Chapter 16, Problem 71GP

Solution

To find: The value of x for which electric felt at point P is zero.

  $(\sqrt{2}+1)d\approx 2.41d$

Given:

Point charge a and a/2 are separated by a distance

  

Formula Used:

Take right directed fields as positive and left directed fields as negative.

The net electric field at the point is given as follows:

  $\begin{array}{l}{E}_{\text{nct}}=E_1-E_2\\ =\frac{k{Q}_1}{{(d+x)}^2}-\frac{k{Q}_2}{x^2}\end{array}$

Equate it to zero to find the value of $x$ where the resultant field is zero.

  $\frac{k{Q}_1}{{(d+x)}^2}-\frac{k{Q}_2}{x^2}=0$

Substitute $Q$ for $Q_1$ and $\frac{\text{Q}}{\text{2}}\text{for}{\text{Q}}_{\text{2}}$

  $\begin{array}{l}\frac{kQ}{{(d+x)}^2}-\frac{k\left(\frac{Q}{2}\right)}{x^2}=0\\ \frac{kQ}{{(d+x)}^2}=\frac{k\left(\frac{Q}{2}\right)}{x^2}\\ \frac{1}{{(d+x)}^2}=\frac{1}{2x^2}\\ 2x^2=x^2+2dx+d^2\\ x^2-2dx-d^2=0\end{array}$

Calculation:

Solve the quadratic equation $x^2-2dx-d^2=0$by comparing it with the general quadratic equation $a{x}^2+bx+c=0$whose roots are $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Then,

  $a=1,b=-2d,\text{ and }c=-d^2$

Thus,

  $\begin{array}{l}x=\frac{-(-2d)\pm \sqrt{{(-2d)}^2-(4)(1)\left(-d^2\right)}}{2(1)}\\ =\frac{2d\pm 2d\sqrt{2}}{2}\\ =d(\sqrt{2}\pm 1)\end{array}$

The value $(\sqrt{2}-1)d$ of $x$is not considered because it is less than d.

Therefore, the possible distance of $x,$ where the net electric field is zero, is $(\sqrt{2}+1)d\approx 2.41d$