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Chapter 23: Problem 55

A particle with charge \(+7.60 \mathrm{nC}\) is in a uniform electric field directed to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest, it moves to the right. After it has moved \(8.00 \mathrm{cm},\) the additional force has done \(6.50 \times 10^{-5} \mathrm{J}\) of work and the particle has \(4.35 \times 10^{-5} \mathrm{J}\) of kinetic energy. (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the end point? (c) What is the magnitude of the electric field?

Short Answer

Expert verified
(a) Work by electric force: \(-2.15 \times 10^{-5} \mathrm{J}\). (b) Potential difference: \(2.83 \times 10^{3} \mathrm{V}\). (c) Electric field: \(3.54 \times 10^{3} \mathrm{N/C}\).

Step by step solution

01

Determine Work Done by Electric Force

To find the work done by the electric force, we use the work-energy principle. The total work done on the particle is the sum of the work done by the electric force and the additional force, which equals the change in kinetic energy.Given:- Work done by additional force, \(W_a = 6.50 \times 10^{-5} \mathrm{J}\)- Change in kinetic energy, \(\Delta KE = 4.35 \times 10^{-5} \mathrm{J}\)Since the sum of the works done by the electric force and the additional force equals the change in kinetic energy, we have:\[ W_e + W_a = \Delta KE \]Rearranging gives:\[ W_e = \Delta KE - W_a = 4.35 \times 10^{-5} - 6.50 \times 10^{-5} = -2.15 \times 10^{-5} \mathrm{J} \]Thus, the work done by the electric force is \(-2.15 \times 10^{-5} \mathrm{J}\).
02

Calculate Potential Difference

To find the potential difference between the starting and ending points, use the definition of work done by an electric field and potential difference:\[ W_e = -q \Delta V \]Where:- \(W_e = -2.15 \times 10^{-5} \mathrm{J}\) (work done by the electric field)- \(q = 7.60 \times 10^{-9} \mathrm{C}\) (charge of the particle)Rearranging gives:\[ \Delta V = -\frac{W_e}{q} = -\frac{-2.15 \times 10^{-5} \mathrm{J}}{7.60 \times 10^{-9} \mathrm{C}} \approx 2.83 \times 10^{3} \mathrm{V} \]Thus, the potential difference is \(2.83 \times 10^{3} \mathrm{V}\), meaning the potential at the starting point is higher by this value compared to the endpoint.
03

Determine Electric Field Magnitude

To find the magnitude of the electric field, we relate work done by an electric force, electric field \(E\), charge \(q\), and displacement \(d\):\[ W_e = - qEd \]Given:- Displacement, \(d = 8.00 \mathrm{cm} = 0.08 \mathrm{m}\)- \(W_e = -2.15 \times 10^{-5} \mathrm{J}\) (work done by the electric field)- \(q = 7.60 \times 10^{-9} \mathrm{C}\) (charge)Rearranging for \(E\) gives:\[ E = -\frac{W_e}{qd} = -\frac{-2.15 \times 10^{-5} \mathrm{J}}{7.60 \times 10^{-9} \mathrm{C} \times 0.08 \mathrm{m}} \approx 3.54 \times 10^{3} \mathrm{N/C} \]So, the magnitude of the electric field is \(3.54 \times 10^{3} \mathrm{N/C}\).

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Key Concepts

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

Work-Energy Principle
The work-energy principle is a fundamental concept in physics that describes the relationship between work done on an object and its kinetic energy change. It states that the work done by all forces acting on an object equals the change in its kinetic energy.
In this problem, the charged particle moves under the influence of both an electric force and an additional force. The total work is the sum of these forces' contributions, changing the particle's kinetic energy. If you know the work done by one of the forces and the change in kinetic energy, you can find the work done by the other force.
For our particle, the work-energy formula is: \[W_e + W_a = \Delta KE\]where:
  • \(W_e\) is the work done by the electric force
  • \(W_a\) is the work done by the additional force
  • \(\Delta KE\) is the change in kinetic energy
Rearranging the formula helps us find the electric force's work: \[W_e = \Delta KE - W_a\]This relationship shows how knowing the forces or kinetic changes allows determination of unknown aspects, central to solving mechanics problems.
Potential Difference
Potential difference (voltage) is crucial when dealing with electric fields and is the work done per unit charge to move a charge between two points. In simple terms, it's the difference in electric potential energy per charge.
The formula for potential difference given the work done by an electric field is: \[\Delta V = -\frac{W_e}{q}\]where:
  • \(W_e\) is the work done by the electric force
  • \(q\) is the charge
In our case, knowing the work done by the electric field and the particle's charge allows us to calculate the potential difference between the start and end points.
This result can be interpreted as the electric potential being greater where the particle starts, relative to where it stops, by the calculated potential difference. Understanding potential difference helps to grasp how electric fields store and transfer energy.
Electric Field Magnitude
The electric field magnitude is a measure of the force per charge exerted by the electric field. It is a crucial quantity when analyzing electric forces in physics.
To find the electric field magnitude, utilize the relationship between work, electric field, charge, and displacement:\[W_e = - qEd\]Given that:
  • \(W_e\) is the work done by the electric field
  • \(q\) is the charge
  • \(d\) is the displacement
Rearranging for the electric field magnitude \(E\) yields:\[E = -\frac{W_e}{qd}\]Plugging in known values allows the calculation of the electric field strength in Newtons per Coulomb (N/C). This measurement is important as it describes how strong the electric field is and how it affects charged particles.
With this understanding, one can predict how charged particles will behave in any field and how much force they will experience over a distance.

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Most popular questions from this chapter

In a certain region of space the electric potential is given by \(V=+A x^{2} y-B x y^{2},\) where \(A=5.00 \mathrm{V} / \mathrm{m}^{3}\) and \(B=\) 8.00 \(\mathrm{V} / \mathrm{m}^{3} .\) Calculate the magnitude and direction of the electric field at the point in the region that has coordinates \(x=2.00 \mathrm{m}\) \(y=0.400 \mathrm{m},\) and \(z=0\).

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 \(q_{1}\) ?

A very long insulating cylindrical shell of radius 6.00 \(\mathrm{cm}\) carries charge of linear density 8.50\(\mu \mathrm{C} / \mathrm{m}\) spread uniformly over its outer surface. What would a voltmeter read if it were connected between (a) the surface of the cylinder and a point 4.00 \(\mathrm{cm}\) above the surface, and (b) the surface and a point 1.00 \(\mathrm{cm}\) from the central axis of the cylinder?

Charge \(Q=+4.00 \mu \mathrm{C}\) is distributed uniformly over the volume of an insulating sphere that has radius \(R=5.00 \mathrm{cm} .\) What is the potential difference between the center of the sphere and the surface of the sphere?

A small sphere with mass \(5.00 \times 10^{-7} \mathrm{kg}\) and charge \(+3.00 \mu C\) is released from rest a distance of 0.400 \(\mathrm{m}\) above a large horizontal insulating sheet of charge that has uniform surface charge density \(\sigma=+8.00 \mathrm{pC} / \mathrm{m}^{2} .\) Using energy methods, calculate the speed of the sphere when it is 0.100 \(\mathrm{m}\) above the sheet of charge?
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