Problem 24 - University Physics with Modern Physics

Question

Two charges, one of 2.50\(\mu \mathrm{C}\) and the other of \(-3.50 \mu \mathrm{C}\) , are placed on the \(x\) -axis, one at the origin and the other at \(x=0.600 \mathrm{m}\) , as shown in Fig. 21.36 . Find the position on the \(x\) -axis where the net force on a small charge \(+q\) would be zero. figure can't copy

Step-by-Step Solution

Answer

The position is approximately 0.219 m from the origin.

Step 1: Understand the Problem

We need to find the position on the x-axis where the forces from the two charges on a positive charge +q will cancel each other out, resulting in a net force of zero.

Step 2: Set Up the Force Equations

The force on charge +q due to the 2.50μC charge (at the origin) is given by Coulomb's law: \[ F_1 = k \frac{|2.50 imes 10^{-6} ext{ C}| imes q}{x^2} \]For the -3.50μC charge (at \(x=0.600 ext{ m}\)), the force is: \[ F_2 = k \frac{|3.50 imes 10^{-6} ext{ C}| imes q}{(0.600 - x)^2} \]The net force on +q is zero when \( F_1 = F_2 \).

Step 3: Set the Force Equations Equal

Equate the magnitudes of the forces:\[ \frac{2.50 imes 10^{-6}}{x^2} = \frac{3.50 imes 10^{-6}}{(0.600 - x)^2} \]This equation results because the forces need to be equal in magnitude for their net to be zero.

Step 4: Solve for x

Cross-multiply to clear the fractions:\[ 2.50(0.600 - x)^2 = 3.50x^2 \]Simplify and rearrange into a standard quadratic form:\[ 1.25 - 3.00x + 1.25x^2 = 1.75x^2 \]\[ 0 = 0.5x^2 + 3.00x - 1.25 \]Use the quadratic formula to solve for \(x\):\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Where \(a = 0.5\), \(b = 3.00\), \(c = -1.25\).Calculate the discriminant and solution.

Step 5: Calculate the Final Answer

Compute using the quadratic formula:\[ b^2 - 4ac = (3.00)^2 - 4(0.5)(-1.25) = 9 + 2.5 = 11.5 \]\[ x = \frac{-3.00 \pm \sqrt{11.5}}{1.00} \]This gives two potential solutions, choose the realistic one for x positioned between 0 and 0.600 meters.

Key Concepts

Coulomb's Law

Coulomb's Law is a fundamental principle in electrostatics that explains the force between two point charges. This law states that the electric force (\( F \)) between two charges is directly proportional to the magnitude of each charge and inversely proportional to the square of the distance between them. The formula for Coulomb's Law is given by:\[ F = k \frac{|q_1 q_2|}{r^2} \]where:

• \( F \) is the force in newtons (N),
• \( k \) is Coulomb's constant (\( 8.9875 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
• \( q_1 \) and \( q_2 \) are the magnitudes of the two charges in coulombs (C),
• \( r \) is the distance between the charges in meters (m).

This principle helps us calculate the forces between any two charges, regardless of their sizes or signs. In the context of our exercise, which involves charges of 2.50μC and -3.50μC placed on the x-axis, Coulomb's Law helps determine the electric force that each charge exerts on the small charge \(+q\), allowing us to set up equations to find where the forces balance out.

Electric Force Equilibrium

Electric Force Equilibrium occurs when the net electric force acting on a charge is zero. In this state, all the forces from different charges balance each other out perfectly. This is important when trying to find a position where a charge feels no net force, as seen in the exercise where we need the small charge \(+q\) to experience no net force along the x-axis. To achieve this equilibrium state, the magnitudes of the electric forces exerted by different charges must be equal, even if their directions differ. This is set up by using the expression from Coulomb's Law for each interacting pair of charges, and equating them:\[ \frac{2.50 \times 10^{-6}}{x^2} = \frac{3.50 \times 10^{-6}}{(0.600 - x)^2} \]Here, we define \( x \) as the distance from one charge to the point where \(+q\) is placed. Solving for \( x \) enables us to determine that special spot where the electric forces cancel out. This setup ensures the net force is zero, perfectly balancing the attractions and repulsions from the surrounding charges.

Quadratic Equation in Physics

Quadratic equations often appear in physics problems, especially when dealing with forces and equilibrium. In our context, the process of equating the forces using Coulomb's Law simplified into a quadratic equation for the position \( x \). The standard form of a quadratic equation is:\[ ax^2 + bx + c = 0 \]To solve this equation, we use the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a \), \( b \), and \( c \) are coefficients derived from our particular equation. For our problem, they translate to:

• \( a = 0.5 \)
• \( b = -3.00 \)
• \( c = -1.25 \)

Plugging these into the quadratic formula helps calculate precise positions on the x-axis where the forces equate to zero, ensuring net neutrality. Understanding this mathematical concept is vital for solving similar problems in physics where forces and positions must align under specific conditions.