Question

A positive charge \(q_{1}=1.00 \mu \mathrm{C}\) is fixed at the origin, and a second charge \(q_{2}=-2.00 \mu \mathrm{C}\) is fixed at \(x=10.0 \mathrm{~cm} .\) Where along the \(x\) -axis should a third charge be positioned so that it experiences no force?

Short answer

Answer: The third charge should be placed at approximately 7.35 cm along the x-axis to experience no net force.

Step-by-step solution

Define the problem and variables

We have two charges: \(q_1 = 1.00\mu C\) at position \(x = 0\) and \(q_2 = -2.00\mu C\) at position \(x = 10.0 \mathrm{~cm}\). Let's denote the third charge as \(q\). We need to find the x-coordinate of \(q\) such that the net force acting on it is zero.

Set up the equation for the net force

The force that charge \(q_1\) exerts on the third charge is given by Coulomb's law: \(F_{1} = k {\frac{q_{1}q}{r_{1}^2}}\), where \(k = \cfrac{8.9875 \times 10^9 N m^2}{C^2}\) is the electrostatic constant, \(r_1\) is the distance between \(q_1\) and the third charge. Similarly, the electrostatic force \(q_2\) exerts on the third charge is \(F_{2} = k \cfrac{q_{2}q}{r_{2}^2}\), where \(r_2\) is the distance between \(q_2\) and the third charge. To find the position where the third charge experiences no net force, we need to equalize these forces: \(F_{1} = F_{2}\).

Rewrite the net force equation in terms of distances

By substituting the force expressions into the net force equation, we get: \(k \cfrac{q_{1}q}{r_{1}^2} = k \cfrac{q_{2}q}{r_{2}^2}\) Now, we can cancel out the constants \(k\) and \(q\), as they are not equal to zero: \(\cfrac{q_{1}}{r_{1}^2} = \cfrac{q_{2}}{r_{2}^2}\) Let \(x_{q}\) be the position of the third charge along the \(x\)-axis. Then, the distances between the charges are \(r_1 = x_q\) and \(r_2 = 10.0 - x_q\). Substitute these values into the equation: \(\cfrac{1.00\mu C}{(x_q)^2} = \cfrac{-2.00\mu C}{(10.0 - x_q)^2}\)

Solve for the position \(x_q\) of the third charge

To solve for the position, we can first simplify the equation by dividing both sides by \(1.00\mu C\). Then we'll obtain: \(\cfrac{1}{(x_q)^2} = -2 \cfrac{1}{(10.0 - x_q)^2}\) Next, take the reciprocal of both sides: \((x_q)^2 = -\cfrac{1}{2}(10.0 - x_q)^2\) We can see that there is a negative sign on the right side of the equation, which means there might be no real solution. However, since electric forces can be attractive or repulsive, the negative sign indicates that we need to place the third charge between the two charges. By considering this fact, we can change the equation to: \((x_q)^2 = 2(10.0 - x_q)^2\) Now, solve the equation for \(x_q\): \(2(10.0 - x_q)^2 = (x_q)^2\) \(2(100 - 20x_q + (x_q)^2) = (x_q)^2\) Simplify: \(200 -40x_q + 2(x_q)^2 = (x_q)^2\) Subtract \((x_q)^2\) from both sides: \(x_q^2 - 40x_q + 200 = 0\) Now, we'll solve the quadratic equation using the quadratic formula: \(x_q = \cfrac{-(-40) \pm \sqrt{(-40)^2 - 4(1)(200)}}{2(1)}\) \(x_q = \cfrac{40 \pm \sqrt{1600 - 800}}{2}\) We have two possible solutions: \(x_q = \cfrac{40}{2} \pm \cfrac{\sqrt{800}}{2}\) \(x_q \approx 20 \pm 12.65\) \(x_q \approx 7.35 \mathrm{~cm}\) or \(x_q \approx 32.65 \mathrm{~cm}\) When we think about the physical meaning of these solutions, we notice that the second solution at 32.65 cm is not between charges \(q_1\) and \(q_2\). Therefore, the correct position for the third charge is at \(x_q \approx 7.35 \mathrm{~cm}\) along the \(x\)-axis.