Question

As shown in the figure, point charge \(q_{1}\) is \(3.979 \mu \mathrm{C}\) and is located at \(x_{1}=-5.689 \mathrm{~m},\) and point charge \(q_{2}\) is \(8.669 \mu \mathrm{C}\) and is located at \(x_{2}=14.13 \mathrm{~m} .\) What is the \(x\) -coordinate of the point at which the net force on a point charge of \(5.000 \mu \mathrm{C}\) will be zero?

Short answer

Answer: The x-coordinate at which the net force on the test charge is zero is approximately -2.14 m.

Step-by-step solution

Identify the given information

The given information is: - Charge \(q_1 = 3.979 \mu C\) and position \(x_1 = -5.689 m\) - Charge \(q_2 = 8.669 \mu C\) and position \(x_2 = 14.13 m\) - Test charge \(\text{q} = 5.000 \mu C\) Our goal is to find the x-coordinate where the net electrostatic force on the test charge is zero.

Set up the Coulomb's law formula

We will use Coulomb's law to find the forces exerted by \(q_1\) and \(q_2\) on the test charge at any x-coordinate: $$F = \frac{k \cdot q_1 \cdot q_2}{r^2}$$ where \(F\) is the electrostatic force, \(k\) is the electrostatic constant (\(8.988\times10^9 Nm^2/C^2\)), \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between the charges.

Find the forces exerted by \(q_1\) and \(q_2\) on the test charge

Let the net force on the test charge be zero at the x-coordinate \(x\). Then the distance between the test charge and charges \(q_1\) and \(q_2\) are \(|x - x_1|\) and \(|x - x_2|\), respectively. The force exerted by \(q_1\) is: $$F_1 = \frac{k \cdot q_1 \cdot \text{q}}{(x - x_1)^2}$$ The force exerted by \(q_2\) is: $$F_2 = \frac{k \cdot q_2 \cdot \text{q}}{(x - x_2)^2}$$

Equate the magnitudes of the forces and solve for x

Since the net force is zero, the magnitudes of \(F_1\) and \(F_2\) must be equal (but in opposite directions). Therefore: $$\frac{k \cdot q_1 \cdot \text{q}}{(x - x_1)^2} = \frac{k \cdot q_2 \cdot \text{q}}{(x - x_2)^2}$$ Solving for x, we simplify the equation by dividing both sides by \(k \cdot \text{q}\): $$\frac{q_1}{(x - x_1)^2} = \frac{q_2}{(x - x_2)^2}$$ Now we can cross-multiply and solve for \(x\): $$q_1(x - x_2)^2 = q_2(x - x_1)^2$$ Plug in the given values: $$3.979 \mu C(x - 14.13 m)^2 = 8.669 \mu C(x - (-5.689 m))^2$$ Solving for \(x\), we get: $$x \approx -2.14 m$$

Final Answer

The x-coordinate at which the net force on the test charge is zero is approximately \(-2.14 m\).