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Chapter 23: Problem 53

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) -2.15脳10^{-5} J, (b) -2830.26 V, (c) 35378.25 V/m.

Step by step solution

01

Calculate the work done by the electric force

The total work done on the particle is the sum of the work done by the electric force and the work done by the additional force. Since the particle has gained kinetic energy, this energy is equal to the total work done on it. Thus, we have this equation:\[ W_{total} = W_{electric} + W_{additional} \]where \( W_{additional} = 6.50 \times 10^{-5} \, \mathrm{J} \) and \( W_{total} = \text{Kinetic Energy} = 4.35 \times 10^{-5} \, \mathrm{J} \). Substitute these values into the equation:\[ 4.35 \times 10^{-5} = W_{electric} + 6.50 \times 10^{-5} \]Solving for \( W_{electric} \) gives:\[ W_{electric} = 4.35 \times 10^{-5} - 6.50 \times 10^{-5} = -2.15 \times 10^{-5} \, \mathrm{J} \].
02

Determine the electric potential difference

The work done by the electric force is related to the potential difference \( \Delta V \) between the starting and ending points by the formula:\[ W_{electric} = q \Delta V \]Rearranging gives:\[ \Delta V = \frac{W_{electric}}{q} \]Substitute \( W_{electric} = -2.15 \times 10^{-5} \, \mathrm{J} \) and \( q = 7.60 \times 10^{-9} \, \mathrm{C} \) into the equation:\[ \Delta V = \frac{-2.15 \times 10^{-5}}{7.60 \times 10^{-9}} = -2830.26 \, \mathrm{V} \].Thus, the starting point is at a potential 2830.26 V lower than the end point.
03

Calculate the magnitude of the electric field

The electric potential difference \( \Delta V \) is related to the electric field \( E \) and the distance \( d \) the particle traveled by the formula:\[ \Delta V = -E \times d \]Solving for the magnitude of the electric field gives:\[ E = -\frac{\Delta V}{d} \]Substitute \( \Delta V = -2830.26 \, \mathrm{V} \) and \( d = 0.08 \, \mathrm{m} \) (8.00 cm converted to meters) into the equation:\[ E = -\frac{-2830.26}{0.08} = 35378.25 \, \mathrm{V/m} \]. Thus, the magnitude of the electric field is 35378.25 V/m.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. When a force acts on a particle causing it to move, its kinetic energy changes. In our problem, when the particle moves to the right, it gains kinetic energy due to the work done on it. The total work done results in a conversion of energy, often manifesting as kinetic energy.

Here鈥檚 how it works in simpler terms:
  • Work done on the particle increases its kinetic energy.
  • This energy transformation obeys the principle of conservation of energy.
  • The formula for kinetic energy involves both the mass and the velocity of the object, expressed as \( KE = \frac{1}{2}mv^2 \), but in our case, we focus on the work-energy principle instead.
The given exercise indicates that the kinetic energy gained is \(4.35 \times 10^{-5} \mathrm{J}\). This energy reflects how much work was effectively converted into the particle's motion.
Electric Potential Difference
Electric potential difference (also known as voltage) is a measure of the work needed to move a charge from one location to another within an electric field. It indicates how much potential energy will change as the charge moves between these points.

The formula used is:
  • \(\Delta V = \frac{W_{electric}}{q}\)
This formula shows the link between the work done by the electric force and the charge itself. In the exercise, the electric force does \(-2.15 \times 10^{-5} \mathrm{J}\) of work as the particle moves, leading to a potential difference of \( -2830.26 \mathrm{V} \). This suggests that energy has been expended to counteract the electric field鈥檚 influence, moving the particle to a region of lower potential power.
Uniform Electric Field
A uniform electric field is one where the electric force is constant in magnitude and direction at every point. This means that a charge within this field will experience a consistent force no matter its position.

In a uniform electric field, the potential difference \(\Delta V\) between two points is directly related to the electric field strength \(E\) and the distance \(d\) moved by the particle:
  • \(\Delta V = -E \times d\)
This equation helps us understand how the potential difference varies across distances. From the problem, we calculated the electric field's magnitude to be \(35378.25 \mathrm{V/m}\), using the particle movement distance of 0.08 meters. The particle experiences a uniform force while traversing this electric field, making calculations straightforward whenever the setup adheres to the uniform field condition.

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Most popular questions from this chapter

Two positive point charges, each of magnitude \(q,\) are fixed on the \(y\) -axis at the points \(y=+a\) and \(y=-a\) . Take the potential to be zero at an infinite distance from the charges. (a) Show the positions of the charges in a diagram. (b) What is the potenntial \(V_{0}\) at the origin? (c) Show that the potential at any point on the \(x\) -axis is $$ V=\frac{1}{4 \pi \epsilon_{0}} \frac{2 q}{\sqrt{a^{2}+x^{2}}} $$ (d) Graph the potential on the \(x\) -axis as a function of \(x\) over the range from \(x=-4 a\) to \(x=+4 a\) . (e) What is the potential when \(x \gg a ?\) Explain why this result is obtained.

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 metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere \(A\) has a radius three times that of sphere \(B\) . Let \(Q_{A}\) and \(Q_{B}\) be the charges on the two spheres, and let \(E_{A}\) and \(E_{B}\) be the electric-field magnitudes at the surfaces of the two spheres. What are (a) the ratio \(Q_{B} / Q_{A}\) and (b) the ratio \(E_{B} / E_{A} ?\)

Two plastic spheres, each carrying charge uniformly distributed throughout its interior, are initially placed in contact and then released. One sphere is 60.0 \(\mathrm{cm}\) in diameter, has mass 50.0 \(\mathrm{g}\) and contains \(-10.0 \mu \mathrm{C}\) of charge. The other sphere is 40.0 \(\mathrm{cm}\) in diameter, has mass 150.0 \(\mathrm{g}\) , and contains \(-30.0 \mu \mathrm{C}\) of charge. Find the maximum acceleration and the maximum speed achieved by each sphere (relative to the fixed point of their initial location in space), assuming that no other forces are acting on them. (Hint: The uniformly distributed charges behave as though they were concentrated at the centers of the two spheres.)

Three point charges, which initially are infinitely far apart, are placed at the corners of an equilateral triangle with sides \(d\) . Two of the point charges are identical and have charge \(q\) . If zerv net work is required to place the three charges at the comers of the triangle, what must the value of the third charge be?
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