Question

A point charge \(+3 q\) is located at the origin, and a point charge \(-q\) is located on the \(x\) -axis at \(D=0.500 \mathrm{~m}\). At what location on the \(x\) -axis will a third charge, \(q_{0}\) experience no net force from the other two charges?

Short answer

Based on the calculations and analysis, there is no location on the x-axis where the charge \(q_0\) would experience no net force from the \(+3q\) and \(-q\) charges.

Step-by-step solution

Define variables and equations

Let \(x_0\) be the location of \(q_0\) on the x-axis. The distance between \(+3q\) and \(q_0\) is \(x_0\), and the distance between \(-q\) and \(q_0\) is \((0.5 - x_0)\). In order to have a net force of zero, the forces between \(+3q\) and \(q_0\) and between \(-q\) and \(q_0\) must be equal and in opposite directions. Let \(F_{+3q\rightarrow q_0}\) be the force on \(q_0\) due to the \(+3q\) charge, and \(F_{-q\rightarrow q_0}\) be the force on \(q_0\) due to the \(-q\) charge. According to Coulomb's Law, we have the following equations: $$F_{+3q\rightarrow q_0} = k \frac{3q\cdot q_0}{x_0^2}$$ $$F_{-q\rightarrow q_0} = k \frac{-q\cdot q_0}{(0.5 - x_0)^2}$$

Set the forces equal and solve for \(x_0\)

Since the net force on \(q_0\) must be zero, \(F_{+3q\rightarrow q_0} = F_{-q\rightarrow q_0}\). Therefore, we can set the equations equal and solve for \(x_0\): $$k \frac{3q\cdot q_0}{x_0^2} = k \frac{-q\cdot q_0}{(0.5 - x_0)^2}$$ We can simplify the equation by multiplying both sides by \(x_0^2(0.5 - x_0)^2\): $$3q\cdot q_0(0.5 - x_0)^2 = -q\cdot q_0x_0^2$$ The \(q_0\) terms will cancel out, and then we can divide both sides by \(-q\): $$-3(0.5 - x_0)^2 = x_0^2$$

Solve the quadratic equation

Now, expand and rearrange the equation to obtain a quadratic equation in the form of \(ax^2 + bx + c = 0\): $$-3(0.25 - x_0 + x_0^2) = x_0^2$$ $$-0.75 + 3x_0 - 3x_0^2 = x_0^2$$ $$4x_0^2 - 3x_0 + 0.75 = 0$$

Find the roots using the quadratic formula

To solve the quadratic equation, we will use the quadratic formula: $$x_0 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Plugging in our values, we get: $$x_0 = \frac{3 \pm \sqrt{(-3)^2 - 4(4)(0.75)}}{2(4)}$$ $$x_0 = \frac{3 \pm \sqrt{9 - 12}}{8}$$ Since the radicand is negative, there is no real solution for this problem. This means that there is no location on the x-axis where the third charge would experience no net force from the other two charges.