91Ó°ÊÓ

Two small charged spheres are placed in vacuum on the \(x\) -axis: \(+3.0 \mu \mathrm{C}\) at \(x=0\) and \(-5.0 \mu \mathrm{C}\) at \(x=40 \mathrm{~cm} .\) Where must a third charge \(q\) be placed if the force it experiences is to be zero? The situation is represented in Fig. 24-4. We know that \(q\) must be placed somewhere on the \(x\) -axis. (Why?) Suppose that \(q\) is positive. When it is placed in interval \(B C\), the two forces on it are in the same direction and cannot cancel. When it is placed to the right of \(C\), the attractive force from the \(-5 \mu \mathrm{C}\) charge is always larger than the repulsion of the \(+3.0 \mu \mathrm{C}\) charge. Therefore, the force on \(q\) cannot be zero in this region. Only in the region to the left of \(B\) can cancellation occur. (Can you show that this is also true if \(q\) is negative?) For \(q\) placed as shown, when the net force on it is zero, we have \(F_{E 3}=F_{E 5}\) and so, for distances in meters, $$ k_{0} \frac{q\left(3.0 \times 10^{-6} \mathrm{C}\right)}{d^{2}}=k_{0} \frac{q\left(5.0 \times 10^{-6} \mathrm{C}\right)}{(0.40 \mathrm{~m}+d)^{2}} $$ After canceling \(q, k_{0}\), and \(10^{-6} \mathrm{C}\) from each side, cross- multiply to obtain $$ 5 d^{2}=3.0(0.40+d)^{2} \text { or } d^{2}-1.2 d-0.24=0 $$ Using the quadratic formula, $$ d=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{1.2 \pm \sqrt{1.44+0.96}}{2}=0.60 \pm 0.775 \mathrm{~m} $$ Two values, \(1.4 \mathrm{~m}\) and \(-0.18 \mathrm{~m}\), are therefore found for \(d\). The first is the correct one; the second gives the point in \(B C\) where the two forces have the same magnitude but do not cancel.

Step by step solution

01

Setting Up Equilibrium Condition

To find where the third charge experiences zero net force, we need the magnitudes of the forces exerted by the two charges to be equal. Let's denote the positions of the "+3.0 \mu C" charge at \(x = 0\) and the "-5.0 \mu C" charge at \(x = 0.40 \text{ m}\). The force exerted by each charge on the third charge, \(q\), must be equal. Thus, we set: \[F_{E3} = F_{E5}\]Resulting in the equation:\[k_{0} \frac{q(3.0 \times 10^{-6} \text{ C})}{d^2} = k_{0} \frac{q(5.0 \times 10^{-6} \text{ C})}{(0.40 \text{ m} + d)^2}\]

02

Simplifying the Equation

Both \(q\) and \(k_{0}\) cancel out from the equation. This leads to a simpler equation:\[\frac{3.0}{d^2} = \frac{5.0}{(0.40 + d)^2}\]

03

Cross-multiplying to Eliminate Fractions

Eliminate the fractions by cross-multiplying to get the equation:\[5d^2 = 3(0.40 + d)^2\]Expand and rearrange this equation to form a quadratic:\[d^2 - 1.2d - 0.24 = 0\]

04

Using the Quadratic Formula to Solve for d

Apply the quadratic formula \(d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve the quadratic equation where \(a = 1\), \(b = -1.2\), \(c = -0.24\).Calculate:\[d = \frac{1.2 \pm \sqrt{1.44 + 0.96}}{2} = 0.60 \pm 0.775 \text{ m}\]This results in two potential solutions for \(d\):\[d = 1.375 \text{ m} \quad \text{and} \quad d = -0.175 \text{ m}\]

05

Identifying Feasible Solution

The position \(d = -0.175 \text{ m}\) doesn't make sense as it lies between the charges on the \(x\)-axis according to the problem. Hence, the correct position of the charge where forces will cancel each other is \(d = 1.375 \text{ m}\), to the left of the +3.0 \mu \text{C} charge.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Charged Spheres

Imagine two small spheres with electric charges placed on the same horizontal line, known as the x-axis. One sphere has a positive charge of +3.0 μC and is located at the origin, or 0 meters. The other sphere has a negative charge of -5.0 μC and is placed 40 centimeters away, along the same line. These charged spheres create an electric field in their vicinity. When we introduce a third charge, our goal is to find the point along the x-axis where this charge experiences no net force. The interaction between the charged spheres and the third charge will depend on their distances from each other. The force experienced by the third charge can be calculated using Coulomb's Law, which states that the electric force between any two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. By finding a spot where the forces from the two original charges exactly cancel each other out, we would achieve electric force equilibrium for the third charge.

Quadratic Formula

The resolution of mathematical equations involving electric charges often leads to quadratic equations, which require more in-depth solving techniques. After simplifying the original step-by-step solution, we are left with a quadratic equation in terms of the position of the third charge, denoted by \(d\). The equation has the form \(d^2 - 1.2d - 0.24 = 0\). To find the position \(d\) where the net force on the third charge is zero, we can use the quadratic formula:

• The quadratic formula is \(d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• In our scenario, \(a = 1\), \(b = -1.2\), and \(c = -0.24\).

Substituting these values into the quadratic formula gives us two solutions for \(d\): 0.60 ± 0.775 m. Thus, \(d\) can be either 1.375 m or -0.175 m. The quadratic formula provides a systematic method for finding potential solutions by considering both positive and negative results, as indicated by the \(\pm\) sign.

Force Cancellation

In the arena of charged particles, force cancellation plays a crucial role in determining equilibrium points. Equilibrium occurs when the sum of forces acting on an object is zero, meaning that any pushes or pulls are perfectly balanced. For the third charge in our exercise, it's essential to identify where the forces due to the -5.0 μC and +3.0 μC charges exactly cancel each other out. This condition is manifested when the forces meet the relationship: \(F_{E3} = F_{E5}\). The calculations show us that:

• Each possible location of the third charge results is analyzed to see if it leads to zero net force.
• The correct position is found at \(d = 1.375 \, \text{m}\), which means this point lies 1.375 meters to the left of the +3.0 μC charge.
• Forces will not cancel in the other solution, \(d = -0.175 \, \text{m}\), given it doesn’t satisfy the equilibrium due to the spatial arrangement and charge magnitudes.

Thus, force cancellation ensures that the third charge remains stationary, perfectly balancing the attractive and repulsive influences from the other spheres.