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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 \(q_{2}=-7.80 \mu \mathrm{C} \quad\) and \(\operatorname{mass}\) \(1.50 \mathrm{g},\) 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 \(\mathrm{m} / \mathrm{s}\) (Fig. E23.5). 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 \(\mathrm{m}\) apart? (b) How close does \(q_{2}\) get to \(q_{1}\) ?

Short Answer

Expert verified
(a) 15.4 m/s, (b) 0.175 m.

Step by step solution

01

Understanding Conservation of Energy

We will use the conservation of energy principle here. The total energy when the spheres are 0.800 m apart is equal to the total energy when they are 0.400 m apart. This includes the kinetic energy and the electric potential energy. The equation is given by:\[ \frac{1}{2}mv_i^2 + k \frac{q_1 q_2}{r_1} = \frac{1}{2}mv_f^2 + k \frac{q_1 q_2}{r_2} \]where \(v_i\) and \(v_f\) are the initial and final velocities, and \(r_1\) and \(r_2\) are the initial and final distances between the charges.
02

Substitute Known Values

Substituting the known values into our energy conservation equation:\[ \frac{1}{2} imes 1.50 imes 10^{-3} \times 22.0^2 + \frac{8.99 \times 10^9 \times (-2.80 \times 10^{-6}) \times (-7.80 \times 10^{-6})}{0.800} = \frac{1}{2} imes 1.50 \times 10^{-3} \times v_f^2 + \frac{8.99 \times 10^9 \times (-2.80 \times 10^{-6}) \times (-7.80 \times 10^{-6})}{0.400} \]
03

Solve for Final Velocity

Simplify and solve the equation for \(v_f^2\):- Calculate the initial kinetic energy: \( \frac{1}{2} imes 1.50 imes 10^{-3} imes 22.0^2 \approx 0.363 \) J.- Calculate the initial potential energy: \( \frac{8.99 \times 10^9 imes (21.84 \times 10^{-12})}{0.800} \approx 0.246 \) J.- Calculate the final potential energy: \( \frac{8.99 \times 10^9 imes (21.84 \times 10^{-12})}{0.400} \approx 0.493 \) J.- Solve for final kinetic energy: \( 0.363 + 0.246 = 0.493 + \frac{1}{2} imes 1.50 \times 10^{-3} \times v_f^2 \).- Solve for \(v_f\): \(v_f = 15.4 \) m/s.
04

Calculate Closest Approach

To find how close \(q_2\) gets to \(q_1\), we set the final velocity \(v_f\) to zero at the closest approach, which gives the minimum potential energy point. Hence, all initial kinetic energy + initial potential energy is converted into potential energy:\[ 0.363 + 0.246 = \frac{8.99 \times 10^9 \times 21.84 \times 10^{-12}}{r_{min}} \]Solving for \(r_{min}\):\(r_{min} \approx 0.175 \) m.

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Key Concepts

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

Electric Potential Energy
Electric potential energy is a form of energy associated with the positions of charged particles relative to each other. When two charges are involved, as with the small metal spheres in our example, the electric potential energy reflects their potential to interact via electrical forces.

This energy arises from the electrostatic forces between the charges. The formula for calculating the electric potential energy between two point charges is \[ U = k \frac{q_1 q_2}{r} \]where:
  • \( U \) is the electric potential energy,
  • \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \),
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
  • \( r \) is the distance between the charges.

In our scenario, we see how this energy changes as the distance between the charged spheres shifts from 0.800 m to 0.400 m. The change in electric potential energy is crucial in determining new velocities of the spheres as they draw closer.
Kinetic Energy
Kinetic energy is the energy of motion. It is defined by how fast an object is moving and its mass. In the context of moving charges, kinetic energy can be transformed into other energy forms, such as electric potential energy, as they interact with each other.

The basic formula for calculating kinetic energy is:\[ KE = \frac{1}{2} mv^2 \]where:
  • \( KE \) is the kinetic energy,
  • \( m \) is the mass of the object,
  • \( v \) is the velocity of the object.

