Question

21.58 ••• two charges are placed on the x - axis: one, of 2.50 μc, at the origin and the other, of - 3.50 μc, at x = 0.600 m (fig. p21.58). find the position on the x - axis where the net force on a small charge +q would be zero.
figure p21.58

Explanation:

Step1: Apply Coulomb's law

The force on the charge $+q$ due to the first charge $Q_1 = 2.50\times10^{- 6}\text{C}$ at the origin is $F_1=k\frac{Q_1q}{x_1^{2}}$, and due to the second charge $Q_2=- 3.50\times10^{-6}\text{C}$ at $x = 0.600\text{m}$ is $F_2 = k\frac{Q_2q}{(x - 0.600)^{2}}$. When the net - force $F_{net}=F_1 + F_2=0$, we have $k\frac{Q_1q}{x_1^{2}}=-k\frac{Q_2q}{(x - 0.600)^{2}}$. Canceling out $k$ and $q$, we get $\frac{Q_1}{x^{2}}=\frac{-Q_2}{(x - 0.600)^{2}}$.

Step2: Substitute the values of charges

Substitute $Q_1 = 2.50\times10^{-6}\text{C}$ and $Q_2=-3.50\times10^{-6}\text{C}$ into the equation $\frac{Q_1}{x^{2}}=\frac{-Q_2}{(x - 0.600)^{2}}$. We have $\frac{2.50\times10^{-6}}{x^{2}}=\frac{3.50\times10^{-6}}{(x - 0.600)^{2}}$. Cross - multiply to get $2.50(x - 0.600)^{2}=3.50x^{2}$.

Step3: Expand and simplify the equation

Expand $(x - 0.600)^{2}=x^{2}-1.2x + 0.36$. Then $2.50(x^{2}-1.2x + 0.36)=3.50x^{2}$. Which is $2.50x^{2}-3x + 0.9 = 3.50x^{2}$. Rearrange to get $x^{2}+3x - 0.9 = 0$.

Step4: Solve the quadratic equation

The quadratic formula for $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 1$, $b = 3$, and $c=-0.9$. First, calculate the discriminant $\Delta=b^{2}-4ac=3^{2}-4\times1\times(-0.9)=9 + 3.6 = 12.6$. Then $x=\frac{-3\pm\sqrt{12.6}}{2}$. We take the positive root since $x$ must be a physical position on the $x$ - axis. $x=\frac{-3+\sqrt{12.6}}{2}\approx\frac{-3 + 3.55}{2}=0.275\text{m}$ is not valid because it is between the two charges where the forces cannot cancel out (the forces are in the same direction). The valid root is $x=\frac{-3-\sqrt{12.6}}{2}\approx - 3.27\text{m}$.

Answer:

$x\approx - 3.27\text{m}$