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Chapter 23: Problem 19

Two point charges \(q_{1}=+2.40 \mathrm{nC} \quad\) and \(\quad q_{2}=\) \(-6.50 \mathrm{nC}\) are 0.100 \(\mathrm{m}\) apart. Point \(A\) is midway between them; point \(B\) is 0.080 \(\mathrm{m}\) from \(q_{1}\) and 0.060 \(\mathrm{m}\) from \(q_{2}\) (Fig. E23.19). Take the electric potential to be zero at infinity. Find (a) the potential at point \(A\); (b) the potential at point \(B ;\) (c) the work done by the electric field on a charge of 2.50 \(\mathrm{nC}\) that travels from point \(B\) to point \(A\) .

Short Answer

Expert verified
(a) Use potential formula for both charges at midpoint; (b) Sum potentials at respective distances for each charge; (c) Calculate work using potential difference and charge.

Step by step solution

01

Calculate Potential at Point A

The electric potential at a point due to a point charge is given by the formula \( V = \frac{kq}{r} \), where \( k = 8.99 \times 10^9 \, \mathrm{N \cdot m^2/C^2} \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the distance from the charge to the point. Since Point A is midway between the charges, the distance from each charge to Point A is \( r = 0.050 \, \mathrm{m}\) (half of 0.100 m). Calculate the potential due to each charge and then sum them: \( V_A = \frac{k q_1}{r} + \frac{k q_2}{r} \).
02

Calculate Potential at Point B

For Point B, calculate the potential separately due to each charge at their respective distances and sum them. The potential at a point due to a charge is \( V = \frac{kq}{r} \). Hence: \( V_B = \frac{k q_1}{0.080} + \frac{k q_2}{0.060} \). Use the given distances 0.080 m for \( q_1 \) and 0.060 m for \( q_2 \).
03

Calculate Work Done from B to A

The work done by the electric field on a charge \( q \) moving from a potential \( V_B \) to \( V_A \) is given by the formula \( W = q(V_A - V_B) \). Use the charge value \( q = 2.50 \, \mathrm{nC} = 2.50 \times 10^{-9} \, \mathrm{C} \) and the potentials calculated in Steps 1 and 2 to find the work done: \( W = 2.50 \times 10^{-9} (V_A - V_B) \).

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Key Concepts

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

Coulomb's Law
Coulomb's law is a fundamental principle in physics that describes the force between two point charges. It states that the force (\( F \) ) between two charges is directly proportional to the product of the magnitudes of the charges (\( q_1 \) and \( q_2 \) ) and inversely proportional to the square of the distance (\( r \) ) between them. The mathematical expression is:\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]Where \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \mathrm{N \cdot m^2/C^2} \) ).
  • The force is attractive if the charges are of opposite signs and repulsive if the same.
  • It is a vector quantity, meaning it has both magnitude and direction.
Understanding Coulomb's law helps to calculate electric potential, as potential is derived from the energy interactions described by this force law.
Point Charge
A point charge is an idealized electric charge that is considered to have no size; it is focused entirely at a single point in space. Point charges are used to simplify the analysis of electrical interactions in space.
  • They are hypothetical concepts that represent charged particles, such as electrons or protons, with negligible size.
  • Their influence on the electric field and potential is calculated using Coulomb's law.
In exercises involving point charges, the calculation of electric fields and potential around them assumes that these charges are not spread out but concentrated at distinct points, making the math easier to handle. This is crucial for understanding how charges interact at various distances.
Electric Field
The electric field is a vector field around a charged particle that represents the force exerted per unit charge at any point in the field. It is a fundamental concept when dealing with charges.The electric field (\( E \) ) created by a point charge (\( q \) ) at a distance (\( r \) ) is given by:\[ E = \frac{kq}{r^2} \]Where \( k \) is the Coulomb's constant.
  • The direction of the electric field due to a positive charge is radially outward, and inward for a negative charge.
  • Its strength decreases with the square of the distance from the charge.
The concept of an electric field allows us to understand how charges can influence other charges even at a distance, without any physical connection.
Work Done by Electric Field
The work done by the electric field is an important concept related to how energy is transferred to or from a charged particle as it moves within an electric field.The work done (\( W \) ) by an electric field on a charge (\( q \) ) as it moves from one point to another is calculated using the change in electric potential (\( \Delta V \) ) between those points:\[ W = q \cdot (V_A - V_B) \]Where \( V_A \) and \( V_B \) are the electric potentials at the initial and final points, respectively.
  • If the charge moves along the direction of the field, the field does positive work and transfers energy to the charge.
  • If the charge moves against the direction of the field, the field does negative work and energy is taken from the charge.
Understanding this concept helps in determining energy conversions in electrostatics, crucial for both theoretical analyses and practical applications.

