C}$

Negative charge, $Q_2=-5\times {10}^{-6}\text{C}$

Formula Used:

The net electric field must be zero, so:

  $\begin{array}{l}{E}_{net}=0\\ E_1+E_2=0\\ E_1=-E_2\end{array}$

Let the point where electric field is zero be x distance away from negative charge.

  $\frac{k{Q}_1}{{\left(2+x\right)}^2}=\frac{k{Q}_2}{x^2}$

Calculation:

The field will be a combination of two fields taking only magnitudes into consideration

  $\begin{array}{l}\frac{k{Q}_1}{{\left(2+x\right)}^2}=\frac{k{Q}_2}{x^2}\\ \frac{Q_1}{Q_2}=\frac{{(2+x)}^2}{x^2}\\ \frac{2.5\times {10}^{-5}}{5\times {10}^{-6}}=\frac{{(2+x)}^2}{x^2}\\ 5=\frac{{(2+x)}^2}{x^2}\\ \frac{2+x}{x}=\pm \sqrt{5}\\ 1+\frac{2}{x}=\pm \sqrt{5}\\ \frac{2}{x}=\pm 2.236-1\\ =1.236,-3.236\\ x=\frac{2}{1.236},\frac{2}{-3.236}\\ x=1.6181,-0.618\end{array}$

Conclusion:

From the origin, the points where electric field would be zero are:

  $x_1=3.62\text{m}$ and $x_2=1.38\text{m}$