Problem 24 - Physics

Question

A positive charge \(+q_{1}\) is located to the left of a negative charge \(-q_{2} .\) On a line passing through the two charges, there are two places where the total potential is zero. The first place is between the charges and is \(4.00 \mathrm{~cm}\) to the left of the negative charge. The second place is \(7.00 \mathrm{~cm}\) to the right of the negative charge. (a) What is the distance between the charges? (b) Find \(q_{1} / q_{2}\), the ratio of the magnitudes of the charges.

Step-by-Step Solution

Answer

Distance = 7 cm; Charge ratio = 7/4.

Step 1: Understanding Electric Potential

Electric potential at a point due to a point charge is given by the formula \( V = \frac{k \cdot q}{r} \), where \( k \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the distance from the charge. The point where the total potential is zero means that the potentials due to the two charges cancel each other out: \( V_1 = -V_2 \).

Step 2: Set Up Equations for Zero Potential Points

We know from the problem statement that there are two places where the total potential is zero: one is \( 4.00 \text{ cm} \) to the left of the negative charge, and the other is \( 7.00 \text{ cm} \) to the right of the negative charge. We set up the equations based on the positions of zero potential: for the point between charges, \( \frac{k \cdot q_1}{a + x} = \frac{k \cdot q_2}{x} \), and for the point to the right, \( \frac{k \cdot q_1}{a + b} = \frac{k \cdot q_2}{b} \), where \( a = 4 \text{ cm} \) and \( b = 7 \text{ cm} \).

Step 3: Relate the Two Equations

For the points where the electric potential is zero, the cancellation of effects due to the opposing charges allows us to eliminate \( k \) from both equations, providing relations: \( q_1 \cdot x = q_2 (a + x) \) and \( q_1 \cdot b = q_2 \cdot (a + b) \).

Step 4: Simplify and Solve for Distance

From the simplified equations, equate \( q_1 \) expressions: \( \frac{x}{a + x} = \frac{b}{a + b} \). Insert the given distances \( x = 4 \text{ cm} \) and \( b = 7 \text{ cm} \) into this equation, and solve for \( a \). The solution is \( a = 3 \text{ cm} \). So, the total distance \( d \) between the charges is \( a + x = 7 \text{ cm} \).

Step 5: Determine Charge Ratio

Use one of the derived equations to find \( q_1/q_2 \). From \( q_1 \cdot x = q_2 (a + x) \): \( q_1 = q_2 \cdot \frac{a + x}{x} \). Substitute \( a + x = 7 \text{ cm} \) and \( x = 4 \text{ cm} \) to get \( q_1 = q_2 \cdot \frac{7}{4} \), thus, \( \frac{q_1}{q_2} = \frac{7}{4} \).

Step 6: Summarize Results

The distance between the charges is \( 7 \text{ cm} \) and the ratio \( \frac{q_1}{q_2} \) is \( \frac{7}{4} \).

Key Concepts

Coulomb's law

Coulomb's law is crucial for understanding the forces between charged objects. It describes how the electric force between two point charges depends on their magnitudes and the distance between them. The formula for Coulomb's law is \[ F = k \frac{{|q_1 \cdot q_2|}}{{r^2}} \] where:

• \( F \) is the magnitude of the force between the charges,
• \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, ext{N m}^2/ ext{C}^2) \),
• \( q_1 \) and \( q_2 \) are the amounts of the two charges,
• \( r \) is the distance between the charges.

This law helps calculate the force but also underpins the concept of electric potential and electric fields. Forces are based on both the amounts of the interacting charges and how far apart they are.

point charge

A point charge refers to a charge that is concentrated at a single point in space. Though real charges have volume, the point charge concept simplifies calculations for electric fields and potential. When solving physics problems, the charge distributions are often treated as point charges if the dimensions of the object are much smaller compared to the distances involved in the problem.
Point charges are essential when considering how electric potential and fields exert effects at various points in their surrounding environment. They also facilitate understanding of the symmetry in electric fields and potentials, as these fields emanate isotropically from point charges unless influenced by nearby charges.

charge ratio

The charge ratio between two charges, such as \( \frac{q_1}{q_2} \), indicates the relative magnitudes of the charges. In electrostatics problems like the one described, determining this ratio helps understand how the electric potential or field strengths compare at certain points.
In the given exercise, by setting up and solving equations at the zero potential points, we found that \( \frac{q_1}{q_2} = \frac{7}{4} \). This implies that the positive charge \( q_1 \) is 1.75 times larger than the negative charge \( q_2 \). Understanding the charge ratio is crucial because it influences where points of zero total potential are found between or around the charges.

electric field

The electric field is a vector field surrounding charges, which represents the force a charge would experience per unit of charge placed in the field's domain. Represented usually by \( \vec{E} \), the electric field due to a single point charge is \[ \vec{E} = k \frac{q}{{r^2}} \hat{r} \] where \( \hat{r} \) is a unit vector pointing from the charge to the point of interest.
Electric fields are directional and radiate outward (for positive charges) or inward (for negative charges). They add vectorially, affecting each other when multiple charges are present. Understanding the implications of electric fields allows us to predict how charges will interact and where neutral or zero potential regions might exist.