Conservation of Electric Charge and the Process of Charging AP Physics 2 FRQ – Exam Style Questions etc.
Conservation of Electric Charge and the Process of Charging AP Physics 2 FRQ
Unit 10: Electric Force, Field, and Potential
Weightage : 15–18%
Exam Style Practice Questions, Conservation of Electric Charge and the Process of Charging AP Physics 2 FRQ
Question
(10 points, suggested time 20 minutes)
Particles A and B each have positive charge +Q and are held fixed at two vertices of an equilateral triangle of side length d, as shown. Point P is located equidistant from each vertex of the triangle.
Students Y and Z discuss the electric field and the electric potential at Point P after a third charged particle is placed at the bottom-right vertex. The students make the following statements.
Student Y: “If a particle with positive charge +2Q is placed at the bottom-right vertex, the magnitude of the electric field will be zero at Point P.”
Student Z: “To make the value of the electric potential zero at Point P, a particle with negative charge −Q should be placed at the bottom-right vertex.”
(a) In a coherent, paragraph-length response, evaluate the accuracy of each student’s statement. If any aspect of either student’s statement is inaccurate, explain how to correct the student’s statement. Support your evaluations using appropriate physics principles.
(b) Particles A and B are once again held in place at two vertices of the equilateral triangle. The students want to represent the electric potential energy of a system of particles when a third charged particle is brought from very far away to the bottom-right vertex. Scenarios 1 and 2 are considered.
i. In Scenario 1, a third particle with positive charge +Q is moved from very far away to the bottom-right vertex and then held in place. A bar is shown on the following chart that represents the electric potential energy \(U_{i1}\) of the system consisting of all three particles when the third particle with positive charge is very far away from the other particles. In the grid provided, complete the bar chart.
• Draw a bar to represent the work \(W_{1}\) required to move the third particle with positive charge from very far away to the bottom-right vertex.
• Draw another bar to represent the electric potential energy \(U_{f1}\) of the system consisting of all three particles when the third particle with positive charge is held in place at the bottom-right vertex.
The height of each bar should be proportional to the energy represented. If the quantity is zero, write a “0” in that column.
ii. In Scenario 2, a particle with negative charge −Q is moved from very far away to the bottom-right vertex and then held in place. A bar is shown on the following chart that represents the electric potential energy \(U_{i2}\) of the system consisting of all three particles when the particle with negative charge is very far away from the other particles. In the grid provided, complete the bar chart.
• Draw a bar to represent the work \(@_{2}\) required to move the particle with negative charge from very far away to the bottom-right vertex.
• Draw another bar to represent the electric potential energy \(U_{f2}\) of the system consisting of all three particles when the particle with negative charge is held in place at the bottom-right vertex.
The height of each bar should be proportional to the energy represented. If the quantity is zero, write a “0” in that column.
▶️Answer/Explanation
1(a) Example Response
Student Y is incorrect. Before the third particle is placed at the bottom-right vertex, the electric field from particles A and B at Point P is down and to the right. The electric field from a positively charged particle placed at the bottom-right vertex is up and to the left. The third particle needs to have charge +Q , rather than +2Q , in order to have the correct magnitude to make the resultant field zero at Point P .
Student Z is incorrect. Before the third particle is placed at the bottom-right vertex, the value of the electric potential at Point P is positive. Because Point P is equidistant from all three particles, the electric potential at Point P is proportional to the total charge of the system. If the total charge of the system is zero, the electric potential at Point P will be zero. This requires the third particle to have charge −2Q .
OR
Student Y is incorrect that a particle with charge +2Q placed at the bottom-right vertex will result in no electric field at Point P . The horizontal component of the electric field from Particle A is less than the horizontal component of the electric field from the particle with charge +2Q . The sum of the vertical components of the fields from particles A and B is less than the vertical component of the field from the particle with charge +2Q . Therefore, the resulting electric field at Point P is nonzero and points in a direction between particles A and B.
Student Z is incorrect. Before the third particle is placed at the bottom-right vertex, the value of the electric potential at Point P is positive. Electric potential is a scalar quantity, so if the third particle has charge −2Q rather than −Q , the electric potential at Point P will be zero.
(b)(i) Example Response
(b)(ii) Example Response
Question
Object I, shown above, has a charge of + 3 x 10–6 coulomb and a mass of 0.0025 kilogram.
a. What is the electric potential at point P, 0.30 meter from object I?
Object II, of the same mass as object I, but having a charge of + 1 x 10–6 coulomb, is brought from infinity to point P, as shown above.
b. How much work must be done to bring the object II from infinity to point P?
c. What is the magnitude of the electric force between the two objects when they are 0.30 meter apart?
d. What are the magnitude and direction of the electric field at the point midway between the two objects?
