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Chapter 23: Problem 8

Three equal \(1.20 \mu \mathrm{C}\) point charges are placed at the corners of an equilateral triangle with sides \(0.400 \mathrm{~m}\) long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)

Short Answer

Expert verified
The potential energy of the system is approximately \( 8.10 Joules.\)

Step by step solution

01

Understand Concept

Potential energy of a system with two point charges is given by \(U = k \cdot \frac{q_1 \cdot q_2 }{r}\), where \(k\) is Coulomb's constant, \(q_1\) and \(q_2\) are the charges and \(r\) is the distance between the charges.
02

Apply Formula

Apply the formula for each pair of charges. Here \(k = 8.99 \times 10^9 N m^2/C^2\), \(q_1 = q_2 = 1.2 \times 10^{-6} C\) and \(r = 0.4m \). Thus, the potential energy between two charges is \(U = 8.99 \times 10^9 \cdot \frac{1.2 \times 10^{-6} \cdot 1.2 \times 10^{-6}}{0.4} = 2.6985 Joules.\)
03

Combine Potentials of Pairs

Since the system contains 3 identical pairs of charges, simply triple the potential calculated in step 2. So, the total potential energy of the system is \(3 \cdot 2.6985 = 8.0955 Joules.\)

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Key Concepts

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

Point Charges
Point charges are fundamental to understanding electrostatics. Imagine each point charge as a tiny particle that holds an electrical charge. This charge can be either positive or negative. In most problems, these charges are considered to be fixed in position and are small enough that the distance between them can be treated as the defining factor of their interactions.

Point charges exert forces on each other. They can either attract or repel based on their charge. Like charges repel, while opposite charges attract. The force between any two point charges depends on their magnitudes and the distance separating them. In problems involving point charges, it is vital to pay attention to these attributes as they play crucial roles in calculating forces and potential energies.
Coulomb's Law
Coulomb's law is a fundamental principle in electrostatics. It describes how the force between two point charges is determined. The law can be expressed with the formula: \[ F = k \cdot \frac{|q_1 \cdot q_2|}{r^2} \]
In this formula:
  • \(F\) is the force between the charges.
  • \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \text{ N m}^2/\text{C}^2\).
  • \(q_1\) and \(q_2\) represent the quantities of the charges.
  • \(r\) is the distance between the centers of the two charges.
Coulomb's law not only quantifies the magnitude of the force but also the direction. If the product \(q_1 \cdot q_2\) is positive, the force is repulsive. Conversely, if the product is negative, the force is attractive. This law is crucial in determining how point charges will behave in any electrostatic setup.
Equilateral Triangle
In the context of geometry, an equilateral triangle is a triangle where all three sides are equal in length. This symmetry is beneficial in problems involving electric charges because it simplifies calculations.

Each angle in an equilateral triangle measures \(60^\circ\). In our specific exercise, the charges are placed at each corner of the triangle, necessitating the calculation of potential energy between each pair of charges. On occasions where geometric symmetry, like that in an equilateral triangle, exists, calculations become more straightforward, and a comprehensive understanding of spatial arrangements aids in dissecting and solving these physics problems easily.
Electrostatics
Electrostatics is the branch of physics that studies electric charges at rest. It involves analyzing static electric forces, electric fields, and potential energy. It鈥檚 vital to include potential energy as it dictates how charges interact and whether they are in stable configurations or not.

When multiple charges exist in a system, like in our exercise, their arrangement determines the potential energy. This is derived from pairing every two charges and computing their potential energy. The overall potential energy of the system is the summation of all individual pair potentials. Understanding electrostatic interactions is key to mastering concepts related to charge distributions and their energy implications.

