Skip to main content
Ch 25: The Electric Potential
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoKnight Calc 5th EditionCh 25: The Electric PotentialProblema 75b
Capitolo 25, Problema 75b

You are given the equation(s) used to solve a problem. Finish the solution of the problem: (9.0×109Nm2/C2)q₁q₂/0.030m =90×10−6J; q₁+q₂=40nC.

Guida verificata passo dopo passo
1
Step 1: Start by identifying the given information. The first equation relates the electric potential energy between two charges to their magnitudes and separation distance: \( U = \frac{k q_1 q_2}{r} \), where \( U = 90 \times 10^{-6} \; \text{J} \), \( k = 9.0 \times 10^9 \; \text{Nm}^2/\text{C}^2 \), and \( r = 0.030 \; \text{m} \). The second equation is \( q_1 + q_2 = 40 \; \text{nC} \).
Step 2: Rearrange the first equation to solve for \( q_1 q_2 \): \( q_1 q_2 = \frac{U r}{k} \). Substitute the known values \( U = 90 \times 10^{-6} \; \text{J} \), \( r = 0.030 \; \text{m} \), and \( k = 9.0 \times 10^9 \; \text{Nm}^2/\text{C}^2 \) into this equation.
Step 3: Use the second equation \( q_1 + q_2 = 40 \; \text{nC} \) to express one charge in terms of the other. For example, let \( q_1 = 40 \; \text{nC} - q_2 \). Substitute this expression for \( q_1 \) into the equation \( q_1 q_2 = \frac{U r}{k} \).
Step 4: Expand the equation \( (40 \; \text{nC} - q_2) q_2 = \frac{U r}{k} \) to form a quadratic equation in terms of \( q_2 \). Simplify the terms to get \( -q_2^2 + 40 \; \text{nC} \cdot q_2 - \frac{U r}{k} = 0 \).
Step 5: Solve the quadratic equation for \( q_2 \) using the quadratic formula \( q_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -1 \), \( b = 40 \; \text{nC} \), and \( c = -\frac{U r}{k} \). Once \( q_2 \) is found, use \( q_1 = 40 \; \text{nC} - q_2 \) to find \( q_1 \).

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Coulomb's Law

Coulomb's Law describes the electrostatic force between two charged objects. It states that the force is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This relationship is mathematically expressed as F = k(q₁q₂/r²), where k is Coulomb's constant. Understanding this law is essential for solving problems involving electric forces and potential energy.
Video consigliato:
Percorso guidato
09:52
Coulomb's Law

Electric Potential Energy

Electric potential energy is the energy stored in a system of charged particles due to their positions relative to each other. It can be calculated using the formula U = k(q₁q₂/r), where U is the potential energy, k is Coulomb's constant, and r is the distance between the charges. This concept is crucial for understanding how energy is transferred in electric fields and is directly related to the work done in moving charges.
Video consigliato:
Percorso guidato
07:51
Electric Potential Energy

Charge Conservation

Charge conservation is a fundamental principle stating that the total electric charge in an isolated system remains constant over time. In the context of the problem, the equation q₁ + q₂ = 40 nC indicates that the sum of the two charges is fixed. This principle is vital for solving problems involving multiple charges, as it allows for the determination of unknown charges when one or more values are provided.
Video consigliato:
Percorso guidato
05:43
Conservation of Charge
Pratica correlata
Domanda del libro di testo

An electric dipole consists of 1.0 g spheres charged to ±2.0 nC at the ends of a 10-cm-long massless rod. The dipole rotates on a frictionless pivot at its center. The dipole is held perpendicular to a uniform electric field with field strength 1000 V/m, then released. What is the dipole's angular velocity at the instant it is aligned with the electric field?

127
views
Domanda del libro di testo

Two 2.0-mm-diameter beads, C and D, are 10 mm apart, measured between their centers. Bead C has mass 1.0 g and charge 2.0 nC. Bead D has mass 2.0 g and charge −1.0 nC. If the beads are released from rest, what are the speeds vC and vD at the instant the beads collide?

1367
views
Domanda del libro di testo

FIGURE P25.72 shows a thin rod with charge Q that has been bent into a semicircle of radius R. Find an expression for the electric potential at the center.

442
views
Domanda del libro di testo

FIGURE P25.70 shows a thin rod of length L and charge Q. Find an expression for the electric potential a distance x away from the center of the rod on the axis of the rod.

307
views
Domanda del libro di testo

Bead A has a mass of 15 g and a charge of −5.0 nC. Bead B has a mass of 25 g and a charge of −10.0 nC. The beads are held 12 cm apart (measured between their centers) and released. What maximum speed is achieved by each bead?

138
views
Domanda del libro di testo

The wire in FIGURE P25.74 has linear charge density λ. What is the electric potential at the center of the semicircle?

182
views