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Chapter 21: Problem 81

As shown in the figure, charge 1 is \(3.94 \mu \mathrm{C}\) and is located at \(x_{1}=-4.7 \mathrm{~m},\) and charge 2 is \(6.14 \mu \mathrm{C}\) and is at \(x_{2}=\) \(12.2 \mathrm{~m} .\) What is the \(x\) -coordinate of the point at which the net force on a point charge of \(0.300 \mu \mathrm{C}\) is zero?

Short Answer

Expert verified
Based on the given problem and the steps in the solution, calculate the point on the x-axis where the net force acting on a point charge of \(0.300 \mu C\) is zero.

Step by step solution

01

Identify relevant formulas

We will use Coulomb's Law to calculate the electric force between the charges: \(F = k \dfrac{q_1q_2}{r^2}\), where \(F\) is the force, \(k\) is Coulomb's constant (\(8.99 \times 10^9 N m^2 C^{-2}\)), \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between them.
02

Set up the equation for the net force on the 0.300 碌C charge

Let's call the \(0.300 \mu C\) charge \(q_3\) and its position \(x\). Then, the force acting on it due to charges 1 and 2 can be given by: \(F_{13} = k \dfrac{q_1q_3}{(x-x_1)^2}\) and \(F_{23} = k \dfrac{q_2q_3}{(x-x_2)^2}\), where \(F_{13}\) and \(F_{23}\) are the forces acting on charge \(q_3\) due to charges 1 and 2, respectively. The net force is zero when \(F_{13}\) is equal in magnitude but opposite in direction to \(F_{23}\): \(F_{13} = -F_{23}\).
03

Substitute the known values and solve for x

Substitute the values of the charges and their positions into the equation from Step 2, and solve for x: \(k \dfrac{q_1q_3}{(x-x_1)^2} = -k \dfrac{q_2q_3}{(x-x_2)^2}\). Note that \(k\) and \(q_3\) can be canceled out from both sides of the equation: \(\dfrac{q_1}{(x-x_1)^2} = -\dfrac{q_2}{(x-x_2)^2}\). Now, substitute the values of \(q_1\), \(q_2\), \(x_1\), and \(x_2\): \(\dfrac{3.94 \times 10^{-6}}{(x -(-4.7))^2} = -\dfrac{6.14 \times 10^{-6}}{(x - 12.2)^2}\).
04

Simplify and solve for x

Rearrange the equation and simplify. You can multiply both sides of the equation by \((x+4.7)^2(x-12.2)^2\): \((3.94\times10^{-6})(x-12.2)^2 = -(6.14\times10^{-6})(x+4.7)^2\) Now, solve for x: \(3.94 (x^2 - 24.4x + 148.84) = -6.14 (x^2 + 9.4x + 22.09)\) Expand and move all terms to one side of the equation: \(3.94x^2 - 6.14x^2 - 96.216x + 57.676x + 585.8006 = 0\) Simplify the equation: \(-2.20x^2 - 38.54x + 585.8006 = 0\)
05

Solve the quadratic equation

Use the quadratic formula to solve for x: \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a=-2.20\), \(b=-38.54\), and \(c=585.8006\). Plugging in these values: \(x = \dfrac{38.54 \pm \sqrt{(-38.54)^2 - 4(-2.20)(585.8006)}}{2(-2.20)}\) Solve for x to find the two possible solutions: \(x_1 \approx -1.05 m\) and \(x_2 \approx 255.36 m\).
06

Choose the correct solution

Only one of these solutions makes physical sense. Since charge 1 is at \(x_1=-4.7 m\) and charge 2 is at \(x_2=12.2 m\), it is more likely that the point where the net force is zero will be between these two charges. Therefore, the correct solution is \(x \approx -1.05 m\).

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Key Concepts

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

Understanding Electric Force in Coulomb's Law
The process of mastering physics concepts often begins with understanding the fundamental forces that govern interactions in the universe. Among these, the electric force is a key player. It's described by Coulomb's Law, which states that the electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

The formula derived from Coulomb's Law is thus:
\[ F = k \frac{q_1q_2}{r^2} \]
where \(k\) represents Coulomb's constant, \(q_1\) and \(q_2\) are the magnitudes of the charges, and \(r\) is the distance between the charges. This relationship is pivotal for solving problems involving electric forces and can be visualized by imagining charges exerting invisible lines of force upon each other, attracting or repelling depending on their polarities. In the context of our problem, we use this law to assess the forces acting on a nominal third charge to determine where it would experience no net force.
Achieving Net Force Zero
Another intriguing scenario in the realm of physics is the concept of net force zero. This occurs when all the forces acting upon an object balance out, resulting in no overall force impacting the object's motion. It's an essential condition for equilibrium in various physical systems.

In the given exercise, we seek the specific point at which a third charge, introduced into the system of two existing charges, experiences equal and opposite forces from these charges, leading to zero net force. To clarify:
\[ F_{13} = -F_{23} \]
This condition indicates that the force due to charge 1 on this third charge (\(F_{13}\)) must be equal in magnitude but opposite in direction to the force due to charge 2 (\(F_{23}\)). Balancing these forces is similar to finding the 'sweet spot' in a tug of war where neither side gains ground鈥攖he spot where the third charge 'feels' both forces equally but remains motionless due to their cancelation.
Solving with the Quadratic Equation
Complex interactions often lead to equations that can't be solved through simple algebra, which is when the quadratic equation comes into play. It is an indispensable tool for solving second-order polynomials of the form:
\[ ax^2 + bx + c = 0 \]
To find the roots of such an equation, which represent the solutions to our problem, the quadratic formula is used:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our exercise, after accounting for Coulomb's Law and setting up an equation for net force zero, we arrive at a quadratic equation that we solve to find potential positions for the third charge. However, not all solutions may make physical sense. Hence, as detailed in the solution steps, one must scrutinize the context to select the appropriate physical solution.

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Most popular questions from this chapter

When a metal plate is given a positive charge, which of the following is taking place? a) Protons (positive charges) are transferred to the plate from another object. b) Electrons (negative charges) are transferred from the plate to another object. c) Electrons (negative charges) are transferred from the plate to another object, and protons (positive charges) are also transferred to the plate from another object. d) It depends on whether the object conveying the charge is a conductor or an insulator.

Two lightweight metal spheres are suspended near each other from insulating threads. One sphere has a net charge; the other sphere has no net charge. The spheres will a) attract each other. b) exert no net electrostatic force on each other. c) repel each other. d) do any of these things depending on the sign of the charge on the one sphere.

Suppose the Earth and the Moon carried positive charges of equal magnitude. How large would the charge need to be to produce an electrostatic repulsion equal to \(1.00 \%\) of the gravitational attraction between the two bodies?

Rubbing a balloon causes it to become negatively charged. The balloon then tends to cling to the wall of a room. For this to happen, must the wall be positively charged?

In the Bohr model of the hydrogen atom, the electron moves around the one- proton nucleus on circular orbits of well-determined radii, given by \(r_{n}=n^{2} a_{\mathrm{B}}\), where \(n=1,2,3, \ldots\) is an integer that defines the orbit and \(a_{\mathrm{B}}=5.29 \cdot 10^{-11} \mathrm{~m}\) is the radius of the first (minimum) orbit, called the Bohr radius. Calculate the force of electrostatic interaction between the electron and the proton in the hydrogen atom for the first four orbits. Compare the strength of this interaction to the gravitational interaction between the proton and the electron.
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