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Chapter 23: Problem 11

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 zero net work is required to place the three charges at the corners of the triangle, what must the value of the third charge be?

Short Answer

Expert verified
The third charge \(Q\) must be \(-\frac{q}{2}\).

Step by step solution

01

Calculate Work for Like Charges

The work required to bring two identical charges \(q\) from infinitely far away to a distance \(d\) apart is given by the formula for electric potential energy between two charges: \[ W_{qq} = \frac{kq^2}{d} \]where \(k\) is Coulomb's constant.
02

Consider Potential Energy of Third Charge

The work done to bring the third charge \(Q\) to the corners with the other two charges \(q\) already placed is \[ 2 \times \frac{kqQ}{d} \]This takes into account the interaction with both of the \(q\) charges.
03

Set Up the Net Work Equation

The total work done should be zero according to the problem statement, so:\[ \frac{kq^2}{d} + 2 \times \frac{kqQ}{d} = 0 \]
04

Solve for the Unknown Charge

Rearrange the equation:\[ \frac{kq^2}{d} + \frac{2kqQ}{d} = 0 \]Simplify:\[ kq^2 + 2kqQ = 0 \]\[ q^2 + 2qQ = 0 \]Solve for \(Q\):\[ Q = -\frac{q}{2} \]

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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 entities with a definite amount of electric charge. They are considered as a single point in space where all the charge is concentrated. In physics problems, point charges are often idealized representations of charged objects.
  • Point charges can be positive or negative, with positive charges repelling each other and negative charges doing the same.
  • However, a positive charge attracts a negative charge, and vice versa.
In the exercise, we work with three point charges. Two have an equal charge of \(q\) and the third charge, denoted as \(Q\), needs to be found. These charges are placed at the corners of an equilateral triangle. Working with point charges allows us to simplify the complex nature of electrical interactions into precise mathematical equations. Understanding the characteristics of point charges is essential to solving problems involving electric potential energy.
Work and Energy in Physics
Work and energy are key concepts in physics that help describe how forces affect the motion of objects. Work is done when a force causes displacement. It is the product of force and the distance moved along the line of the force's direction.Energy can take many forms, potential energy being one of them. Electric potential energy is the energy stored due to the positions of charged particles. It depends on the configuration of the charges and the distances between them.In the context of the exercise, bringing two charges together requires work. This work is stored as electric potential energy. Calculating this energy for identical point charges \(q\) brought to a distance \(d\) apart plays a crucial part in solving the exercise:\[W_{qq} = \frac{kq^2}{d}\]Then, when another charge \(Q\) is introduced, additional work is required. However, the total work needed to place all charges into position cannot be more than zero, requiring careful consideration of how forces due to each charge act against each other.
Coulomb's Law
Coulomb's Law is a fundamental principle that describes the electric force between two point charges. It states that the magnitude of the electrostatic force attracted by two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between their centers.The formula is:\[F = \frac{k |q_1 q_2|}{r^2}\]
  • \(F\) is the magnitude of the force between the charges.
  • \(k\) is Coulomb's constant, approximately \(8.9875 \times 10^9 \, \text{N m}^2/\text{C}^2\).
  • \(q_1\) and \(q_2\) are the magnitudes of the charges.
  • \(r\) is the distance between the charges.
In our exercise, Coulomb's Law helps us derive the expression for the electric potential energy, which is a critical step in determining the configuration of the charges such that no net work is required. By analyzing the force balance and potential energy, we can determine that the third charge \(Q\) must be \(-\frac{q}{2}\) to balance the system and satisfy the condition of zero total work.

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Most popular questions from this chapter

A ring of diameter 8.00 \(\mathrm{cm}\) is fixed in place and carries a charge of \(+5.00 \mu \mathrm{C}\) uniformly spread over its circumference. (a) How much work does it take to move a tiny \(+3.00-\mu C\) charged ball of mass 1.50 \(\mathrm{g}\) from very far away to the center of the ring? (b) Is it necessary to take a path along the axis of the ring? Why? (c) If the ball is slightly displaced from the center of the ring, what will it do and what is the maximum speed it will reach?

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?

Nuclear Fusion in the Sun. The source of the sun's energy is a sequence of nuclear reactions that occur in its core. The first of these reactions involves the collision of two protons, which fuse together to form a heavier nucleus and release energy. For this process, called nuclear fusion, to occur, the two protons must first approach until their surfaces are essentially in contact. (a) Assume both protons are moving with the same speed and they collide head-on. If the radius of the proton is \(1.2 \times 10^{-15} \mathrm{m},\) what is the minimum speed that will allow fusion to occur? The charge distribution within a proton is spherically symmetric, so the electric field and potential outside a proton are the same as if it were a point charge. The mass of the proton is \(1.67 \times 10^{-27} \mathrm{kg} .\) (b) Another nuclear fusion reaction that occurs in the sun's core involves a collision between two helium nuclei, each of which has 2.99 times the mass of the proton, charge \(+2 e\) , and radius \(1.7 \times 10^{-15} \mathrm{m}\) . Assuming the same collision geometry as in part (a), what minimum speed is required for this fusion reaction to take place if the nuclei must approach a center-to-center distance of about \(3.5 \times 10^{-15} \mathrm{m}\) ? As for the proton, the charge of the helium nucleus is uniformly distributed throughout its volume. (c) In Section 18.3 it was shown that the average translational kinetic energy of a particle with mass \(m\) in a gas at absolute temperature \(T\) is \(\frac{3}{2} k T\) , where \(k\) is the Boltzmann constant (given in Appendix F). For two protons with kinetic energy equal to this avprage value to be able to undergo the process described in part (a), what absolute temperature is required? What absolute temperature is required for two average helium nuclei to be able to undergo the process described in part (b)? (At these temperatures, atoms are completely ionized, so nuclei and electrons move separately.) (d) The temperature in the sun's core is about 1.5 \(\times 10^{7}\) K. How does this compare to the temperatures calculated in part (c)? How can the reactions described in parts (a) and (b) occur at all in the interior of the sun? (Hint: See the discussion of the distribution of molecular speeds in Section \(18.5 . )\)

In a certain region of space the electric potential is given by \(V=+A x^{2} y-B x y^{2},\) where \(A=5.00 \mathrm{V} / \mathrm{m}^{3}\) and \(B=\) 8.00 \(\mathrm{V} / \mathrm{m}^{3} .\) Calculate the magnitude and direction of the electric field at the point in the region that has coordinates \(x=2.00 \mathrm{m}\) \(y=0.400 \mathrm{m},\) and \(z=0\).

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)?
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