AP Physics 2- 10.6 Capacitors- Exam Style questions - FRQs- New Syllabus
Capacitors AP Physics 2 FRQ
Unit 10: Electric Force, Field, and Potential
Weightage : 15–18%
Question
The apparatus shown in the figure above consists of two oppositely charged parallel conducting plates, each with area \(A=0.25\,\text{m}^2\), separated by a distance \(d=0.010\,\text{m}\). Each plate has a hole at its center through which electrons can pass. High velocity electrons produced by an electron source enter the top plate with speed \(v_0=5.40\times10^6\,\text{m/s}\), take \(1.49\,\text{ns}\) to travel between the plates, and leave the bottom plate with speed \(v_f=8.02\times10^6\,\text{m/s}\).
Most-appropriate topic codes (AP Physics 2):
• Topic \(10.6\) — Capacitors (Part \( \mathrm{(c)} \))
• Topic \(12.2\) — Magnetism and Moving Charges (Part \( \mathrm{(d)} \))
▶️ Answer/Explanation
(a)
The electrons speed up as they move downward between the plates, so the electric force on the electrons is downward.
Since electrons have negative charge, the electric force on them is opposite the direction of the electric field. Therefore, the electric field must point upward.
Electric field lines point from the positive plate toward the negative plate. Since the field points upward, the bottom plate is positive and the top plate is negative.
\(\boxed{\text{The top plate is negatively charged.}}\)
(b)
First calculate the acceleration of the electron between the plates:
\(a=\dfrac{v_f-v_0}{t}\)
\(a=\dfrac{8.02\times10^6\,\text{m/s}-5.40\times10^6\,\text{m/s}}{1.49\times10^{-9}\,\text{s}}\)
\(a\approx1.76\times10^{15}\,\text{m/s}^2\)
The electric force provides this acceleration:
\(F=ma\)
Also,
\(F=eE\)
Therefore,
\(E=\dfrac{ma}{e}\)
\(E=\dfrac{\left(9.11\times10^{-31}\,\text{kg}\right)\left(1.76\times10^{15}\,\text{m/s}^2\right)}{1.60\times10^{-19}\,\text{C}}\)
\(E\approx1.0\times10^4\,\text{N/C}\)
\(\boxed{E=1.0\times10^4\,\text{N/C}}\)
(c)
For parallel plates,
\(E=\dfrac{Q}{\varepsilon_0 A}\)
Solve for \(Q\):
\(Q=\varepsilon_0AE\)
\(Q=\left(8.85\times10^{-12}\,\text{C}^2/\text{N}\cdot\text{m}^2\right)\left(0.25\,\text{m}^2\right)\left(1.0\times10^4\,\text{N/C}\right)\)
\(Q=2.2\times10^{-8}\,\text{C}\)
\(\boxed{Q=2.2\times10^{-8}\,\text{C}}\)
(d)(i)
The path should curve smoothly from the bottom plate to point \(X\), with no sharp corner.
The magnetic force is always perpendicular to the velocity of the electron. A force that is always perpendicular to the velocity changes the direction of the velocity but not the speed, producing a curved, circular path.
\(\boxed{\text{The electron follows a curved path because the magnetic force is centripetal.}}\)
(d)(ii)
As the electron enters the magnetic field, its velocity is downward. To curve toward point \(X\), the magnetic force must initially point to the right.
For a positively charged particle moving downward, a magnetic field out of the page would produce a force to the left. However, the electron is negatively charged, so the force direction is opposite.
Therefore, for the electron to experience a force to the right, the magnetic field must be directed out of the page.
\(\boxed{\text{The magnetic field is directed out of the page.}}\)