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Chapter 23: Problem 19

Two point charges \(q_{1}=\) \(+2.40 \mathrm{nC}\) and \(q_{2}=-6.50 \mathrm{nC}\) 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. E.23.19). Take the clectric potential to be zero at infinity. Find (a) the potential at point \(A ;\) (b) the potential at point \(B ;(\mathrm{c})\) 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
The electric potential at point A can be found by summing the potentials due to each charge at point A, using the formula \(V=\frac{kQ}{r}\) and a similar method can be used for point B. The work done by the electric field on a charge moving from \(B\) to \(A\) can be found using the formula \(W=q\Delta V\), where \(\Delta V=V_{A}-V_{B}\)

Step by step solution

01

Find the electric potential at point A

Point \(A\) is midway between \(q_1\) and \(q_2\), so the distance from each charge to point \(A\) is \(\frac{0.100}{2}=0.050 m\). Then, apply \(V=\frac{kQ}{r}\) for each of the two charges to find their potentials at point \(A\) and add these together. \[V_{A}=\frac{kq_{1}}{0.05 \mathrm{m}}+\frac{kq_{2}}{0.05 \mathrm{m}}\]
02

Find the electric potential at point B

The distances from each charge to point \(B\) are given as 0.080 m and 0.060 m for \(q_1\) and \(q_2\) respectively. Again use the formula \(V=\frac{kQ}{r}\) for each charge and add the potentials.\[V_{B}=\frac{kq_{1}}{0.08 \mathrm{m}}+\frac{kq_{2}}{0.06 \mathrm{m}}\]
03

Compute the work done by the electric field

The work done by the electric field on a charge of \(q = 2.50 nC\) moving from \(B\) to \(A\) is given by \(W=q\Delta V\), where \(\Delta V=V_{A}-V_{B}\).\[W = q(V_{A} - V_{B})\]

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Key Concepts

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

Point Charges
Imagine the world as a stage where charged particles are actors, and point charges are the main characters in the story of electric fields. A point charge is an idealized model of charged particles with their size being infinitely small. Envision it like a tiny dot that packs a wallop 鈥� its influence radiates out into the surroundings, creating an electric field that can exert forces on other charges.

Now, think of point charges as tiny but mighty sources of power. For instance, in our exercise scenario, we have a pair of point charges with distinct personalities: while one is positive (\( q_{1} = +2.40 \text{nC} \)), the other is its negative counterpart (\( q_{2} = -6.50 \text{nC} \)). Despite their minuscule size, these charges affect the space around them, including points A and B. The interaction of these point charges with their environment is essential for understanding electric potential and the work done by fields, much like how characters in a play interact and shape the plot.
Electric Potential Formula
Electric potential, often referred to as voltage, is all about location, location, location. Just like real estate, it can tell us a lot about the 'value' of a position in an electric field, only here the currency is energy per unit charge. The electric potential formula is a tool to measure this value, given by the expression \( V = \frac{kQ}{r} \), where \(V\) is the electric potential, \(k\) is Coulomb's constant, \(Q\) represents the amount of charge, and \(r\) is the distance from the charge to the point in question.

For layman's terms, think of it like the 'electric pressure' at a point in space due to a charge. The potential created by our point charges at locations A and B in our exercise takes into account the respective distances from the charges. Adding up the potentials due to individual point charges gives you the total electric potential at that location 鈥� it's like adding up the influence from multiple cell towers on your phone's signal strength.
Work Done by Electric Field
The work done by an electric field is the act of moving a charge from one point to another within the field. It's akin to pushing a shopping cart up and down the aisles of a grocery store - the path matters, and so does the effort required to do the pushing.

In physics terms, work is the product of the charge and the change in potential (\( W = q \Delta V \)). It's calculated by taking the difference in electric potential between two points (like A and B from our story) and multiplying it by the charge that's being transported. The resulting work is the energy transferred during the move; positive work is done when a charge moves against the electric field, and negative work is done when it moves with the field. In our exercise, we're like cashiers calculating the total bill by multiplying the price difference by the quantity - only our product is energy, and our 'prices' are the potentials at points A and B.

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Most popular questions from this chapter

A very long uniform line of charge with charge per unit length \(\lambda=+5.00 \mu \mathrm{C} / \mathrm{m}\) lies along the \(x\) -axis, with its midpoint at the origin. A very large uniform sheet of charge is parallel to the \(x y\) -plane; the center of the sheet is at \(z=+0.600 \mathrm{~m}\). The sheet has charge per unit area \(\sigma=+8.00 \mu \mathrm{C} / \mathrm{m}^{2}\), and the center of the sheet is at \(x=0\). \(y=0 .\) Point \(A\) is on the \(z\) -axis at \(z=+0.300 \mathrm{~m}\), and point \(B\) is on the \(z\) -axis at \(z=-0.200 \mathrm{~m}\). What is the potential difference \(V_{A B}=V_{A}-V_{B}\) between points \(A\) and \(B ?\) Which point, \(A\) or \(B,\) is at higher potential?

(a) An electron is to be accelerated from \(3.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) to \(8.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from \(8.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) to a halt?

\( \cdot\) (a) If a spherical raindrop of radius \(0.650 \mathrm{~mm}\) carries a charge of \(-3.60 \mathrm{pC}\) uniformly distributed over its volume, what is the potential at its surface? (Take the potential to be zero at an infinite distance from the raindrop.) (b) Two identical raindrops, each with radius and charge specified in part (a), collide and merge into one larger raindrop. What is the radius of this larger drop, and what is the potential at its surface, if its charge is uniformly distributed over its volume?

A helium nucleus, also known as an \(\alpha\) (alpha) particle, consists of two protons and two neutrons and has a diameter of \(10^{-15} \mathrm{~m}\) \(=1 \mathrm{fm} .\) The protons, with a charge of te, are subject to a repulsive Coulomb force. since the neutrons have zero charge, there must be an attractive force that counteracts the electric repulsion and keeps the protons from flying apart. This so-called strong force plays a central role in particle physics. (a) As a crude model, assume that an \(\alpha\) particle consists of two pointlike protons attracted by a Hooke's-law spring with spring constant \(k,\) and ignore the neutrons. Assume further that in the absence of other forces, the spring has an equilibrium separation of zero. Write an expression for the potential energy when the protons are separated by distance \(d\). (b) Minimize this potential to find the equilibrium separation \(d_{0}\) in terms of \(e\) and \(k .\) (c) If \(d_{0}=1.00 \mathrm{fm}\), what is the value of \(k ?\) (d) How much energy is stored in this system, in terms of electron volts? (e) A proton has a mass of \(1.67 \times 10^{-27} \mathrm{~kg}\). If the spring were to break, the \(\alpha\) particle would disintegrate and the protons would fly off in opposite directions. What would be their ultimate speed?

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 \(25.0 \mathrm{~m} / \mathrm{s} ?\)
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