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Chapter 19: Problem 11

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 \(\mathrm{A}\) 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 \( \frac{q_{B}}{q_{A}} \) is 2.39.

Step by step solution

01

Understanding Electrical Potential

Electrical potential from a point charge is calculated using the formula \( V = \frac{kq}{r} \) where \( k \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the distance from the charge to the point. For this problem, we need to compare the potentials at the spot due to charges A and B.
02

Setting Up the Equation for Equal Potentials

Since the potentials at the spot are equal due to both charges, we can write the equation: \( \frac{k q_{A}}{0.18} = \frac{k q_{B}}{0.43} \). Notice that \( k \) cancels out on both sides.
03

Simplifying and Solving for the Ratio

With \( k \) cancelled out, the equation simplifies to \( \frac{q_{A}}{0.18} = \frac{q_{B}}{0.43} \). Rearrange this to solve for the ratio \( \frac{q_{B}}{q_{A}} \): \( q_{B} \times 0.18 = q_{A} \times 0.43 \). This gives us \( \frac{q_{B}}{q_{A}} = \frac{0.43}{0.18} \).
04

Calculating the Ratio

Divide 0.43 by 0.18 to find the ratio: \( \frac{q_{B}}{q_{A}} = 2.39 \). This means charge B is 2.39 times the magnitude of charge A at the given spot.

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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 describes the force between two charges. In essence, it tells us how strongly two charges pull or push on each other. The formula is given by: \[F = \frac{kq_1q_2}{r^2}\]where \( F \) is the force between the charges, \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the values of the two charges, and \( r \) is the distance between them.
What makes Coulomb's law so powerful is its application in understanding electric potential and fields. It's crucial to remember that this law only applies precisely in a vacuum, but it can be approximated for other mediums too.
  • Positive charges repel each other
  • Negative charges attract positive charges
This law lays the foundation for understanding how charges interact, which is central to calculating electric potential and potential energy.
charge ratio calculation
When we calculate the ratio of two charges, we're comparing their magnitudes. In the exercise, we are determining the ratio \( \frac{q_B}{q_A} \). This means how many times charge B's magnitude is compared to charge A's.
The key step here is to set up an equation for equal potentials. Given the equal potentials created by different distances, we used the formula from the electric potential concept: \( V = \frac{kq}{r} \). Since the potentials are equal, we equate the two expressions for potential:
\[\frac{kq_A}{0.18} = \frac{kq_B}{0.43}\]By canceling the constant \( k \) and solving, we rearrange to find the ratio:
  • \( q_B \times 0.18 = q_A \times 0.43 \) (cross-multiply to simplify)
  • \( \frac{q_B}{q_A} = \frac{0.43}{0.18} \)
This gives the result of \( \frac{q_B}{q_A} = 2.39 \), showing that charge B is 2.39 times larger than charge A.
electric potential formula
Electric potential is a vital concept when discussing electric charges. It represents the potential energy per unit charge at a specific point in space due to a charge or a distribution of charges.
The formula to calculate electric potential \( V \) due to a point charge is:\[V = \frac{kq}{r}\]where \( V \) is the potential, \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, \mathrm{N} \, \mathrm{m}^2/\mathrm{C}^2) \), \( q \) is the charge, and \( r \) is the distance from the charge.
  • Potential is high near the charge and decreases with distance.
  • The unit of electric potential is volts \( (V) \).
In the exercise, this concept helps us determine how the potential at the spot is influenced by both charges. By equating the potentials due to both charges and cancelling common factors, we are able to solve for unknowns like the ratio of the charges. Understanding this formula helps in many applications, ranging from circuit analysis to understanding electric fields.

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Most popular questions from this chapter

Two hollow metal spheres are concentric with each other. The inner sphere has a radius of \(0.1500 \mathrm{~m}\) and a potential of \(85.0 \mathrm{~V}\). The radius of the outer sphere is \(0.1520 \mathrm{~m}\) and its potential is \(82.0 \mathrm{~V}\). If the region between the spheres is filled with Teflon, find the electric energy contained in this space.

A particle has a charge of \(+1.5 \mu \mathrm{C}\) and moves from point \(A\) to point \(B\), a distance of \(0.20 \mathrm{~m}\). The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at \(A\) and \(B\) is \(\mathrm{EPE}_{A}-\mathrm{EPE}_{B}=+9.0 \times 10^{-4} \mathrm{~J}\). (a) Find the magnitude and direction of the electric force that acts on the particle. (b) Find the magnitude and direction of the electric field that the particle experiences.

The membrane that surrounds a certain type of living cell has a surface area 8 of \(5.0 \times 10^{-9} \mathrm{~m}^{2}\) and a thickness of \(1.0 \times 10^{-8} \mathrm{~m}\). Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of \(5.0 .\) (a) The potential on the outer surface of the membrane is \(+60.0 \mathrm{mV}\) greater than that on the inside surface. How much charge resides on the outer surface? (b) If the charge in part (a) is due to \(\mathrm{K}^{+}\) ions (charge \(+e\) ), how many such ions are present on the outer surface?

Go Concept Questions Capacitor A and capacitor B both have the same voltage across their plates. However, the energy of capacitor A can melt \(m\) kilograms of ice at \(0^{\circ} \mathrm{C}\), while the energy of capacitor B can boil away the same amount of water at \(100{ }^{\circ} \mathrm{C}\). (a) Which requires more energy, melting the ice or boiling the water? (b) Which capacitor has the greater capacitance? Explain your answers. The capacitance of capacitor \(A\) is \(9.3 \mu F .\) What is the capacitance of capacitor B? Be sure that your answer is consistent with your answers to the Concept Questions.

A positive charge of \(+q_{1}\) is located \(3.00 \mathrm{~m}\) to the left of a negative charge \(-q_{2}\). The charges have different magnitudes. On the line through the charges, the net electric field is zero at a spot \(1.00 \mathrm{~m}\) to the right of the negative charge. On this line there are also two spots where the potential is zero. Locate these two spots relative to the negative charge.
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