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Chapter 23: Problem 1

A point charge \(q_{1}=+2.40 \mu C\) is held stationary at the origin. A second point charge \(q_{2}=-4.30 \mu C\) moves from the point \(x=0.150 \mathrm{m}, y=0\) to the point \(x=0.250 \mathrm{m}, y=0.250 \mathrm{m} .\) How much work is done by the electric force on \(q_{2} ?\)

Short Answer

Expert verified
The work done by the electric force on \( q_2 \) is -0.357 J (negative sign indicates work extracted from the system).

Step by step solution

01

Understanding the Formula for Work Done by Electric Forces

The work done by the electric force when moving a charge in an electric field is calculated by the change in the potential energy between two positions: \[ W = -\Delta U = U_i - U_f \]where \( U_i \) and \( U_f \) are the initial and final potential energies respectively. It's important to note the negative sign convention indicates work done on the system by an internal conservative force.
02

Calculating the Initial Potential Energy

The formula for the potential energy between two point charges is given by:\[ U = k \frac{{q_1 q_2}}{r} \]Substituting the initial values gives:\[ U_i = k \frac{{(+2.40 \times 10^{-6} \, \mathrm{C})(-4.30 \times 10^{-6} \, \mathrm{C})}}{0.150 \, \mathrm{m}} \]Using \( k = 8.99 \times 10^9 \, \mathrm{N \cdot m^2/C^2} \), \( U_i \approx -0.618 \, \mathrm{J} \).
03

Calculating the Final Potential Energy

To find the final potential energy when the charge moves to the point (0.250 m, 0.250 m), calculate the distance from the origin using the distance formula:\[ r_f = \sqrt{0.250^2 + 0.250^2} = 0.354 \, \mathrm{m} \]Then, substitute into the potential energy formula:\[ U_f = k \frac{{(+2.40 \times 10^{-6} \, \mathrm{C})(-4.30 \times 10^{-6} \, \mathrm{C})}}{0.354 \, \mathrm{m}} \approx -0.261 \, \mathrm{J} \]
04

Finding the Change in Potential Energy

The change in potential energy \( \Delta U \) is given by:\[ \Delta U = U_f - U_i = (-0.261 \, \mathrm{J}) - (-0.618 \, \mathrm{J}) = 0.357 \, \mathrm{J} \]
05

Calculating the Work Done

Finally, the work done by the electric force on \( q_2 \) is:\[ W = -\Delta U = -0.357 \, \mathrm{J} \]This signifies 0.357 Joules of work is done by the electric force in moving the charge from the initial to the final position.

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Key Concepts

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

Point Charge
A point charge is a very small particle which carries an electric charge. It can either have a positive charge, like a proton, or a negative charge, like an electron. In the context of our exercise, we have two point charges: one at the origin with a charge of +2.40 碌C, and another charge (-4.30 碌C) moving in the electric field created by the stationary charge.

Since point charges are idealized as having no size, we assume their charge is concentrated at a single point. This makes it easier to use equations to calculate forces and potential energies without considering the physical dimensions of the charge.

Key points about point charges:
  • They interact with each other according to Coulomb's law, which defines the force between two charges.
  • The force between them is proportional to the product of their charges and inversely proportional to the square of the distance between them.
  • Point charges are used in theory to simplify complex systems and calculations by removing variables related to dimensions of real-life objects.
Electric Potential Energy
Electric potential energy is the energy a charge possesses due to its position in an electric field. It is similar to gravitational potential energy, which depends on an object's height in Earth's gravitational field.

When point charges are involved, electric potential energy (U) between two charges depends on:
  • The magnitude of each charge.
  • The distance between the charges.
  • The electrostatic constant ( k = 8.99 imes 10^9 N\cdot m^2/C^2).
Mathematically, electric potential energy (in joules, J) is given by:\[ U = k \frac{q_1 q_2}{r} \] where q_1 and q_2 are the point charges and r is the distance between them.

In the exercise:
  • The initial potential energy (U_i) is calculated when the moving charge is at position (0.150 m, 0).
  • The final potential energy (U_f) is calculated when the charge moves to (0.250 m, 0.250 m).
Understanding electric potential energy helps explain why certain movements of charges either require energy or release energy, much like lifting or dropping objects in a gravitational field.
Work Done by Electric Force
The work done by an electric force refers to the energy needed to move a charge in an electric field. In our exercise, it explains how much energy is involved in moving charge q_2 between two specific positions in the field created by q_1.

This concept is associated with changes in electric potential energy. The relationship is:\[ W = -\Delta U = U_i - U_f \] where W is the work done, \Delta U is the change in potential energy, and the negative sign indicates work done by the field.The work done is negative when the electric field does the work on the charge, often resulting in energy being released or potential energy decreasing. Conversely, a positive value would mean energy is added to move the charge.

Key points related to work done by electric force:
  • It is dependent on the movement of the charge in the field.
  • Understanding how work relates to potential energy changes helps in predicting charge movements in the field.
  • In our specific exercise about moving charge q_2, the calculated work done is -0.357 J, emphasizing the energy involved in this process.

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Most popular questions from this chapter

In a certain region of space, the electric potential is \(V(x, y, z)=A x y-B x^{2}+C y,\) where \(A, B,\) and \(C\) are positive constants. (a) Calculate the \(x-y\) - and \(z\) -components of the electric field. (b) At which points is the electric field equal to zero?

Alpha particles \(\left(\text { mass }=6.7 \times 10^{-27} \mathrm{kg}, \text { charge }=+2 e\right)\) are shot directly at a gold foil target. We can model the gold nucleus as a uniform sphere of charge and assume that the gold does not move. (a) If the radius of the gold nucleus is \(5.6 \times\) \(10^{-15} \mathrm{m}\) , what minimum speed do the alpha particles need when they are far away to reach the surface of the gold nucleus? (lanore relativistic effects.) (b) Give good physical reasons why we can ignore the effects of the orbital electrons when the alpha particle is (i) outside the electron orbits and (ii) inside the electron orbits.

Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by \(45.0 \mathrm{mm},\) and the potential difference between them is 360 \(\mathrm{V}\) . (a) What is the magnitude of the electric field (assumed to be uniform) in the region between the plates? (b) What is the magnitude of the force this field exerts on a particle with charge \(+2.40 \mathrm{nC} ?\) (c) Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower. (d) Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.

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 conducting 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.

The electric field at the surface of a charged, solid, copper 0sphere with radius 0.200 \(\mathrm{m}\) is 3800 \(\mathrm{N} / \mathrm{C}\) , directed toward the center of the sphere. What is the potential at the center of the sphere, if we take the potential to be zero infinitely far from the sphere?
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