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Chapter 23: Problem 1

A point charge \(q_{1}=+2.40 \mu \mathrm{C}\) is held stationary at the origin. A second point charge \(q_{2}=-4.30 \mu \mathrm{C}\) 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 charge \(q_2\) is \(0.99 x 10^-4 J\).

Step by step solution

01

Identify Starting and Ending Points

The initial point is \(i: (0.150m,0)\) and the final point is \(f: (0.250m, 0.250m)\). Calculate the distance between the origin and these points using Pythagoras' theorem. The distance of the initial point \(r_i\) is simply \(0.150m\) as it lies on the x-axis. For the final point, the distance \(r_f = \sqrt{{(0.250)^2+(0.250)^2}} = 0.354m\).
02

Calculate Potential Energy at Starting and Ending Points

We can use the principle of superposition to calculate the electric potential energy at initial and final points separately. The general formula for electric potential energy is \(U = \frac{{kqQ}}{r}\), where \(k = 8.99 x 10^9 Nm^2/C^2\) is Coulomb鈥檚 constant. At the initial point, the potential energy \(U_i = \frac{{kq_1q_2}}{r_i} = 2.15 x 10^-4 J\). At the final point, the potential energy \(U_f = \frac{{kq_1q_2}}{r_f} = 1.16 x 10^-4 J\).
03

Calculate work done by the Electric Force

The work done by a conservative force \(W = - \Delta U = U_i - U_f = 2.15 x 10^-4 J - 1.16 x 10^-4 J = 0.99 x 10^-4 J\)

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Key Concepts

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

Work Done by Electric Force
The concept of 'work done by electric force' is a fundamental aspect of electromagnetism that describes the energy transferred by the force from one point to another. In simple terms, when a charged particle moves in the presence of an electric field, work is done by the field as the particle experiences a force and moves a certain distance. The amount of work done is calculated based on the change in electric potential energy of the particle, since electric forces are conservative.

In mathematical terms, the work done by the electric force on a charge is the negative of the change in electric potential energy (\( W = - \triangle U \) ), expressed in joules (J). This equation tells us that if the electric potential energy decreases, the work done by the electric force is positive, indicating that energy is given to the particle. Conversely, if energy is taken from the particle, the work will be negative. This principle is key to solving problems where charges move in electric fields, such as the textbook exercise provided.
Coulomb's Law
Understanding Coulomb's law is crucial when we discuss charges and electric force. This fundamental law, discovered by Charles-Augustin de Coulomb, tells us how the magnitude of the electric force between two stationary, point charges changes with distance.

The equation is elegantly simple: \( F = k \frac{ |q_1q_2| }{r^2} \) , where:
  • \( F \) is the magnitude of the electric force,
  • \( k \) is the Coulomb's constant (\( 8.99 \times 10^9 N \times m^2/C^2 \)),
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
  • \( r \) is the distance between the charges.

The equation shows the force is directly proportional to the product of the charge magnitudes and inversely proportional to the square of the distance between them. This inverse-square law behavior is a common thread linking other fundamental forces in physics, such as gravity.
Point Charge
A 'point charge' is an idealized model used in electrostatics to represent a charge with no spatial extent; it is considered to be concentrated at a single point in space. This simplification allows for the application of Coulomb's law and calculation of potential energies in a straightforward manner.

In practice, no real charge is strictly a point charge, since all particles have some physical size. However, for charges that are very small or very far apart compared to their sizes, modeling them as point charges is an accurate approximation. It鈥檚 this model that enables us to perform calculations for the exercise in question, where we consider the point charges and their movement within an electric field.
Electric Force
The 'electric force' is one of the most essential concepts in electromagnetism. This force arises from the electric field generated by a charge and is felt by another charge within that field. The direction of the force is dependent on the nature of the charges: like charges repel, while opposite charges attract each other.

The magnitude of the electric force can be computed using Coulomb's law, as mentioned earlier. A key property of electric force is its conservativeness. This means that the work done by an electric force on a charge is independent of the path taken by the charge and depends only on its initial and final positions. This is why we can calculate a definitive amount of work done when a charge moves from one point to another in an electric field, as seen in the textbook exercise provided.

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Most popular questions from this chapter

A total clectric charge of \(3.50 \mathrm{nC}\) is distributed uniformly over the surface of a metal sphere with a radius of \(24.0 \mathrm{~cm}\). If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) \(48.0 \mathrm{~cm}\) (b) \(24.0 \mathrm{~cm}\) (c) \(12.0 \mathrm{~cm}\)

A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the accumulation of charge near the cathode, the electric potential between the electrodes is given by \(V(x)=C x^{4 / 3}\) where \(x\) is the distance from the cathode and \(C\) is a constant, characteristic of a particular diode and operating conditions. Assume that the distance between the cathode and anode is \(13.0 \mathrm{~mm}\) and the potential difference between electrodes is \(240 \mathrm{~V} .\) (a) Determine the value of \(C\). (b) Obtain a formula for the electric field between the electrodes as a function of \(x\). (c) Determine the force on an electron when the electron is halfway between the electrodes.

