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Chapter 23: Problem 5

A small metal sphere, carrying a net charge of \(q_{1}=-2.80 \mu \mathrm{C}\). is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of \(q_{2}=-7.80 \mu \mathrm{C}\) and mass \(1.50 \mathrm{~g}\) is projected toward \(q_{1}\). When the two spheres are \(0.800 \mathrm{~m}\) apart, \(q_{2}\) is moving toward \(q_{1}\) with speed \(22.0 \mathrm{~m} / \mathrm{s} \quad\) (Fig. \(\quad \mathbf{E} 23.5\) ). Assume that the two spheres can be treated as point charges. You can ignore the force of gravity. (a) What is the speed of \(q_{2}\) when the spheres are 0.400 \mathrm{~m} \text { apart? (b) How close does } q_{2} \text { get to } q_{1} ?

Short Answer

Expert verified
The speed of \(q_{2}\) when the spheres are 0.400 m apart can be found by solving the conservation of energy equation formed in part a. The closest distance between the two spheres can be found in a similar manner by solving the energy conservation equation for when \(q_{2}\) comes to a stop, as formulated in part b.

Step by step solution

01

Identify the known values

From the problem, \(q_{1}=-2.80 \mu \mathrm{C}\), \(q_{2}=-7.80 \mu \mathrm{C}\), mass of \(q_{2}\) is 1.50g, initial separation is 0.800m, initial speed of \(q_{2}\) is 22.0 m/s, and the desired separation in part a is 0.400m.
02

Apply conservation of energy for part (a)

The electric potential energy is given by \(U=k \frac{|q_1q_2|}{r}\), and the kinetic energy is given by \(K=\frac{1}{2} mv^2\), where \(k\) is Coulomb's constant. At the starting point, the total energy \(E_{initial}\) is given by \(E_{initial} = U_{initial} + K_{initial} = k\frac{|q_1q_2|}{r_{initial}} + \frac{1}{2} mv_{initial}^2\). At the desired point, the total energy \(E_{final}\) is given by \(E_{final} = U_{final} + K_{final}= k\frac{|q_1q_2|}{r_{final}} + \frac{1}{2} mv_{final}^2\). Because energy is conserved, set \(E_{initial} = E_{final}\) to solve for \(v_{final}\).
03

Solve for the final speed in part (a)

After simplifying the equation obtained in step 2, the final speed \(v_{final}\) of \(q_{2}\) when the spheres are 0.400m apart can be solved for.
04

Apply conservation of energy for part (b)

\(q_{2}\) will stop moving when it comes the closest to \(q_{1}\), so the kinetic energy at that point is zero. The total energy therefore is only the potential energy at that point. Set the initial total energy and the final total energy equal to each other again, this time with \(v_{final}=0\), to solve for the final separation \(r_{final}\), which, in this case, represents the closest distance between the two spheres.
05

Solve for the closest distance in part (b)

After simplifying the equation obtained in step 4, the closest distance \(r_{final}\) can be solved for. Be aware of the negative sign of the charges, which indicates their repulsion.

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Key Concepts

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

Conservation of Energy
In physics, the conservation of energy principle is a fundamental concept that states that the total energy in a closed system remains constant over time. This means energy cannot be created or destroyed, only transformed from one form to another. In the context of electrostatics, this principle is crucial for understanding the behavior of charged objects.
For example, in our problem, the two charged spheres interact through their electric fields. When sphere \(q_{2}\) approaches \(q_{1}\), the system's electric potential energy changes. However, by the conservation of energy, the total mechanical energy (the sum of kinetic and electric potential energy) stays consistent.
  • Initially, sphere \(q_{2}\) has kinetic energy due to its motion and electric potential energy due to its position relative to \(q_{1}\).
  • As sphere \(q_{2}\) moves closer to \(q_{1}\), its speed changes while still conserving the total mechanical energy by exchanging kinetic energy and electric potential energy.
By setting the initial energy equal to the final energy, we can solve for unknowns like the speed of \(q_{2}\) at different positions or the closest approach. Through this principle, we can predict how the motion of charged objects unfolds under electrostatic forces.
Kinetic Energy
Kinetic energy is a form of energy associated with the motion of objects. It's expressed as \(K = \frac{1}{2} mv^2\), where \(m\) is mass and \(v\) is velocity. For our moving charged sphere \(q_{2}\), knowing its kinetic energy helps us understand how fast it's moving at any point during its motion towards \(q_{1}\).
Initially, as \(q_{2}\) approaches \(q_{1}\) from a distance of 0.800 m, it has a defined velocity and hence a specific kinetic energy. As it comes closer to \(q_{1}\), the kinetic energy changes because its velocity changes due to the interaction of electric forces between the spheres.
  • At each point in its journey, the kinetic energy represents the amount of work \(q_{2}\) can perform due to its motion.
  • To find the new velocity at a closer distance, we calculate the positions where kinetic energy has converted to electric potential energy.
Ultimately, kinetic energy plays a pivotal role in analyzing the speed of the sphere at different separations from \(q_{1}\), helping us utilize the conservation of energy principle to predict its behavior accurately.
Coulomb's Law
Coulomb's Law explains the force between two point charges. It鈥檚 defined as \(F = k \frac{|q_1 q_2|}{r^2}\), where \(k\) is Coulomb's constant, \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between them. This law is essential in this exercise as it underpins the electric potential energy calculations.
Electric potential energy \(U\), derived from Coulomb's Law, is expressed as \(U = k \frac{|q_1 q_2|}{r}\). In this scenario, the potential energy between the charged spheres changes as their distance \(r\) varies. This change is pivotal to determine the motion and interaction of \(q_{2}\) with \(q_{1}\).
  • When \(q_{2}\) moves closer to \(q_{1}\), the potential energy increases due to a decrease in distance \(r\), as they have like charges (both negative) that repel.
  • The electric force as per Coulomb鈥檚 Law helps us understand why \(q_{2}\) changes speed as it moves through different distances. This is compensated by changes in its kinetic energy, showcasing the conservation principle.
Understanding Coulomb's Law gives us a real insight into how distance alteration affects force and energy between charged entities, helping solve problems related to electrical interactions effectively.

