P19R2015 – Three Point Charge Distribution
Three Point Charge Distribution
Three point charges $q_1$, $q_2$ and $q_3$ are held fixed as shown below. A positive test charge $q$ placed at point $P$ experiences a zero net electric force.
1. A negative test charge $-q$ is placed at point $P$ and released. What happens?
View answerHide answerWe first draw the forces acting on the positive test charge $q$. The positive charges $q_1$ and $q_2$ will repel $q$ while the negative charge $q_3$ will attract $q$ as shown below. These three forces cancel out overall: the $x$-component of $F_3$ cancels $F_1$ and the $y$-component of $F_3$ cancels $F_2$.
If a negative test charge $-q$ is placed at point $P$ then all three forces still have the same magnitude (because $-q$ and $q$ carry the same charge in magnitude) but their directions are flipped as shown below. The positive charges $q_1$ and $q_2$ will attract $-q$ while the negative charge $q_3$ will repel $-q$.
These three forces will still cancel out overall: the $x$-component of $F_3$ cancels $F_1$ and the $y$-component of $F_3$ cancels $F_2$.
Therefore, the negative test charge experiences a zero net electric force and remains in equilibrium.
2. Assuming $q_1=+Q$, what are $q_2$ and $q_3$?
View answerHide answerSince both scenarios are equivalent, let us consider the case of the positive test charge $q$ below.
We then apply the equilibrium condition and write
\begin{aligned}
\left\{ \begin{array}{c}
F_{net\ x}=0 \\
F_{net\ y}=0 \end{array}
\right.\ \ \ \ &\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
-F_1+F_3{\mathrm{cos} \left(\theta \right)\ }=0 \\
\\
F_2-F_3{\mathrm{sin} \left(\theta \right)\ }=0 \end{array}
\right. \\
\\
&\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{-\frac{kqq_1}{16d^2}+\frac{4kq\left|q_3\right|}{125d^2}=0} \\
\\
\displaystyle{\frac{kqq_2}{9d^2}-\frac{3kq\left|q_3\right|}{125d^2}=0} \end{array}
\right. \\
\\
&\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{-\frac{q_1}{16}+\frac{4\left|q_3\right|}{125}=0} \\
\\
\displaystyle{\frac{q_2}{9}-\frac{3\left|q_3\right|}{125}=0} \end{array}
\right.
\end{aligned}Substituting $q_1=+Q$ into the system of equations, we solve for $q_2$ and $q_3$ as follows
\begin{aligned}
\left\{ \begin{array}{c}
\displaystyle{-\frac{q_1}{16}+\frac{4\left|q_3\right|}{125}=0} \\
\\
\displaystyle{\frac{q_2}{9}-\frac{3\left|q_3\right|}{125}=0} \end{array}
\right.\ \ \ \ &\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{-\frac{Q}{16}+\frac{4\left|q_3\right|}{125}=0} \\
\\
\displaystyle{\frac{q_2}{9}-\frac{3\left|q_3\right|}{125}=0} \end{array}
\right. \\
\\
&\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{\left|q_3\right|=\frac{125Q}{64}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
\\
\displaystyle{q_2=\frac{27\left|q_3\right|}{125}=\frac{27Q}{64}} \end{array}
\right.
\end{aligned}Finally, recalling that $q_3$ is a negative point charge, we conclude that the $q_2$ and $q_3$ are given by
\boxed{q_2=\frac{27}{64}Q\ \ \ \ \ and\ \ \ \ \ q_3=-\frac{125}{64}Q}