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Chapter 23: Problem 15

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) 0.000748 J; (c) -0.00206 J.

Step by step solution

01

Understanding the Relationship

The work done by an electric force on a charge in a uniform electric field is given by the formula: \( W = qEd \cos \theta \), where \( q \) is the charge, \( E \) is the electric field strength, \( d \) is the displacement, and \( \theta \) is the angle between the field and the displacement.
02

Calculate Work for Part (a)

In part (a), the charge moves 0.450 m to the right. Since the electric field is vertical, the angle \( \theta \) is 90 degrees. The cosine of 90 degrees is 0, hence the work done \( W = qEd \cos 90^\circ = 0 \).
03

Calculate Work for Part (b)

For part (b), the charge moves 0.670 m upward. Since the movement is in the direction of the field, \( \theta = 0 \) degrees. Thus, \( \cos 0 = 1 \). Therefore, the work done is \( W = (28.0 \times 10^{-9} \mathrm{C})(4.00 \times 10^4 \mathrm{V/m})(0.670 \mathrm{m}) \cdot 1 = 0.000748 \mathrm{J} \).
04

Calculate Work for Part (c)

In part (c), the charge moves at 45 degrees downward from the horizontal; therefore, the angle with the vertical upward field is \( 135^\circ \). Calculate \( \cos 135^\circ = -\frac{\sqrt{2}}{2} \). Plugging in the values, \( W = (28.0 \times 10^{-9} \mathrm{C})(4.00 \times 10^4 \mathrm{V/m})(2.60 \mathrm{m})(-\frac{\sqrt{2}}{2}) \approx -2.06 \times 10^{-3} \mathrm{J} \).

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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 charge placed within it. It arises due to the interaction between the electric field and the electric charge. The formula for electric force is given by \( F = qE \), where
  • \( F \) stands for the electric force,
  • \( q \) denotes the electric charge,
  • \( E \) represents the magnitude of the electric field.
Electric forces can either attract or repel charges, depending on whether the charges are like or unlike. This force is a vector, meaning it has both magnitude and direction. Understanding the direction is crucial because it determines how a charge will move when placed in an electric field.
Work Done by Electric Force
The work done by electric force is the amount of energy transferred by this force as a charge moves through an electric field. It is expressed in joules (J) and calculated using the equation \( W = qEd \cos \theta \). This equation involves several components:
  • \( q \) is the charge in coulombs,
  • \( E \) is the electric field strength in volts per meter (V/m),
  • \( d \) is the displacement in meters,
  • \( \theta \) is the angle between the electric field direction and the path of displacement.
The term \( \cos \theta \) helps identify the angle of the displacement with respect to the field. When the movement is parallel to the field, \( \theta = 0 \) and the work is maximal. When the displacement is perpendicular, \( \theta = 90^\circ \), the cos value is zero, meaning no work is done.
Uniform Electric Field
A uniform electric field has a constant strength and direction at every point within that field. This is often visualized as equally spaced, parallel lines. The uniformity means
  • every charge experiences the same amount of electric force, regardless of its position within the field,
  • the calculations remain consistent across the field.
In a physics problem, like the original exercise, a uniform electric field simplifies understanding the behavior of moving charges. The vector nature of the field directly influences how work and force are computed, particularly since directionality becomes a less complex issue.
Displacement in Electric Field
Displacement in an electric field refers to the movement of a charge from one point to another within the field. This concept is crucial in calculating the work done by an electric force. Displacement is a vector quantity, having both magnitude and direction.When considering displacement:
  • The path taken by the charge affects the work done.
  • The angle \( \theta \) between the displacement direction and electric field direction is key for calculating work.
  • The displacement magnitude affects the total energy expended or gained by the charge.
Remember, only the component of displacement in the direction of the electric field contributes to the work. It's essential to analyze each path a charge can take separately, as seen in rightward, upward, or angled movements鈥攁nalyzing their unique angles with the field adjusts the calculated work accordingly.

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Most popular questions from this chapter

In a certain region, a charge distribution exists that is spherically symmetric but nonuniform. That is, the volume charge density \(\rho(r)\) depends on the distance \(r\) from the center of the distribution but not on the spherical polar angles \(\theta\) and \(\phi .\) The electric potential \(V(r)\) due to this charge distribution is $$V(r)=\left\\{\begin{array}{l}{\frac{\rho_{0} a^{2}}{18 \epsilon_{0}}\left[1-3\left(\frac{r}{a}\right)^{2}+2\left(\frac{r}{a}\right)^{3}\right] \text { for } r \leq a} \\ {0} & {\text { for } r \geq a}\end{array}\right.$$ where \(\rho_{0}\) is a constant having units of \(\mathrm{C} / \mathrm{m}^{3}\) and \(a\) is a constant having units of meters. (a) Derive expressions for \(\vec{E}\) for the regions \(r \leq a\) and \(r \geq a .[\)Hint\(:\) Use Eq. \((23.23) .]\) Explain why \(\vec{\boldsymbol{E}}\) has only a radial component. (b) Derive an expression for \(\rho(r)\) in each of the two regions \(r \leq a\) and \(r \geq a .[\)Hint t: Use Gauss's law for two spherical shells, one of radius \(r\) and the other of radius \(r+d r .\) The charge contained in the infinitesimal spherical shell of radius \(d r\) is \(d q=4 \pi r^{2} \rho(r) d r . ](c)\) Show that the net charge contained in the volume of a sphere of radius greater than or equal to \(a\) is zero. [Hint: Integrate the expressions derived in part (b) for \(\rho(r)\) over a spherical volume of radius greater than or equal to \(a.\)) Is this result consistent with the electric field for \(r>a\) that you calculated in part (a)?

Two charges of equal magnitude \(Q\) are held a distance \(d\) apart. Consider only points on the line passing through both charges. (a) If the two charges have the same sign, find the location of all points (if there are any) at which (i) the potential (relative to infinity is zero (is the electric field zero at these points? (ii) the electric field is zero (is the potential zero at these points?). (b) Repeat part (a) for two charges having opposite signs.

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?

The electric field at the surface of a charged, solid, copper sphere 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?

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.)
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