Question

Two point charges lie on the \(x\) -axis. If one point charge is \(6.0 \mu C\) and lies at the origin and the other is \(-2.0 \mu C\) and lies at \(20.0 \mathrm{~cm}\), at what position must a third charge be placed to be in equilibrium?

Short answer

Answer: The third charge must be placed at \(x = 10 \ \mathrm{cm}\) along the x-axis for it to be in equilibrium.

Step-by-step solution

Understand Coulomb's Law and forces involved

We will use Coulomb's Law to calculate the electric forces exerted by each of the two given point charges. Coulomb's Law is given by: $$F = \frac{kq_1q_2}{r^2}$$ where \(F\) is the electric force between two charges, \(k\) is Coulomb's constant \((8.99 \times 10^{9} Nm^2/C^2)\), \(q_1\) and \(q_2\) are the magnitudes of the charges, and \(r\) is the distance between the charges. There will be two forces acting on the third charge \(q\): 1. The force exerted by the \(6.0 \mu C\) charge at the origin, which we'll call \(F_1\). 2. The force exerted by the \(-2.0 \mu C\) charge at \(20.0 cm\), which we'll call \(F_2\).

Define a variable for the position of the third charge

Let's call the position of the third charge along the x-axis as \(x\) (\(0<x<20\)).

Calculate the forces exerted on the third charge

The force exerted by the \(6.0 \mu C\) charge at the origin (\(F_1\)) on the third charge can be written using Coulomb's Law as: $$F_1 = \frac{k\cdot6.0\mu C\cdot q}{x^2}$$ And the force exerted by the \(-2.0 \mu C\) charge at \(20.0 \mathrm{~cm}\) (\(F_2\)) on the third charge is: $$F_2 = \frac{k\cdot(-2.0\mu C)\cdot q}{(20-x)^2}$$

Equate the forces for equilibrium

For the third charge to be in equilibrium, the two forces \(F_1\) and \(F_2\) must cancel each other out. Therefore, we can equate the magnitudes of the forces: $$\frac{k\cdot6.0\mu C\cdot q}{x^2} = \frac{k\cdot2.0\mu C\cdot q}{(20-x)^2}$$ Note that \(k\) and \(q\) will cancel each other out.

Solve for the position of the third charge

Solve the above equation for the position \(x\) of the third charge: $$\frac{6.0\mu C}{x^2} = \frac{2.0\mu C}{(20-x)^2}$$ Divide both sides by \(2.0 \mu C\): $$\frac{3}{x^2} = \frac{1}{(20-x)^2}$$ Now, we'll cross-multiply: $$3(20-x)^2 = x^2$$ Expanding and simplifying the equation: $$3(400-40x+x^2) = x^2$$ $$1200 - 120x + 3x^2 = x^2$$ $$2x^2 - 120x + 1200 = 0$$ To solve this quadratic equation, we can divide by 2: $$x^2 - 60x + 600 = 0$$ We can use the quadratic formula to solve for \(x\): $$x = \frac{-(-60) \pm \sqrt{(-60)^2 - 4(1)(600)}}{2(1)}$$ Upon solving, we find two solutions for \(x\): \(x = 10\) and \(x = 60\). However, since the \(-2.0 \mu C\) charge is at \(20.0 \mathrm{~cm}\) and the equilibrium position must be between the two charges, the only valid solution is: \(x = 10 \ \mathrm{cm}\)

Key concepts

Coulomb's Law

Coulomb's Law is a fundamental principle in electrostatics that describes the electric force between two charged objects. The law is expressed by the equation:

• \( F = \frac{kq_1q_2}{r^2} \)

where:

• \( F \) is the magnitude of the force between the charges,
• \( k \) is Coulomb's constant with a value of \( 8.99 \times 10^{9} \, \text{Nm}^2 \text{C}^{-2} \),
• \( q_1 \) and \( q_2 \) are the charges, and
• \( r \) is the distance separating them.

According to Coulomb's Law, the force is attractive if charges have opposite signs and repulsive if they have the same sign. This law is analogous to Newton's law of universal gravitation, but it applies to electric charges rather than masses. Understanding Coulomb's Law is crucial in predicting how charged objects will interact.

Equilibrium of Charges

In electrostatics, the equilibrium position for a charged object is where it experiences no net electric force. This occurs when the forces acting on the charge balance each other out. In the context of the given problem, we need to find the position on the x-axis where the forces exerted by two other charges cancel each other.
To achieve equilibrium, the sum of forces acting on the third charge must be zero. So, if there's a positive force on one side, there must be an equal negative force on the other side to balance it out. This balance leads to a charge finding a stable point where it neither moves towards nor away from other charges. Identifying this equilibrium involves setting the forces equal to each other so that their magnitudes are the same, but their directions opposing.

Electric Force Calculation

Calculating the electric force involves using Coulomb's Law to find the force exerted on a point charge by other nearby charges. In scenarios where multiple charges are involved, such as our exercise, it's essential to individually calculate forces from each charge and consider their directions.

• For the first charge located at the origin: \( F_1 = \frac{k \times 6.0 \mu C \times q}{x^2} \).
• For the second charge located at 20 cm: \( F_2 = \frac{k \times (-2.0 \mu C) \times q}{(20-x)^2} \).

The net force is the vector sum of these individual forces. To ensure the third charge is in equilibrium, you'd set these forces equal to each other, solve for \( x \), ensuring they are tracked based on their directions, one being positive and the other negative due to their nature as repulsive and attractive forces.

Quadratic Equation in Physics

In this problem, reaching an equilibrium position for the charge requires solving a quadratic equation. This equation arises from equating the magnitudes of the forces acting on the charge.

• Initially, we derived: \( \frac{6.0 \mu C}{x^2} = \frac{2.0 \mu C}{(20-x)^2} \),
• which simplifies to the quadratic equation: \( x^2 - 60x + 600 = 0 \).

Once in this standard quadratic form, it can be solved using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

where \( a \), \( b \), and \( c \) are the coefficients of the equation. In physics, solving quadratic equations can determine key points such as equilibrium, maximum height, or even time intervals. When solving, ensure that the final solution logically fits within the physical constraints of the problem, like distance in this case.