Chapter 2, Problem 2.1P

Three point charges of equal magnitude q, that will yield a zero net electric field at the origin.

To determine

The coordinates of fourth point charge.

Answer to Problem 2.1P

The fourth point charge is at $\left(1.19,1.19\right)$.

Explanation of Solution

Given Information:

The magnitude of all four point charges is q. Three charges are at $x=-2$ , $y=+2$ and $y=-\sqrt{2}$ . The total electric field at origin is zero.

Calculation:

The electric field at origin due to first point charge,

   $\begin{array}{l}{\text{E<mo>→</mo>}}_1=\frac{q}{4\pi {\epsilon }_0{\left(2\right)}^2}{\text{a<mo>^</mo>}}_x\\ =\frac{q}{16\pi {\epsilon }_0}{\text{a<mo>^</mo>}}_x\end{array}$

The electric field at origin due to second point charge,

   $\begin{array}{l}{\text{E<mo>→</mo>}}_2=\frac{q}{4\pi {\epsilon }_0{\left(2\right)}^2}\left(-{\text{a <mo>^</mo>}}_y\right)\\ =-\frac{q}{16\pi {\epsilon }_0}{\text{a<mo>^</mo>}}_y\end{array}$

The electric field at origin due to third point charge,

   $\begin{array}{l}{\text{E<mo>→</mo>}}_3=\frac{q}{4\pi {\epsilon }_0\left(2\right)}{\text{a<mo>^</mo>}}_y\\ =\frac{q}{8\pi {\epsilon }_0}{\text{a<mo>^</mo>}}_y\end{array}$

Let ${\text{E<mo>→</mo>}}_4$ be the electric field due to fourth point charge. Then,

   $\begin{array}{l}{\text{E<mo>→</mo>}}_1+{\text{E<mo>→</mo>}}_2+{\text{E<mo>→</mo>}}_3+{\text{E<mo>→</mo>}}_4=0\\ \Rightarrow \left(\frac{q}{16\pi {\epsilon }_0}{\text{a <mo>^</mo>}}_x\right)+\left(-\frac{q}{16\pi {\epsilon }_0}{\text{a <mo>^</mo>}}_y\right)+\left(\frac{q}{8\pi {\epsilon }_0}{\text{a <mo>^</mo>}}_y\right)+{\text{E<mo>→</mo>}}_4=0\\ \Rightarrow \frac{q}{16\pi {\epsilon }_0}\left({\text{a <mo>^</mo>}}_x+{\text{a <mo>^</mo>}}_y\right)+{\text{E<mo>→</mo>}}_4=0\\ \Rightarrow {\text{E<mo>→</mo>}}_4=-\frac{q}{16\pi {\epsilon }_0}\left({\text{a <mo>^</mo>}}_x+{\text{a <mo>^</mo>}}_y\right)\\ =\frac{q}{4\pi {\epsilon }_0\left(4\right)}\left(-{\text{a <mo>^</mo>}}_x-{\text{a <mo>^</mo>}}_y\right)\\ =\frac{q}{4\pi {\epsilon }_0\left(4/\sqrt{2}\right)}\left(-\frac{1}{\sqrt{2}}{\text{a <mo>^</mo>}}_x-\frac{1}{\sqrt{2}}{\text{a <mo>^</mo>}}_y\right)\\ \Rightarrow r^2=4/\sqrt{2}=2\sqrt{2}\\ \Rightarrow r=1.68\end{array}$

From this calculation, the x coordinate and y coordinate are equal to positive sign. The distance of this point from origin is 1.68. So, the position of fourth point charge is,

   $\begin{array}{l}=\left(\frac{1.68}{\sqrt{2}},\frac{1.68}{\sqrt{2}}\right)\\ =\left(1.19,1.19\right)\end{array}$

Conclusion:

The fourth point charge is at $\left(1.19,1.19\right)$.