/*! 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 19 Two point charges \(q_1 = +\)2.4... [FREE SOLUTION] | 91影视

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Chapter 23: Problem 19

Two point charges \(q_1 = +\)2.40 nC and \(q_2 = -\)6.50 nC are 0.100 m apart. Point \(A\) is midway between them; point \(B\) is 0.080 m from \(q_1\) and 0.060 m from \(q_2\) (\(\textbf{Fig. E23.19}\)). Take the electric potential to be zero at infinity. Find (a) the potential at point \(A\); (b) the potential at point \(B\); (c) the work done by the electric field on a charge of 2.50 nC that travels from point \(B\) to point \(A\).

Short Answer

Expert verified
Potential at A is the sum of potentials from both charges at the midpoint. Potential at B considers each charge's distance at B. Work is the charge times the potential difference (A-B).

Step by step solution

01

Understand the Problem

We have two charges, \( q_1 = +2.40 \text{ nC} \) and \( q_2 = -6.50 \text{ nC} \), 0.100 m apart. We need to find the electric potential at two points: Point \( A \), which is halfway between the charges, and Point \( B \), which is 0.080 m from \( q_1 \) and 0.060 m from \( q_2 \). We also need to determine the work done by the electric field on a charge moving from \( B \) to \( A \). Electric potential should be considered zero at infinity.
02

Calculate the Potential at Point A

Electric potential due to a point charge \( q \) is given by the formula \( V = \frac{kq}{r} \), where \( k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \) is Coulomb's constant, and \( r \) is the distance from the charge. At midpoint \( A \), the distance \( r \) from both \( q_1 \) and \( q_2 \) is 0.050 m (half of 0.100 m). Calculate the potential contributions from both charges and add them:\[V_A = V_{q_1} + V_{q_2} = \frac{k q_1}{0.050} + \frac{k q_2}{0.050}\]Substituting in:\[V_A = \frac{8.99 \times 10^9 \times 2.40 \times 10^{-9}}{0.050} + \frac{8.99 \times 10^9 \times (-6.50) \times 10^{-9}}{0.050}\]Calculate each term and then combine them to find \( V_A \).
03

Calculate the Potential at Point B

For Point \( B \), use the same formula, but with the distances given for each charge's influence at \( B \): 0.080 m for \( q_1 \) and 0.060 m for \( q_2 \):\[V_B = \frac{k q_1}{0.080} + \frac{k q_2}{0.060}\]Substituting in:\[V_B = \frac{8.99 \times 10^9 \times 2.40 \times 10^{-9}}{0.080} + \frac{8.99 \times 10^9 \times (-6.50) \times 10^{-9}}{0.060}\]Calculate each term and combine them to obtain \( V_B \).
04

Calculate the Work Done Moving from B to A

The work done by the electric field moving a charge \( q = 2.50 \text{ nC} \) from point \( B \) to point \( A \) is calculated using the change in electric potential: \( W = q \cdot (V_A - V_B) \).First, find the potential difference:\[ \Delta V = V_A - V_B \]Then use:\[W = (2.50 \times 10^{-9}) \times \Delta V\]Substitute the calculated potentials from Step 2 and Step 3 to find the work done.

