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Ch 26: Potential and 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 26: Potential and FieldProblem 49e
Chapter 26, Problem 49e

Two positive point charges q are located on the y-axis at y = ±a. Your answer to part d shows that an electron experiences a linear restoring force, so it will undergo simple harmonic motion. What is the oscillation frequency in GHz for an electron moving between two 1.0 nC charges separated by 2.0 mm?

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Step 1: Begin by understanding the setup of the problem. Two positive point charges, each with charge \( q = 1.0 \, \text{nC} \), are located on the y-axis at \( y = +a \) and \( y = -a \). The separation between the charges is \( 2a = 2.0 \, \text{mm} \), so \( a = 1.0 \, \text{mm} \). The electron is placed near the midpoint and experiences a linear restoring force due to the electric field created by the charges.
Step 2: Recall that the restoring force \( F \) acting on the electron is proportional to its displacement \( y \) from the equilibrium position (the midpoint). This is characteristic of simple harmonic motion. The force can be expressed as \( F = -k y \), where \( k \) is the effective spring constant. To find \( k \), calculate the net electric field at a small displacement \( y \) and relate it to the force \( F = q_e E \), where \( q_e \) is the charge of the electron.
Step 3: Use Coulomb's law to calculate the electric field \( E \) at a small displacement \( y \) from the midpoint. The electric field contributions from each charge can be approximated for small \( y \) using a Taylor expansion. Combine the fields from both charges to find the net field, and express \( k \) in terms of \( q \), \( a \), and fundamental constants.
Step 4: Once \( k \) is determined, use the formula for the angular frequency of simple harmonic motion: \( \omega = \sqrt{\frac{k}{m}} \), where \( m \) is the mass of the electron. Substitute the value of \( k \) and the mass of the electron \( m = 9.11 \times 10^{-31} \, \text{kg} \) to find \( \omega \).
Step 5: Convert the angular frequency \( \omega \) to the oscillation frequency \( f \) using \( f = \frac{\omega}{2\pi} \). Finally, convert the frequency from Hz to GHz by dividing by \( 10^9 \). This gives the oscillation frequency in GHz for the electron moving between the two charges.

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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 is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This principle is essential for calculating the force experienced by the electron due to the two point charges.
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Coulomb's Law

Simple Harmonic Motion (SHM)

Simple Harmonic Motion refers to the oscillatory motion of an object where the restoring force is directly proportional to the displacement from an equilibrium position. In this context, the electron experiences a linear restoring force due to the electric field created by the point charges, leading to oscillations around a central point, which can be analyzed using SHM principles.
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Frequency of Oscillation

The frequency of oscillation in SHM is the number of complete cycles per unit time, typically measured in Hertz (Hz). For an electron in an electric field, the frequency can be derived from the mass of the electron and the effective spring constant determined by the forces acting on it. In this case, converting the frequency to gigahertz (GHz) involves scaling the result by a factor of 10^-9.
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Related Practice
Textbook Question

The electric potential is 40 V at point A near a uniformly charged sphere. At point B, 2.0 μm farther away from the sphere, the potential has decreased by 0.16 mV. How far is point A from the center of the sphere?

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Textbook Question

The electric field in a region of space is E→=(800xı^−600yȷ^)\(\overrightarrow{E}\)=(800xî-600yĵ) V/m , where x and y are in m. The zero of electric potential is at the origin. What are (a) the electric field and (b) the electric potential at the point (x,y)=(2.0 m, 1.0 m)? Hint: The potential difference is the same along any path connecting two points.

Textbook Question

Two positive point charges q are located on the y-axis at y = ±a. Symmetry dictates that the electric field along the x-axis has only an x-component: Ey=Ez=0. Find an expression for Ex if x≪a.

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Textbook Question

Metal sphere 1 has a positive charge of 6.0 nC. Metal sphere 2, which is twice the diameter of sphere 1, is initially uncharged. The spheres are then connected together by a long, thin metal wire. What are the final charges on each sphere?

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Textbook Question

Two 2.0 cm×2.0 cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 V battery. The wires are disconnected, and insulated handles are used to pull the plates apart to a new spacing of 2.0 mm. What are the charge on each electrode and the potential difference between them?

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Textbook Question

Two positive point charges q are located on the y-axis at y = ±a. Write an expression for the electric potential at position x on the x-axis.

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