The two objects are then released simultaneously and move apart due to the electric force between them. No other forces act on the objects.
e. What is the speed of object I when the objects are very far apart?
Answer/Explanation
Ans:
a. V = kQ/r = 9 × 104 V
b. W = qΔV (where V at infinity is zero) = 0.09 J
c. F = kqQ/r2 = 0.3 N
d. Between the two charges, the fields from each charge point in opposite directions, making the resultant field the difference between the magnitudes of the individual fields.
E = kQ/r2 gives EI = 1.2 × 106 N/C to the right and EII = 0.4 × 106 N/C to the left
The resultant field is therefore E = EI – EII = 8 × 105 N/C to the right
e. From conservation of momentum mIvI = mIIvII and since the masses are equal we have vI = vII. Conservation of energy gives U = K = 2(½ mv2) = 0.09 J giving v = 6 m
Question
A scientist constructs a device shown in the figure. Region I consists of a charged parallel plate capacitor with vertical plates separated by a distance x. The left plate has a small opening that accelerated particles can pass through. Region II has two large horizontal capacitor plates with a separation of y and a magnetic field created by current-carrying solenoid coils, which are not pictured in the figure. The magnetic field is directed upward out of the page. In an experiment a single proton P is placed at the launch point near the right plate in Region I. The proton accelerates to the left through the hole and continues on a straight path through Region II as seen in the figure.
(A) What is the direction of the electric field in Region I? Justify your answer.
(B) Which capacitor plate has the higher potential in Region II? Justify your answer.
(C) The experiment is repeated, replacing the proton with an alpha particle that has a mass approximately four times larger than the proton and a charge two times larger than the proton. The alpha particle is released from the launch point and passes through the hole in the left plate. Compare the motion of the alpha particle to the motion of the proton through Regions I and II. Justify your reason mg. The Region I capacitor plates have a potential difference of 5400 V and a plate separation x of 0.14 m. The Region II capacitor plates have a separation y of 0.060 m and a magnetic field of 0.50 T.
(D) A proton is again placed at the launch point near the right plate in Region I. Derive an algebraic expression for the velocity of the proton as it passes through the hole in the left plate. Use your expression to calculate the numerical value of the velocity.
(E) Using your work from (D), derive an algebraic expression for the potential difference that must be applied to the capacitor in Region II so that the proton 1noves in a straight line through the region. Use your expression to calculate the numerical value of the potential difference.
(F) Keeping the potential difference the same as the calculated value from (E), the scientist places an unknown particle at the launch point and observes that it travels straight through both Regions I and II just as the proton did. Discuss what the scientist can deduce about the unknown particle. Justify your answer, making appropriate reference to the algebraic expressions derived in (D) and (E).
Answer/Explanation
Ans:
Part (A)
Electric field is to the left. The electric field points in the direction of force on a positive charge. The proton accelerates to the left; therefore, the force is to the left and the electric field is also to the left.
Part (B)
The upper plate must have a higher potential. Utilizing the right-hand rule for forces on moving positive charges, and showing that the proton receives a magnetic force upward from the magnetic field. The electric force is equal and opposite in direction to the magnetic force so that the proton will travel in a straight line. Electric fields point from higher potential to lower potential. Thus, the upper plate is a higher potential.
Part (C)
The alpha particle will curve downward in Region II. The alpha particle will exit Region I with a slower velocity. The electric potential energy has doubled, but the mass has quadrupled leading to a smaller exit velocity. (Or, in Region I the electric force on the charge has doubled, but the mass has quadrupled, leading to a smaller acceleration and exit velocity.) The electric force will now be larger than the magnetic force causing the alpha particle to arc downward. The electric force will double as the change doubles: \(B_{E}=q_{E}\). The magnetic force will not increase as much as the electric force. Even though the charge doubles, the velocity is now slower: \(P_{M}=qvB\).
Part (D)
Conservation of energy: \(AR_{E}=K\)
Part (E)
To move in a straight line: \(B_{E}=P_{M}\)
Magnetic force equal to electric force: Eq=qvB
\(E=\frac{\Delta V}{y}=vB\)
Equations with substitutions:
AK= yvB — (0.060)(1.02 x \(10^{6}\) m/s)(0.50 T)
AK= 30,600 V
Part (F)
From the equation derived in (E) AK= yvB, we know that the velocity of the particle is the same because AK, y, and B are all the same.
From the equation in (D) \(v=\sqrt{\frac{2\Delta Vq}{m}}\), we know that both v and AT are the same. Thus, we can deduce the charge to mass ratio q/m of the unknown particle.