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Most popular questions from this chapter

A heart cell can be modeled as a cylindrical shell that is \(100 \mu \mathrm{m}\) long, with an outer diameter of \(20.0 \mu \mathrm{m}\) and a cell wall thickness of \(1.00 \mu \mathrm{m}\), as shown in Fig. \(\mathrm{P} 23.81 .\) Potassium ions move across the cell wall, depositing positive charge on the outer surface and leaving a net negative charge on the inner surface. During the so-called resting phase, the inside of the cell has a potential that is \(90.0 \mathrm{mV}\) lower than the potential on its outer surface. (a) If the net charge of the cell is zero, what is the magnitude of the total charge on either cell wall membrane? Ignore edge effects and treat the cell as a very long cylinder. (b) What is the magnitude of the electric field just inside the cell wall? (c) In a subsequent depolarization event, sodium ions move through channels in the cell wall, so that the inner membrane becomes positively charged. At the cnd of this event, the inside of the cell has a potential that is \(20.0 \mathrm{mV}\) higher than the potential outside the cell. If we model this event by charge moving from the outer membrane to the inner membrane, what magnitude of charge moves across the cell wall during this event? (d) If this were done entirely by the motion of sodium ions, \(\mathrm{Na}^{+}\), how many ions have moved?

Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by \(2.20 \mathrm{~cm}\). (a) If the surface charge density for each plate has magnitude \(47.0 \mathrm{nC} / \mathrm{m}^{2}\), what is the magnitude of \(\dot{E}\) in the region between the plates? (b) What is the potential difference between the two plates? (c) If the separation between the plates is doubled while the surface charge density is kept constant at the value in part (a), what happens to the magnitude of the electric field and to the potential difference?

Point charges \(q_{1}=+2.00 \mu \mathrm{C}\) and \(q_{2}=-2.00 \mu \mathrm{C}\) are placed at adjacent corners of a square for which the length of each side is \(3.00 \mathrm{~cm}\). Point \(a\) is at the center of the square, and point \(b\) is at the empty corner closest to \(q_{2}\). Take the electric potential to be zero at a distance far from both charges. (a) What is the electric potential at point \(a\) due to \(q_{1}\) and \(q_{2} ?\) (b) What is the electric potential at point \(b ?\) (c) A point charge \(q_{3}=-5.00 \mu \mathrm{C}\) moves from point \(a\) to point \(b .\) How much work is done on \(q_{3}\) by the electric forces exerted by \(q_{1}\) and \(q_{2} ?\) Is this work positive or negative?

23.78 e DATA A small, stationary sphere carries a net charge \(Q .\) You perform the following experiment to measure Q: From a large distance you fire a small particle with mass \(m=4.00 \times 10^{-4} \mathrm{~kg}\) and charge \(q=5.00 \times 10^{-8} \mathrm{C}\) directly at the center of the sphere. The apparatus you are using measures the particle's speed \(v\) as a function of the distance \(x\) from the sphere. The sphere's mass is much greater than the mass of the projectile particle, so you assume that the sphere remains at rest. All of the measured values of \(x\) are much larger than the radius of cither object, so you treat both objects as point particles. You plot your data on a graph of \(v^{2}\) versus \((1 / x)\) (Fig. \(\mathbf{P} 23.78\) ). The straight line \(v^{2}=400 \mathrm{~m}^{2} / \mathrm{s}^{2}-\left[\left(15.75 \mathrm{~m}^{3} / \mathrm{s}^{2}\right) / x\right]\) gives a good fit to the data points. (a) Explain why the graph is a straight line. (b) What is the initial speed \(v_{0}\) of the particle when it is very far from the sphere? (c) What is \(Q ?\) (d) How close does the particle get to the sphere? Assume that this distance is much larger than the radii of the particle and sphere, so continue to treat them as point particles and to assume that the sphere remains at rest.

(a) An electron is to be accelerated from \(3.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) to \(8.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from \(8.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) to a halt?
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