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Two charges of equal magnitude \(Q\) are beld 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?). and (b) Repeat part (a) for two charges having opposite signs.

Short Answer

Expert verified
For like charges: (i) No potential zero; (ii) Field zero at midpoint, not potential. For opposite charges: (i) Potential zero at midpoint; (ii) Field nonzero at potential zero.

Step by step solution

01

Understanding Electric Potential and Field between Charges

The electric potential due to a point charge \(Q\) at a distance \(r\) is given by \(V = \frac{kQ}{r}\), where \(k\) is Coulomb's constant. The electric field due to this charge is \(E = \frac{kQ}{r^2}\). For two charges situated at a distance \(d\) apart, the potential at any point on the line connecting the two is the algebraic sum of potentials due to each charge. Similarly, the field is the vector sum.
02

Analyze for Charges with Same Sign

For two charges \(+Q\), the potential at a point between the charges at a distance \(x\) from one charge is given by \(V = \frac{kQ}{x} + \frac{kQ}{d-x}\). Setting this equal to zero, i.e., \(\frac{kQ}{x} + \frac{kQ}{d-x} = 0\), shows there are no solutions since the expression cannot be zero.For the electric field between the charges, it will be zero at a point \(x\) if \(\frac{kQ}{x^2} = \frac{kQ}{(d-x)^2}\), giving \(x = \frac{d}{2}\). At this point, the potentials due to each charge add up and are not zero.
03

Analyze for Charges with Opposite Signs

For two charges \(+Q\) and \(-Q\), the potential at a point between them becomes \(V = \frac{kQ}{x} - \frac{kQ}{d-x}\). Setting this to zero, we get \(\frac{1}{x} = \frac{1}{d-x}\), which simplifies to \(x = \frac{d}{2}\). At this midpoint, potentials due to each charge cancel out to zero.For the electric field, it is given by \(E = \frac{kQ}{x^2} + \frac{kQ}{(d-x)^2}\), and setting this to zero gives no solution. The field is nonzero where potential is zero.

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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 is a region around a charged particle where a force would be exerted on any other charges in that region. It is a vector quantity, which means it has both a magnitude and a direction.

The electric field strength due to a single point charge can be calculated using the formula:
  • \( E = \frac{kQ}{r^2} \)
where:
  • \( E \) is the electric field strength,
  • \( k \) is Coulomb's constant \( (9 \times 10^9 \text{ N m}^2/\text{C}^2) \),
  • \( Q \) is the charge,
  • \( r \) is the distance from the charge.
For two point charges, the net electric field at any point is determined by adding up the fields due to each charge vectorially.

If the charges have the same sign, the field will never be zero between them because each field points away from the similar charges, and at any point, one field will always be stronger. However, if the charges are opposite, the fields between them will point in opposite directions, potentially canceling out at the midpoint.
Coulomb's Law
Coulomb's Law describes the force between two point charges. The law states that the electric force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. This is given by:
  • \( F = \frac{k |Q_1 Q_2|}{r^2} \)
where:
  • \( F \) is the magnitude of force between the charges,
  • \( Q_1 \) and \( Q_2 \) are the amounts of the two charges,
  • \( r \) is the distance separating the charges,
  • \( k \) is Coulomb's constant.
The force is attractive if the charges are of opposite signs and repulsive if they have the same sign.

Using Coulomb's Law, you can predict the behavior of the electric field in different configurations, such as finding points where the forces might balance each other. This ties into the idea of equilibrium points in the electric field, where the net electric force on a test charge would be zero.
Point Charges
Point charges are an idealized model in physics representing charged objects that are infinitesimally small in size, allowing their charge to be considered as concentrated at a single point in space.

Point charges are often used to simplify complex problems in electrostatics because their electric field and potential can be calculated easily via straightforward formulas. For a single point charge, the electric potential \( V \) at a distance \( r \) is:
  • \( V = \frac{kQ}{r} \)
Point charges facilitate the understanding of electric potential and fields because they allow the consideration of simple, symmetric situations.

In exercises, the calculation of potential and field can be complex when multiple point charges are involved, as their individual potentials and fields combine. The understanding of how they interact, such as assessing whether their potential or field values add up to zero at certain points, is a key skill in mastering electrostatics.

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Most popular questions from this chapter

Three equal \(1.20-\mu \mathrm{C}\) point charges are placed at the corners of an equilateral triangle whose sides are 0.500 \(\mathrm{m}\) 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)

Before the advent of solid-state electronics, vacuum tubes were widely used in radios and other devices. A simple type of vacuum tube known as a diode consists essentially of two electrodes within a highly evacuated enclosure. One electrode, the cathode, is maintained at a high temperature and emits electrons from its surface. A potential difference of a few hundred volts is maintained between the cathode and the other electrode, known as the anode, with the anode at the higher potential. Suppose that in a particular vacuum tube the potential of the anode is 295 \(\mathrm{V}\) higher than that of the cathode. An electron leaves the surface of the cathode with zero initial speed. Find its speed when it strikes the anode.

A very long insulating cylinder of charge of radius 2.50 \(\mathrm{cm}\) carries a uniform linear density of 15.0 \(\mathrm{nC} / \mathrm{m}\) . If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads 175 \(\mathrm{V} ?\)

A point charge has a charge of \(2.50 \times 10^{-11} \mathrm{C}\) . At what distance from the point charge is the electric potential (a) 90.0 \(\mathrm{V}\) and (b) 30.0 \(\mathrm{V}\) ? Take the potential to be zero at an infinite distance from the charge.

A thin insulating rod is bent into a semicircular are of radius \(a,\) and a total electric charge \(Q\) is distributed uniformly along the rod. Calculate the potential at the center of curvature of the are if the potential is assumed to be zero at infinity.

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