Table 5 Qualitative analysis of memos posted by students

Lecture focus (L)

Student 1—“what is exact meaning of flux?; How to derive Coulomb’s law from Gauss law”; “Why is there change in magnetic flux” “What are disadvantages of Coulombs law”; How Coulomb’s law equation equal to gravitational equation ?”

Student 2—“How is the flux zero in zero position”; “Why is the potential difference between the two charges is independent from the path h taken?”; “Why is the gradient operator not a vector in itself?”

Problem focus (P)

Problem Statement: Consider a new oxide dielectric material having an electric permittivity value of 1.74 × 10^–10 C²/N-m². Determine its dielectric constant and electric susceptibility

Student 1 Solution: Dielectric constant, k = 19.66 susceptibility = 165.141 * 10⁻¹²

Student 2 Solution: 58.05 * 10⁻¹² c²/n-m²

LeD focus—L

Student 3 – why does the recombination of an electron and a hole generate light in direct band gap semiconductors and generate heat in indirect band gap semiconductors?

LeD focus—P

Student 2—“What will be the effect on Hall voltage if the direction of magnetic field is reversed?” “what happens when photodiode is low light illuminated?. does it works? What should we do to make photodiode work under this condition?” “How does charge accumulation balance Lorentz force?”; “How can we tell how sensitive a photodiode is?”

Reflection Spot (Conceptual): Why are direct band gap semiconductors preferred to make light emitting diodes (LEDs)?

Student 3—“In direct semiconductors the momentum vector k is aligned along the CB and VB which would make it easy for the LED to pass as much as current into it.”

Student 2—LEDs are mostly made from direct semiconductors because no change in momentum is required for an electron in the conduction band to recombine with a hole in the valence band

Reflection Spot (Problem): When an electric field of 160 V/m is applied to a semiconductor sample whole type is unknown. The sample exhibits a Hall coefficient of value—0.0125 m³/C

(i) Whether the semiconductor is N-type or P-type?

(ii) Determine the current density in the sample, assuming the mobility of electrons (μ_e = 0.6 m²/V s

Student 3 solution—(i) “As the hall coefficient and the hall voltage is negative the semiconductor is N-type” (ii) J = − 7680 A/m²

Student 2 solution—“Charge carrier concentration (n) = 2.0862 * 10²⁵ carriers/m³ ii. drift voltage (Vd) = 0.12 m/s; hall voltage (Vh) = 2.66 * 10⁻⁶ V or 2.66 microvolts.”