In the exercise, when the metal sphere with charge \( q_2 \) moves towards \( q_1 \) with an initial speed, it possesses kinetic energy. As it approaches the other charge, some of this kinetic energy is transformed into potential energy, reducing its speed until it reaches its closest point of approach.
Point Charges
Point charges are theoretical charges that act as if they are concentrated at a single point in space. This simplification allows us to ignore the actual size of the objects and focus solely on the effects of charge and distance.

In physics problems like the one at hand, treating objects as point charges lets us focus on the fundamental principles of charge interaction without worrying about more complex spatial properties.

This approximation works well for many problems, offering a good understanding of electrostatic interactions between charged bodies by simplifying calculations. Thus, in our exercise, it is assumed that both spheres are point charges, making the calculations manageable while accurately illustrating the core principles of electrostatics.
Electrostatics
Electrostatics is a branch of physics dealing with stationary or slow-moving electric charges. This field focuses on the forces and fields created by these charges, which are foundational to understanding many principles of electricity and magnetism.

Culminating in Coulomb's Law, electrostatics describes how charged objects exert forces on each other based on their charges and separation distance. In our example:
  • The two charged spheres repel each other because like charges repel.
  • Electrostatic forces result in changes in kinetic and potential energy as the distance between charges varies.

Understanding electrostatics involves comprehending conservation of energy principles, especially when dealing with mechanical energies such as kinetic and electric potential energy. This understanding allows us to predict motion and energy changes as charges interact.

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Most popular questions from this chapter

A point charge \(q_{1}=+5.00 \mu \mathrm{C}\) is held fixed in space. From a horizontal distance of \(6.00 \mathrm{cm},\) a small sphere with mass \(4.00 \times 10^{-3} \mathrm{kg}\) and charge \(q_{2}=+2.00 \mu \mathrm{C}\) is fired toward the fixed charge with an initial speed of 40.0 \(\mathrm{m} / \mathrm{s} .\) Gravity can be neglected. What is the acceleration of the sphere at the instant when its speed is 25.0 \(\mathrm{m} / \mathrm{s} ?\)

Electrical Sensitivity of Sharks. Certain sharks can detect an electric field as weak as 1.0\(\mu \mathrm{V} / \mathrm{m} .\) To grasp how weak this field is, if you wanted to produce it between two parallel metal plates by connecting an ordinary 1.5-V AA battery across these plates, how far apart would the plates have to be?

A metal sphere with radius \(R_{1}\) has a charge \(Q_{1}\) . Take the electric potential to be zero at an infinite distance from the sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a long, thin conduct- ing wire to another sphere of radius \(R_{2}\) that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on each sphere; (c) the electric potential at the surface of each sphere; (d) the electric field at the surface of each sphere? Assume that the amount of charge on the wire is much less than the charge on each sphere.

A point charge \(q_{1}\) is held stationary at the origin. A second charge \(q_{2}\) is placed at point \(a,\) and the electric potential energy of the pair of charges is \(+5.4 \times 10^{-8} \mathrm{J} .\) When the second charge is moved to point \(b,\) the electric force on the charge does \(-1.9 \times 10^{-8} \mathrm{J}\) of work. What is the electric potential energy of the pair of charges when the second charge is at point \(b\) ?

Determining the Size of the Nucless. When radium-226 decays radioactively, it emits an alpha particle (the nucleus of helium), and the end product is radon-222. We can model this decay by thinking of the radium-226 as consisting of an alpha particle emitted from the surface of the spherically symmetric radon-222 nucleus, and we can treat the alpha particle as a point charge. The energy of the alpha particle has been measured in the laboratory and has been found to be 4.79 MeV when the alpha particle is essentially infinitely far from the nucleus. Since radon is much heavier than the alpha particle, we can assume that there is no appreciable recoil of the radon after the decay. The radon nucleus contains 86 protons, while the alpha particle has 2 protons and the radium nucleus has 88 protons. (a) What was the electric potential energy of the alpha-radon combination just before the decay, in MeV and in joules? (b) Use your result from part (a) to calculate the radius of the radon nucleus.

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