CP Deflecting Plates of an Oscilloscope. The vertical deflecting plates of a typical classroom oscilloscope are a pair of parallel square metal plates carrying cqual but opposite charges. Typical dimensions are about \(3.0 \mathrm{~cm}\) on a side, with a separation of about \(5.0 \mathrm{~mm} .\) The potential difference between the plates is \(25.0 \mathrm{~V}\). The plates are close enough that we can ignore fringing at the cnds. Under these conditions: (a) how much charge is on each plate, and (b) how strong is the electric field between the plates? (c) If an electron is ejected at rest from the negative plate, how fast is it moving when it reaches the positive plate?

The Millikan Oil-Drop Experiment. The charge of an electron was first measured by the American physicist Robert Millikan during \(1909-1913 .\) In his experiment, oil was sprayed in very fine drops (about \(10^{-4} \mathrm{~mm}\) in diameter) into the space between two parallel horizontal plates separated by a distance \(d\). A potential difference \(V_{A B}\) was maintained between the plates, causing a downward electric field between them. Some of the oil drops acquired a negative charge because of frictional effects or because of ionization of the surrounding air by x rays or radioactivity. The drops were observed through a microscope. (a) Show that an oil drop of radius \(r\) at rest between the plates remained at rest if the magnitude of its charge was \(q=\frac{4 \pi}{3} \frac{\rho r^{3} g d}{V_{A B}}\) where \(\rho\) is oil's density. (Ignore the buoyant force of the air.) By adjusting \(V_{A B}\) to keep a given drop at rest, Millikan determined the charge on that drop. provided its radius \(r\) was known. (b) Millikan's oil drops were much too small to measure their radii directly. Instead, Millikan determincd \(r\) by cutting off the electric field and measuring the terminal speed \(u_{\mathrm{i}}\) of the drop as it fell. (We discussed terminal speed in Section \(\left.5.3 .\right)\) The viscous force \(F\) on a sphere of radius \(r\) moving at speed \(v\) through a fluid with viscosity \(\eta\) is given by Stokes's law: \(F=6 \pi \eta r v .\) When a drop fell at \(v_{1}\), the viscous force just balanced the drop's weight \(w=m g\). Show that the magnitude of the charge on the drop was \(q=18 \pi \frac{d}{V_{A B}} \sqrt{\frac{\eta^{3} v_{1}^{3}}{2 \rho g}}\) (c) You repeat the Millikan oil-drop experiment. Four of your measured values of \(V_{A B}\) and \(v_{1}\) are listed in the table: $$ \begin{array}{lcccc} \text { Drop } & 1 & 2 & 3 & 4 \\ \hline V_{A B}(\mathrm{~V}) & 9.16 & 4.57 & 12.32 & 6.28 \\ v_{1}\left(10^{-5} \mathrm{~m} / \mathrm{s}\right) & 2.54 & 0.767 & 4.39 & 1.52 \end{array} $$ In your apparatus, the separation \(d\) between the horizontal plates is \(1.00 \mathrm{~mm}\). The density of the oil you use is \(824 \mathrm{~kg} / \mathrm{m}^{3}\). For the viscosity \(\eta\) of air, use the value \(1.81 \times 10^{-5} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\). Assume that \(g=9.80 \mathrm{~m} / \mathrm{s}^{2}\). Calculate the charge \(q\) of each drop. (d) If electric charge is quantized (that is, exists in multiples of the magnitude of the charge of an electron), then the charge on each drop is -ne, where \(n\) is the number of excess electrons on each drop. (All four drops in your table have negative charge.) Drop 2 has the smallest magnitude of charge observed in the experiment, for all 300 drops on which measurements were made, so assume that its charge is due to an excess charge of one electron. Determine the number of excess electrons \(n\) for each of the other three drops. (e) Use \(q=-n e\) to calculate \(e\) from the data for each of the four drops, and average these four values to get your best experimental value of \(e .\)

A very large plastic sheet carries a uniform charge density of \(-6.00 \mathrm{nC} / \mathrm{m}^{2}\) on one face. (a) As you move away from the sheet along a line perpendicular to it, does the potential increase or decrease? How do you know, without doing any calculations? Does your answer depend on where you choose the reference point for potential? (b) Find the spacing between equipotential surfaces that differ from each other by \(1.00 \mathrm{~V}\). What type of surfaces are these?
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