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Chapter 23: Problem 17

A charge of 28.0 \(\mathrm{nC}\) is placed in a uniform electric field that is directed vertically upward and has a magnitude of \(4.00 \times 10^{4} \mathrm{V} / \mathrm{m}\) . What work is done by the electric force when the charge moves (a) 0.450 \(\mathrm{m}\) to the right; (b) 0.670 \(\mathrm{m}\) upward; (c) 2.60 \(\mathrm{m}\) at an angle of \(45.0^{\circ}\) downward from the horizontal?

Short Answer

Expert verified
(a) 0 J, (b) 0.00075 J, (c) 0.00026 J.

Step by step solution

01

Understanding Work Done by Electric Force

Work done by an electric force on a charge in an electric field is calculated using the formula \[ W = qEd \cos \theta \]where \( W \) is the work done,\( q \) is the charge,\( E \) is the electric field magnitude,\( d \) is the distance moved by the charge,and \( \theta \) is the angle between the direction of the electric field and the direction of movement.
02

Convert Charge Units

The charge given is 28.0 nC (nanocoulombs). Convert it to coulombs by multiplying by \( 10^{-9} \). Thus,\[ q = 28.0 \times 10^{-9} \text{ C} \]
03

Calculate Work Done in Rightward Movement (a)

For movement to the right, \( \theta = 90^{\circ} \) because the electric field is vertical. Thus, \[ \cos 90^{\circ} = 0 \].Substitute in the formula\[ W = (28.0 \times 10^{-9} \text{ C})(4.00 \times 10^4 \text{ V/m})(0.450 \text{ m}) \cdot 0 = 0 \text{ J} \].No work is done since the charge moves perpendicularly to the field.
04

Calculate Work Done in Upward Movement (b)

When the charge moves upward, it is moving in the same direction as the electric field. Here, \( \theta = 0^{\circ} \) and \( \cos 0^{\circ} = 1 \).Use the formula:\[ W = (28.0 \times 10^{-9} \text{ C})(4.00 \times 10^4 \text{ V/m})(0.670 \text{ m}) \cdot 1 \]\[ W = 0.00075 \text{ J} \].

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Key Concepts

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

Electric Force
Electric force is the force exerted by an electric field on a charged particle. It's similar to how gravitational force acts on a mass. This force can be calculated using Coulomb's law, which states that the force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.
Electric force is always directed along the line connecting the two charges. When we place a charge in an electric field, like the 28.0 nC charge mentioned, the field exerts a force on it. The strength and direction of this force are determined by both the magnitude of the charge and the electric field.
  • The direction of the electric force depends on the nature of the charge: positive charges experience force in the direction of the field, while negative charges experience force in the opposite direction.
  • The magnitude of this force can be thought of as the product of the electric field and the charge's magnitude (F = qE).
Work Done
Work is done when a force moves an object over a distance. In the context of electric fields, work done refers to the energy transferred by the electric force when a charge is moved within the field.
This can be calculated with the formula:\[ W = qEd \cos \theta \]where \( W \) is the work done, \( q \) is the charge, \( E \) is the electric field magnitude, \( d \) is the distance moved by the charge, and \( \theta \) is the angle between the direction of the electric field and the movement of the charge.It's essential to understand that:
  • If the charge moves in the same direction as the electric field, maximum work is performed because \( \theta = 0^{\circ} \) and \( \cos 0^{\circ} = 1 \). This means the entire force is used to move the charge.
  • If the charge moves perpendicular to the electric field, no work is done because \( \theta = 90^{\circ} \) and \( \cos 90^{\circ} = 0 \). Here, the force doesn't contribute to the movement in the direction of the field.
Charge Movement
The movement of a charge through an electric field depends on the direction and magnitude of the forces acting upon it. Using the given question, let's break down how different movements affect the work done:
  • **Rightward movement:** Here, the charge moves perpendicular to the field (\( \theta = 90^{\circ} \)). Thus, no work is done as the movement does not align with the electric field direction.
  • **Upward movement:** This is in the same direction as the electric field, making \( \theta = 0^{\circ} \). The force fully contributes to moving the charge, resulting in maximum work done.
  • **Movement downward at an angle:** This scenario involves an angle of \( 45^{\circ} \) down from the horizontal. Calculating work involves determining the effective component of the force in the electric field's direction using \( \cos 45^{\circ} = \frac{1}{\sqrt{2}} \).

In each scenario, the angle \( \theta \) plays a crucial role in determining how much of the energy the electric force transfers into moving the charge along the path.
Uniform Electric Field
A uniform electric field has the same strength and direction at every point within it. This is an ideal model often used for simplifying calculations in physics and is akin to having a constant and evenly distributed force across an area.
One way to create a uniform electric field is between two parallel plates with a constant voltage applied.In this uniform field:
  • All charges experience the same magnitude of force regardless of their position within the field.
  • Movement of charges directly along or directly opposite the field lines results in clear calculations using the formula \( W = qEd \cos \theta \).
  • Understanding the uniformity of the field helps predict the behavior of charges and allows for easy computation of work done, as seen in the original exercise.
This predictable environment is usually depicted with parallel equally spaced lines, indicating that the electric field strength does not change with position.

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Most popular questions from this chapter

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?

A very long cylinder of radius 2.00 \(\mathrm{cm}\) carries a uniform charge density of 1.50 \(\mathrm{nC} / \mathrm{m}\) . (a) Describe the shape of the equipotential surfaces for this cylinder. (b) Taking the reference level for the zero of potential to be the surface of the cylinder, find the radius of equipotential surfaces having potentials of \(10.0 \mathrm{V}, 20.0 \mathrm{V},\) and 30.0 \(\mathrm{V}\) . (c) Are the equipotential surfaces equally spaced? If not, do they get closer together or farther apart as \(r\) increases?

The \(\mathbf{H}_{2}^{+}\) Ion. The \(\mathbf{H}_{2}^{+}\) ion is composed of two protons, each of charge \(+e=1.60 \times 10^{-19} \mathrm{C},\) and an electron of charge \(-e\) and mass \(9.11 \times 10^{-31} \mathrm{kg} .\) The separation between the pro- tons is \(1.07 \times 10^{-10} \mathrm{m}\) . The protons and the electron may be treated as point charges. (a) Suppose the electron is located at the point midway between the two protons. What is the potential energy of the interaction between the electron and the two protons? (Do not include the potential energy due to the interaction between the two protons.) (b) Suppose the electron in part (a) has a velocity of magnitude \(1.50 \times 10^{6} \mathrm{m} / \mathrm{s}\) in a direction along the perpendicular bisector of the line connecting the two protons. How far from the point midway between the two protons can the electron move? Because the masses of the protons are much greater than the clectron mass, the motions of the protons are very slow and can be ignored. (Note: A realistic description of the electron motion requires the use of quantum mechanics, not Newtonian mechanics.)

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 \(q_{2} ?\)

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.
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