/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Two point charges of equal magni... [FREE SOLUTION] | 91Ó°ÊÓ

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Two point charges of equal magnitude \(Q\) are held a distance \(d\) apart. Consider only points on the line passing through both charges. (a) If the two charges have the same sign, find the location of all points (if there are any) at which (i) the potential (relative to infinity) is zero (is the electric field zero at these points?), and (ii) the electric field is zero (is the potential zero at these points?). (b) Repeat part (a) for two point charges having opposite signs.

Short Answer

Expert verified
(a) For same sign charges: No potential zero; field zero at midpoint. (b) Opposite sign charges: Potential zero at midpoint; no electric field zero.

Step by step solution

01

Understanding the Setup for Same Sign Charges (a)

Two charges, each with magnitude \(Q\), are separated by a distance \(d\) and have the same sign. The electric potential due to a single charge \(Q\) at a distance \(r\) is given by \(V = \frac{kQ}{r}\), and the electric field is given by \(E = \frac{kQ}{r^2}\), where \(k\) is Coulomb's constant.
02

Finding the Potential Zero for Same Sign Charges (i)

The total potential at any point on the line between the charges is \(V_{total} = \frac{kQ}{x} + \frac{kQ}{d-x}\), where \(x\) is the distance from one charge. This simplifies to \(V_{total} = \frac{kQ(d-x)+kQx}{x(d-x)} = \frac{kQd}{x(d-x)}\). Since \(kQd\) is non-zero, \(V_{total}\) cannot be zero unless one of the denominators is infinite, which is not possible between two finite charges. Thus, there are no points where the potential is zero.
03

Analyzing Electric Field Zero for Same Sign Charges (ii)

The electric field at a point is zero if the forces from the charges cancel out. Set \(\frac{kQ}{x^2} = \frac{kQ}{(d-x)^2}\), and solving gives \(x = \frac{d}{2}\). The electric field is zero at the midpoint, but the potential is not.
04

Re-evaluating for Opposite Sign Charges (b)

With opposite signs, the potentials add up to zero at \(x = \frac{d}{2}\) (midpoint), since \(V_{total} = \frac{kQ}{x} - \frac{kQ}{d-x} = 0\) when both terms have equal magnitude. The electric field is not zero here as both forces from charges do not cancel.
05

Electric Field Analysis for Opposite Sign Charges

For opposite sign, finding zero electric field requires setting \(\frac{kQ}{x^2} = \frac{kQ}{(d-x)^2}\), which gives no valid real solutions outside the charges. The electric field is never zero on this line for opposite charges.

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Key Concepts

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

Electric Field
The electric field concept is fundamental in understanding how forces act between charged objects. An electric field is a region around a charged particle that can exert a force on another charged particle placed within this field. The strength and direction of the field depend on both the magnitude of the charge and the distance from it.

The electric field (\(E\)) created by a point charge (\(Q\)) can be expressed by the equation:\[E = \frac{kQ}{r^2}\]where:
  • \(E\) is the electric field strength.
  • \(k\) is Coulomb's constant, approximately \(8.988 \times 10^9 \, \text{Nm}^2/\text{C}^2\).
  • \(r\) is the distance from the charge to the point of interest.
The direction of the field is away from positive charges and toward negative charges. When two charges are involved, like in the exercise, fields from each charge interact, sometimes reinforcing and sometimes canceling each other out depending on their relative directions and magnitudes.

With equal charges, the electric field becomes zero at the midpoint if the charges are equal in magnitude and sign, because the forces cancel out perfectly. However, for opposite sign charges, such a balance point outside the individual fields is not possible, thus no zero field point exists.
Coulomb's Law
Coulomb's Law helps us understand the force interaction between point charges. It states that the force (\(F\)) between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance (\(r\)) between them. The mathematical representation is:\[F = \frac{k |Q_1 Q_2|}{r^2}\]where:
  • \(Q_1\) and \(Q_2\) are the magnitudes of the two charges.
  • \(r\) is the distance between the centers of the two charges.
  • The force is repulsive for like charges and attractive for opposite charges.
Using this law, we can divide the forces into components and find conditions such as zero electric field. The law is the foundation for analyzing point charge interactions, including predicting behavior like the exercise asks – where zero field or potential points occur. When charges have the same sign, like in the exercise part (a), the movements aim to avoid overlap, while in part (b) with opposite charges, the field’s influence provides potential balancing points at exact midpoints.
Point Charges
Point charges are theoretical charges that assume all charge is concentrated in a single, infinitely small point in space. This simplification allows physicists and engineers to apply mathematical principles like Coulomb's Law and electric field equations without the complications of real-world charge distributions.

In the exercise, you see point charges used to simplify the calculation of electric fields and potentials. Two primary scenarios arise:
  • Charges of the same sign, where the potential cannot be zero but the field can balance at midpoint.
  • Charges of opposite signs, where the potential can be zero at a balance point, but the electric field is never zero.
For calculations, such charges make it easier to find exact values as their influence only changes with distance, not with geometric complexities. Point charges enable straightforward application of inverse square law dynamics as captured by Coulomb’s Law, facilitating clear analysis and predictions for electric fields and potentials along the line of charges, as shown in the steps from the exercise.

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Most popular questions from this chapter

Two plastic spheres, each carrying charge uniformly distributed throughout its interior, are initially placed in contact and then released. One sphere is 60.0 cm in diameter, has mass 50.0 g, and contains \(-\)10.0 \(\mu\)C of charge. The other sphere is 40.0 cm in diameter, has mass 150.0 g, and contains \(-\)30.0 \(\mu\)C of charge. Find the maximum acceleration and the maximum speed achieved by each sphere (relative to the fixed point of their initial location in space), assuming that no other forces are acting on them. (\(Hint:\) The uniformly distributed charges behave as though they were concentrated at the centers of the two spheres.)

(a) How much work would it take to push two protons very slowly from a separation of \(2.00 \times 10^{-10}\) m (a typical atomic distance) to \(3.00 \times 10^{-15}\) m (a typical nuclear distance)? (b) If the protons are both released from rest at the closer distance in part (a), how fast are they moving when they reach their original separation?

A gold nucleus has a radius of 7.3 \(\times 10^{-15}\) m and a charge of \(+79e\). Through what voltage must an alpha particle, with charge \(+2e\), be accelerated so that it has just enough energy to reach a distance of 2.0 \(\times 10^{-14}\) m from the surface of a gold nucleus? (Assume that the gold nucleus remains stationary and can be treated as a point charge.)

A small metal sphere, carrying a net charge of \(q_1 = -\)2.80 \(\mu\)C, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of \(q_2 = -\)7.80 \(\mu\)C and mass 1.50 g, is projected toward \(q_1\). When the two spheres are 0.800 m apart, \(q_2\), is moving toward \(q_1\) with speed 22.0 m\(/\)s (\(\textbf{Fig. E23.5}\)). Assume that the two spheres can be treated as point charges. You can ignore the force of gravity. (a) What is the speed of \(q_2\) when the spheres are 0.400 m apart? (b) How close does \(q_2\) get to \(q_1\)?

A charge of 28.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.00 \(\times 10^4\) V\(/\)m. What work is done by the electric force when the charge moves (a) 0.450 m to the right; (b) 0.670 m upward; (c) 2.60 m at an angle of 45.0\(^\circ\) downward from the horizontal?

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