AP Physics C- E&M- 10.2 Redistribution of Charge between Conductors - Exam Style questions - FRQs- New Syllabus
Question
_____ \(y_{2} > y_{1}\) _____ \(y_{2} < y_{1}\) _____ \(y_{2} = y_{1}\)
Justify your answer.
Most-appropriate topic codes (AP Physics C: Electricity and Magnetism):
• Topic \(10.1\) — Electrostatics with Conductors (Part \( \mathrm{d} \))
• Topic \(10.2\) — Redistribution of Charge Between Conductors (Part \( \mathrm{d} \))
▶️ Answer/Explanation
(a)
Three forces act along the vertical vertical axis on Sphere B:
The spring force points upwards because the spring is compressed.
The downward gravitational force acts on the mass of the sphere.
The downward electrostatic repulsion points away from the positive charge of Sphere A.
Therefore, your diagram must feature a distinct upward vector labeled \(F_{N}\) (or \(F_{s}\)), a downward vector labeled \(F_{g}\), and another downward vector labeled \(F_{E}\).
(b)
Establish a force balance from static equilibrium rules:
\(\Sigma F_{y} = 0\)
\(F_{s} – F_{E} – F_{g} = 0\)
Substitute Hooke’s Law for the spring compression force:
\(F_{s} = k_{s}y\)
Substitute Coulomb’s law for the electrostatic force between point charges:
\(F_{E} = \dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qq}{H^{2}}\)
Substitute the weight equation for the gravitational force:
\(F_{g} = Mg\)
Combine these expressions into the main equilibrium equation:
\(k_{s}y – \dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qq}{H^{2}} – Mg = 0\)
Isolate the variable \(y\) by dividing the parameters through by \(k_{s}\):
\(y = \dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qq}{k_{s}H^{2}} + \dfrac{Mg}{k_{s}}\)
(c)(i)
Draw a straight best-fit trend line that splits the plotted coordinates evenly, maintaining an equal balance of data points above and below your line.
(c)(ii)
Find the line slope by evaluating two distinct coordinates on the trend line:
\(\text{Slope} = \dfrac{\Delta y}{\Delta (1/H^{2})} = \dfrac{0.15 – 0.12}{20 – 0} = 0.0015\,\text{m}^{3}\)
Link this experimental slope directly to the derived functional coefficients:
\(\text{Slope} = \dfrac{Qq}{4\pi\epsilon_{0}k_{s}}\)
Rearrange the terms algebraically to solve for the vacuum permittivity constant:
\(\epsilon_{0} = \dfrac{Qq}{4\pi k_{s}(\text{Slope})}\)
Substitute the given parameter values into your expression:
\(\epsilon_{0} = \dfrac{(2.00\times10^{-6})^{2}}{4\pi(25)(0.0015)} = 8.5\times10^{-12}\,\text{C}^{2}/(\text{N}\cdot\text{m}^{2})\)
\(\boxed{\epsilon_{0} = 8.5\times10^{-12}\,\text{C}^{2}/(\text{N}\cdot\text{m}^{2})}\)
(c)(iii)
Locate where the line crosses the vertical axis to find the vertical intercept value:
\(\text{y-intercept} = 0.12\,\text{m}\)
Equate this value to the constant offset term from your part (b) derivation:
\(\text{y-intercept} = \dfrac{Mg}{k_{s}}\)
Isolate the mass term and substitute the gravitational acceleration and spring constant constants:
\(M = \dfrac{(0.12)(25)}{9.8} = 0.30\,\text{kg}\)
\(\boxed{M = 0.30\,\text{kg}}\)
(d)(i)
Correct selection:
\(\boxed{y_{2} < y_{1}}\)
Mobile charge carriers within the conducting Sphere C rearrange freely due to the field of Sphere A.
Electrons polarize by shifting toward the top surface near Sphere A, leaving the bottom surface positive.
This polarization increases the separation distance between the effective charge centers of the spheres.
The net repulsive force drops because of this longer separation distance, which leads to less spring compression.
(d)(ii)
Correct direction:
\(\boxed{\text{Upward arrow}}\)
Grounding the conductor lets electrons travel up from the earth to neutralize the excess positive charge on the bottom side of Sphere C. As a result, the net positive charge on the sphere drops, which diminishes the repulsive electrostatic force and permits the compressed spring to move the platform upward immediately.