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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

A solid sphere of radius \(R\) contains a total charge \(Q\) distributed uniformly throughout its volume. Find the energy needed to assemble this charge by bringing infinitesimal charges from far away. This energy is called the "self- energy" of the charge distribution. (\(\textit{Hint:}\) After you have assembled a charge q in a sphere of radius \(r\), how much energy would it take to add a spherical shell of thickness \(dr\) having charge \(dq\)? Then integrate to get the total energy.)

Two stationary point charges \(+\)3.00 nC and \(+\)2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at a point midway between the two charges and moves along the line connecting the two charges. What is the speed of the electron when it is 10.0 cm from the \(+\)3.00-nC charge?

(a) Calculate the potential energy of a system of two small spheres, one carrying a charge of 2.00 \(\mu\)C and the other a charge of \(-\)3.50 \(\mu\)C, with their centers separated by a distance of 0.180 m. Assume that \(U = 0\) when the charges are infinitely separated. (b) Suppose that one sphere is held in place; the other sphere, with mass 1.50 g, is shot away from it. What minimum initial speed would the moving sphere need to escape completely from the attraction of the fixed sphere? (To escape, the moving sphere would have to reach a velocity of zero when it is infinitely far from the fixed sphere.)

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 thin spherical shell with radius \(R_1 =\) 3.00 cm is concentric with a larger thin spherical shell with radius \(R_2 =\) 5.00 cm. Both shells are made of insulating material. The smaller shell has charge \(q_1 = +\)6.00 nC distributed uniformly over its surface, and the larger shell has charge \(q_2 = -\)9.00 nC distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. (a) What is the electric potential due to the two shells at the following distance from their common center: (i) \(r =\) 0; (ii) \(r =\) 4.00 cm; (iii) \(r =\) 6.00 cm? (b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at higher potential: the inner shell or the outer shell?

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