Skip to main content
Ch 23: The Electric Field
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 23: The Electric FieldProblem 10
Chapter 23, Problem 10

A small glass bead charged to +6.0 nC is in the plane that bisects a thin, uniformly charged, 10-cm-long glass rod and is 4.0 cm from the rod's center. The bead is repelled from the rod with a force of 840 μN. What is the total charge on the rod?

Verified step by step guidance
1
Identify the relevant physical principle: The force between the charged bead and the charged rod is due to the electric field produced by the rod. The force on the bead is given by \( F = q \cdot E \), where \( F \) is the force, \( q \) is the charge of the bead, and \( E \) is the electric field at the bead's location.
Determine the electric field produced by the rod: Since the rod is uniformly charged and the bead is located in the plane that bisects the rod, the electric field at the bead's location can be calculated using the formula for the electric field due to a uniformly charged finite line of charge. The formula is \( E = \frac{k \cdot \lambda \cdot L}{r \cdot \sqrt{r^2 + (L/2)^2}} \), where \( k \) is Coulomb's constant, \( \lambda \) is the linear charge density, \( L \) is the length of the rod, and \( r \) is the perpendicular distance from the rod to the bead.
Relate the linear charge density \( \lambda \) to the total charge \( Q \): The linear charge density is defined as \( \lambda = \frac{Q}{L} \), where \( Q \) is the total charge on the rod and \( L \) is the length of the rod. Substitute \( \lambda \) into the electric field formula to express \( E \) in terms of \( Q \).
Use the force equation to solve for \( Q \): Substitute the expression for \( E \) into \( F = q \cdot E \). This gives \( F = q \cdot \frac{k \cdot Q}{L \cdot r \cdot \sqrt{r^2 + (L/2)^2}} \). Rearrange this equation to solve for \( Q \): \( Q = \frac{F \cdot L \cdot r \cdot \sqrt{r^2 + (L/2)^2}}{k \cdot q} \).
Substitute the known values into the equation: Use \( F = 840 \ \mu\text{N} = 840 \times 10^{-6} \ \text{N} \), \( q = 6.0 \ \text{nC} = 6.0 \times 10^{-9} \ \text{C} \), \( L = 10 \ \text{cm} = 0.10 \ \text{m} \), \( r = 4.0 \ \text{cm} = 0.04 \ \text{m} \), and \( k = 8.99 \times 10^9 \ \text{N·m}^2/\text{C}^2 \). Substitute these values into the formula for \( Q \) to calculate the total charge on the rod.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
7m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Coulomb's Law

Coulomb's Law describes the electrostatic force between two charged objects. It states that the force (F) is directly proportional to the product of the magnitudes of the charges (q1 and q2) and inversely proportional to the square of the distance (r) between them. Mathematically, it is expressed as F = k * (|q1 * q2| / r^2), where k is Coulomb's constant. This law is fundamental for calculating the forces between charged particles.
Recommended video:
Guided course
09:52
Coulomb's Law

Electric Field

An electric field is a region around a charged object where other charges experience a force. It is defined as the force per unit charge experienced by a positive test charge placed in the field. The electric field (E) due to a point charge can be calculated using E = k * (|q| / r^2), where q is the charge creating the field and r is the distance from the charge. Understanding electric fields is crucial for analyzing the interactions between charged objects.
Recommended video:
Guided course
03:16
Intro to Electric Fields

Superposition Principle

The superposition principle states that the total electric field or force acting on a charge due to multiple other charges is the vector sum of the individual fields or forces produced by each charge. This principle allows us to analyze complex charge configurations by breaking them down into simpler components. In the context of the problem, it helps in determining the total charge on the rod by considering the contributions from the bead and the rod's charge.
Recommended video:
Guided course
03:32
Superposition of Sinusoidal Wave Functions
Related Practice
Textbook Question

CALC A 12-cm-long thin rod has the nonuniform charge density λ(x)=(2.0nC/cm)e−∣x∣/(6.0cm)\(\lambda\)(x)=(2.0\,\(\text{nC/cm}\))e^{-|x|/(6.0\,\(\text{cm}\))}, where x is measured from the center of the rod. What is the total charge on the rod? Hint: This exercise requires an integration. Think about how to handle the absolute value sign.

1
views
Textbook Question

A 10-cm-long thin glass rod uniformly charged to +10 nC and a 10-cm-long thin plastic rod uniformly charged to −10 nC are placed side by side, 4.0 cm apart. What are the electric field strengths E1 to E3 at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods?

Textbook Question

Two 10-cm-diameter charged rings face each other, 20 cm apart. The left ring is charged to −20 nC and the right ring is charged to +20 nC. What is the electric field Ē, both magnitude and direction, at the midpoint between the two rings?

1
views
Textbook Question

An electric dipole is formed from two charges, ±q, spaced 1.0 cm apart. The dipole is at the origin, oriented along the y-axis. The electric field strength at the point (x, y)=(0 cm, 10 cm) is 360 N/C. What is the charge q? Give your answer in nC.

1
views
Textbook Question

An electret is similar to a magnet, but rather than being permanently magnetized, it has a permanent electric dipole moment. Suppose a small electret with electric dipole moment 1.0×10−7 C m is 25 cm from a small ball charged to +25 nC, with the ball on the axis of the electric dipole. What is the magnitude of the electric force on the ball?

Textbook Question

Two 10-cm-long thin glass rods uniformly charged to +10 nC are placed side by side, 4.0 cm apart. What are the electric field strengths E1 to E3 at distances 1.0 cm, 2.0 cm, and 3.0 cm to the right of the rod on the left along the line connecting the midpoints of the two rods?