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Chapter 19: Problem 15

Two charges \(A\) and \(B\) are fixed in place, at different distances from a certain spot. At this spot the potentials due to the two charges are equal. Charge A is 0.18 \(\mathrm{m}\) from the spot, while charge \(\mathrm{B}\) is \(0.43 \mathrm{m}\) from it. Find the ratio \(q_{\mathrm{B}} / q_{\mathrm{A}}\) of the charges.

Short Answer

Expert verified
The ratio \( q_B/q_A \) is approximately 2.39.

Step by step solution

01

Understanding the Relationship between Potential and Charge

The electric potential due to a charge \( q \) at a distance \( r \) is given by \( V = \frac{kq}{r} \), where \( k \) is Coulomb's constant. The problem states that the potentials due to charges \( A \) and \( B \) are equal at a specific spot.
02

Setting up the Equation for Equal Potentials

Since the potentials are equal, we can set the potential expressions equal to each other: \( \frac{kq_A}{r_A} = \frac{kq_B}{r_B} \). The \( k \) cancels out, simplifying the equation to \( \frac{q_A}{r_A} = \frac{q_B}{r_B} \).
03

Solving for the Ratio of Charges

We want the ratio \( \frac{q_B}{q_A} \). From the simplified equation, we can solve: \( q_B = q_A \frac{r_B}{r_A} \), which gives us \( \frac{q_B}{q_A} = \frac{r_B}{r_A} \).
04

Substituting the Given Distances

Substitute \( r_A = 0.18 \) meters and \( r_B = 0.43 \) meters into the equation \( \frac{q_B}{q_A} = \frac{r_B}{r_A} \):\[ \frac{q_B}{q_A} = \frac{0.43}{0.18} \].
05

Calculating the Ratio

Calculate the division: \( \frac{0.43}{0.18} = 2.3889 \). Therefore, the ratio \( \frac{q_B}{q_A} \) is approximately 2.39.

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Key Concepts

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

Coulomb's Law
Coulomb's Law is a fundamental principle that explains the electric force between two point charges. The force magnitude between these charges is given by the equation:
  • \( F = k \frac{|q_1 q_2|}{r^2} \)
Here, \( F \) denotes the force between the charges, \( q_1 \) and \( q_2 \) are the amounts of the charges, \( r \) is the distance separating them, and \( k \) is Coulomb's constant, valued at approximately \( 8.99 \times 10^9 \mathrm{Nm}^2/\mathrm{C}^2 \).
This law helps us understand interactions between charges, predicting both the magnitude and direction of the force. For charges with the same sign, the force is repulsive, while it's attractive for charges with opposite signs.
Using this law, we relate electric potential to forces and distances between charges, which is crucial in analyzing the potentials in complex systems and finding quantities like the charge ratio.
Charge Ratio
Charge Ratio is the proportional relationship between the magnitudes of two point charges. When comparing charges in a problem, like in the given exercise, it involves setting the electric potentials equal to solve for this ratio.Electric potential \( V \), due to a charge \( q \) at a distance \( r \), is represented by:
  • \( V = \frac{kq}{r} \)
By setting the potentials from two charges equal at a point, we derive the relationship:
  • \( \frac{q_A}{r_A} = \frac{q_B}{r_B} \)
This simplifies to the ratio \( \frac{q_B}{q_A} = \frac{r_B}{r_A} \), showing how one charge compares to the other in terms of distance from the spot.
The problem solution calculates this with given distances, resulting in a practical example showing how varying distances affect the charge requirement to maintain identical potentials at a spot.
Distance in Electric Fields
Distance in electric fields plays a significant role because it influences the strength and potential produced by a charge. In physics, especially electrostatics, the potential \( V \) decreases with increased distance \( r \) from the charge. This is mathematically expressed in the potential formula:
  • \( V = \frac{kq}{r} \)
The further a point is from a charge, the lower the potential at that point. In the exercise, different distances affect the potentials produced by charges \( A \) and \( B \).

At the specific spot mentioned where the two potentials are equal, the distance helps balance the potential difference caused by the different charge amounts. This balance allows for the calculation of a charge ratio, illustrating the interplay between distance and potential in electric fields.
Understanding this concept helps explain how electric fields work and how adjusting distances results in changes in potential, reflecting the effects of electric field interactions across distances.

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Most popular questions from this chapter

What is the capacitance of a capacitor that stores \(4.3 \mu \mathrm{C}\) of charge on its plates when a voltage of \(1.5 \mathrm{V}\) is applied between them?

Two equipotential surfaces surround a \(+1.50 \times 10^{-8} \mathrm{C}\) point charge. How far is the \(190-\mathrm{V}\) surface from the \(75.0-\mathrm{V}\) surface?

The inner and outer surfaces of a cell membrane carry a negative and a positive charge, respectively. Because of these charges, a potential difference of about \(0.070 \mathrm{V}\) exists across the membrane. The thickness of the cell membrane is \(8.0 \times 10^{-9} \mathrm{m} .\) What is the magnitude of the electric field in the membrane?

During a lightning flash, there exists a potential difference of \(V_{\text {cloud }}-V_{\text {ground }}=1.2 \times 10^{9} \mathrm{V}\) between a cloud and the ground. As a result, a charge of \(-25 \mathrm{C}\) is transferred from the ground to the cloud. (a) How much work \(W_{\text {ground-cloud }}\) is done on the charge by the electric force? (b) If the work done by the electric force were used to accelerate a 1100 -kg automobile from rest, what would be its final speed? (c) If the work done by the electric force were converted into heat, how many kilograms of water at \(0^{\circ} \mathrm{C}\) could be heated to \(100^{\circ} \mathrm{C} ?\)

At a distance of \(1.60 \mathrm{m}\) from a point charge of \(+2.00 \mu \mathrm{C},\) there is an equipotential surface. At greater distances there are additional equipotential surfaces. The potential difference between any two successive surfaces is \(1.00 \times 10^{3} \mathrm{V} .\) Starting at a distance of \(1.60 \mathrm{m}\) and moving radially outward, how many of the additional equipotential surfaces are crossed by the time the electric field has shrunk to one-half of its initial value? Do not include thenstarting surface.
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