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Most popular questions from this chapter

A heart cell can be modeled as a cylindrical shell that is \(100 \mu \mathrm{m}\) long, with an outer diameter of \(20.0 \mu \mathrm{m}\) and a cell wall thickness of \(1.00 \mu \mathrm{m}\), as shown in Fig. \(\mathrm{P} 23.81 .\) Potassium ions move across the cell wall, depositing positive charge on the outer surface and leaving a net negative charge on the inner surface. During the so-called resting phase, the inside of the cell has a potential that is \(90.0 \mathrm{mV}\) lower than the potential on its outer surface. (a) If the net charge of the cell is zero, what is the magnitude of the total charge on either cell wall membrane? Ignore edge effects and treat the cell as a very long cylinder. (b) What is the magnitude of the electric field just inside the cell wall? (c) In a subsequent depolarization event, sodium ions move through channels in the cell wall, so that the inner membrane becomes positively charged. At the cnd of this event, the inside of the cell has a potential that is \(20.0 \mathrm{mV}\) higher than the potential outside the cell. If we model this event by charge moving from the outer membrane to the inner membrane, what magnitude of charge moves across the cell wall during this event? (d) If this were done entirely by the motion of sodium ions, \(\mathrm{Na}^{+}\), how many ions have moved?

23.78 e DATA A small, stationary sphere carries a net charge \(Q .\) You perform the following experiment to measure Q: From a large distance you fire a small particle with mass \(m=4.00 \times 10^{-4} \mathrm{~kg}\) and charge \(q=5.00 \times 10^{-8} \mathrm{C}\) directly at the center of the sphere. The apparatus you are using measures the particle's speed \(v\) as a function of the distance \(x\) from the sphere. The sphere's mass is much greater than the mass of the projectile particle, so you assume that the sphere remains at rest. All of the measured values of \(x\) are much larger than the radius of cither object, so you treat both objects as point particles. You plot your data on a graph of \(v^{2}\) versus \((1 / x)\) (Fig. \(\mathbf{P} 23.78\) ). The straight line \(v^{2}=400 \mathrm{~m}^{2} / \mathrm{s}^{2}-\left[\left(15.75 \mathrm{~m}^{3} / \mathrm{s}^{2}\right) / x\right]\) gives a good fit to the data points. (a) Explain why the graph is a straight line. (b) What is the initial speed \(v_{0}\) of the particle when it is very far from the sphere? (c) What is \(Q ?\) (d) How close does the particle get to the sphere? Assume that this distance is much larger than the radii of the particle and sphere, so continue to treat them as point particles and to assume that the sphere remains at rest.

A metal sphere with radius \(r_{a}\) is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius \(\eta_{b}\). There is charge \(+q\) on the inner sphere and charge \(-q\) on the outer spherical shell. (a) Calculate the potential \(V(r)\) for (i) \(rn\). (Hint: The net potential is the sum of the potentials due to the individual spheres.) Take \(V\) to be zero when \(r\) is infinite. (b) Show that the potential of the inner sphere with respect to the outer is $$ V_{a b}=\frac{q}{4 \pi \epsilon_{0}}\left(\frac{1}{r_{a}}-\frac{1}{n_{b}}\right) $$ (c) Use Eq. (23.23) and the result from part (a) to show that the electric field at any point between the spheres has magnitude \(E(r)=\frac{V_{a b}}{\left(1 / r_{a}-1 / r_{b}\right)} \frac{1}{r^{2}}\) (d) Use Eq. (23.23) and the result from part (a) to find the electric field at a point outside the larger sphere at a distance \(r\) from the center, where \(r>r_{b} .\) (e) Suppose the charge on the outer sphere is not \(-q\) but a negative charge of different magnitude, say \(-Q .\) Show that the answers for parts (b) and (c) are the same as before but the answer for part (d) is different

Charge \(Q=5.00 \mu \mathrm{C}\) is distributed uniformly over the volume of an insulating sphere that has radius \(R=12.0 \mathrm{~cm} .\) A small sphere with charge \(q=+3.00 \mu \mathrm{C}\) and mass \(6.00 \times 10^{-5} \mathrm{~kg}\) is projected toward the center of the large sphere from an initial large distance. The large sphere is held at a fixed position and the small sphere can be treated as a point charge. What minimum speed must the small sphere have in order to come within \(8.00 \mathrm{~cm}\) of the surface of the large sphere?

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 \(e .\)
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