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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 vector field surrounding an electric charge that exerts a force on other charges, attracting or repelling them. The direction of the electric field is always in the direction a positive test charge would move if placed within the field. To visualize, imagine lines radiating outward from a positive charge, each one indicating the force vector. The density of these lines represents the field's strength. As you move further from the charge, the electric field weakens.
To calculate the electric field (E) produced by a point charge (q), we use the formula: \[ E = \frac{k |q|}{r^2} \]where \( E \)is the magnitude of the electric field, \( r \) is the distance from the charge, and \( k \) is Coulomb's constant. The force experienced by another charge \( q_0 \) placed a certain distance from the original point charge is given by \[ F = q_0 \times E \]. This formula is incredibly useful in determining how a charged particle will behave in the presence of others, deriving directly from Coulomb's law.
Point Charges
Point charges are idealized charges that are infinitely small yet possess a significant amount of charge value. In the real world, no charge can truly be a point, but this model simplifies calculations and theoretical physics studies.
In exercises involving point charges, like the one described, we assume each charge is at a precise location with no physical dimensions. This allows us to use formulas for electric fields and potentials more straightforwardly, as the geometry of the system is simplified.
  • For a single point charge, the electric potential (V) at a certain distance (r) can be calculated using: \[ V = \frac{kq}{r} \]where \( k \) is Coulomb's constant, and \( q \) is the charge's magnitude.
  • In a system with multiple charges, the superposition principle helps determine the total potential or field, calculated by summing contributions from all individual charges.
This principle implies that each charge impacts the surroundings independently, making it easier to calculate complex systems' overall effects by adding individual contributions.
Coulomb's Constant
Coulomb's constant, denoted as \( k \), is a fundamental value used in various equations describing electric forces between charges. It derives from Coulomb's Law, which states that the force (\( F \)) between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Coulomb's constant quantifies this relationship, with a standard value of \( k = 8.99 imes 10^{9} ext{ Nm}^2/ ext{C}^2 \). This constant helps ensure that our units for charge, distance, and force are consistent, enabling accurate calculations in electrostatics problems.
  • It's essential when using formulas for electric potential and fields, like \( V = \frac{kq}{r} \)and \( E = \frac{k |q|}{r^2} \), respectively.
  • The constant simplifies to a value convenient for calculations when the charges are measured in coulombs and the distance in meters, a standard in physics.
Understanding \( k \) is crucial in solving problems involving electric interactions, as it connects theoretical predictions with measurable phenomena.

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Most popular questions from this chapter

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work is needed to move the alpha particle to the midpoint of one of the sides of the square?

A small sphere with mass 5.00 \(\times 10^{-7}\) kg and charge \(+\)7.00 \(\mu\)C is released from rest a distance of 0.400 m above a large horizontal insulating sheet of charge that has uniform surface charge density \(\sigma = +\)8.00 pC\(/\)m\(^2\). Using energy methods, calculate the speed of the sphere when it is 0.100 m above the sheet.

A ring of diameter 8.00 cm is fixed in place and carries a charge of \(+\)5.00 \(\mu\)C uniformly spread over its circumference. (a) How much work does it take to move a tiny \(+\)3.00-\(\mu\)C charged ball of mass 1.50 g from very far away to the center of the ring? (b) Is it necessary to take a path along the axis of the ring? Why? (c) If the ball is slightly displaced from the center of the ring, what will it do and what is the maximum speed it will reach?

A hollow, thin-walled insulating cylinder of radius \(R\) and length \(L\) (like the cardboard tube in a roll of toilet paper) has charge \(Q\) uniformly distributed over its surface. (a) Calculate the electric potential at all points along the axis of the tube. Take the origin to be at the center of the tube, and take the potential to be zero at infinity. (b) Show that if \(L \ll R\), the result of part (a) reduces to the potential on the axis of a ring of charge of radius \(R\). (See Example 23.11 in Section 23.3.) (c) Use the result of part (a) to find the electric field at all points along the axis of the tube.

A helium ion (He\(^{++}\)) that comes within about 10 fm of the center of the nucleus of an atom in the sample may induce a nuclear reaction instead of simply scattering. Imagine a helium ion with a kinetic energy of 3.0 MeV heading straight toward an atom at rest in the sample. Assume that the atom stays fixed. What minimum charge can the nucleus of the atom have such that the helium ion gets no closer than 10 fm from the center of the atomic nucleus? (1 fm = 1 \(\times\) 10\(^{-15}\) m, and \(e\) is the magnitude of the charge of an electron or a proton.) (a) 2\(e\); (b) 11\(e\); (c) 20\(e\); (d) 22\(e\).
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