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Most popular questions from this chapter

An infinitely long line of charge has linear charge density \(5.00 \times 10^{-12} \mathrm{C} / \mathrm{m} .\) A proton \(1.67 \times 10^{-27} \mathrm{kg}\) , charge \(+1.60 \times 10^{-19} \mathrm{C} )\) is 18.0 \(\mathrm{cm}\) from the line and moving directly toward the line at \(1.50 \times 10^{3} \mathrm{m} / \mathrm{s}\) . (a) Calculate the proton's initial kinetic energy. (b) How close does the proton get to the line of charge?

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 \(\mathrm{V}\) and \(12.0 \mathrm{V} / \mathrm{m},\) respectively. (Take the potential to be zero at infinity.) (a) What is the distance to the point charge? (b) What is the magnitude of the charge? (c) Is the electric field directed toward or away from the point charge?

Axons. Neurons are the basic units of the nervous system. They contain long tubular structures called axons that propagate electrical signals away from the ends of the neurons. The axon contains a solution of potassium \(\left(\mathrm{K}^{+}\right)\) ions and large negative organic ions. The axon membrane prevents the large ions from leaking out, but the smaller \(\mathrm{K}^{+}\) ions are able to penetrate the membrane to some degree (Fig. E23.37). This leaves an excess negative charge on the inner surface of the axon membrane and an excess positive charge on the outer surface, resulting in a potential difference across the membrane that prevents further \(K^{+}\) ions from leaking out. Measurements show that this potential difference is typically about 70 \(\mathrm{mV}\) . The thickness of the axon membrane itself varies from about 5 to \(10 \mathrm{nm},\) so we'll use an average of 7.5 \(\mathrm{nm}\) . We can model the membrane as a large sheet having equal and opposite charge densities on its faces. (a) Find the electric field inside the axon membrane, assuming (not too realistically) that it is filled with air. Which way does it point: into or out of the axon? (b) Which is at a higher potential: the inside surface or the outside surface of the axon membrane?

A point charge \(q_{1}=+5.00 \mu \mathrm{C}\) is held fixed in space. From a horizontal distance of \(6.00 \mathrm{cm},\) a small sphere with mass \(4.00 \times 10^{-3} \mathrm{kg}\) and charge \(q_{2}=+2.00 \mu \mathrm{C}\) is fired toward the fixed charge with an initial speed of 40.0 \(\mathrm{m} / \mathrm{s} .\) Gravity can be neglected. What is the acceleration of the sphere at the instant when its speed is 25.0 \(\mathrm{m} / \mathrm{s} ?\)

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 is sprayed in very fine drops (around \(10^{-4} \mathrm{mm}\) in diameter) into the space between two parallel horizontal plates separated by a distance \(d . \mathrm{A}\) potential difference \(V_{A B}\) is maintained between the parallel plates, causing a downward electric field between them. Some of the oil drops acquire a negative charge because of frictional effects or because of ionization of the surrounding air by \(x\) rays or radioactivity. The drops are observed through a microscope. (a) Show that an oil drop of radius \(r\) at rest between the plates will remain at rest if the magnitude of its charge is $$q=\frac{4 \pi}{3} \frac{\rho r^{3} g d}{V_{A B}}$$ where \(\rho\) is the density of the oil. (Ignore the buoyant force of the air.) By adjusting \(V_{A B}\) to keep a given drop at rest, the charge on that drop can be determined, provided its radius is known. (b) Millikan's ooil drops were much too small to measure their radii directly. Instead, Millikan determined \(r\) by cutting off the electric field and measuring the terminal speed \(v_{t}\) of the drop as it fell. (We discussed the concept of terminal speed in Section \(5.3 . )\) The viscous force \(F\) on a sphere of radius \(r\) moving with speed \(v\) through a fluid with viscosity \(\eta\) is given by Stokes's law: \(F=6 \pi \eta r v .\) When the drop is falling at \(v_{1}\) , the viscous force just balances the weight \(w=m g\) of the drop. Show that the magnitude of the charge on the drop is $$q=18 \pi \frac{d}{V_{A B}} \sqrt{\frac{\eta^{3} v_{t}^{3}}{2 \rho g}}$$ Within the limits of their experimental error, every one of the thousands of drops that Millikan and his coworkers measured had a charge equal to some small integer multiple of a basic charge \(e\) . That is, they found drops with charges of \(\pm 2 e, \quad \pm 5 e,\) and so on, but none with values such as 0.76\(e\) or 2.49\(e .\) A drop with charge \(-e\) has acquired one extra electron; if its charge is \(-2 e,\) it has acquired two extra electrons, and so on. (c) A charged oil drop in a Millikan oil-drop apparatus is observed to fall 1.00 mm at constant speed in 39.3 s if \(V_{A B}=0 .\) The same drop can be held at rest between two plates separated by 1.00 \(\mathrm{mm}\) if \(V_{A B}=9.16 \mathrm{V} .\) How many excess electrons has the drop acquired, and what is the radius of the drop? The viscosity of air is \(1.81 \times 10^{-5} \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2},\) and the density of the oil is 824 \(\mathrm{kg} / \mathrm{m}^{3} .\)
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