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Electric Potential AP Physics 2 FRQ

Electric Potential AP  Physics 2 FRQ – Exam Style Questions etc.

Electric Potential AP  Physics 2 FRQ

Unit 10: Electric Force, Field, and Potential

Weightage : 15–18%

AP Physics 2 Exam Style Questions – All Topics

Exam Style Practice Questions, Electric Potential AP  Physics 2 FRQ

Question 

(10 points, suggested time 20 minutes)
Two particles, 1 and 2, have different mass and charge as described by the following.

• Particle 1 has mass M and negative charge −Q.
• Particle 2 has mass \(\frac{M}{2}\) and positive charge +2Q.

In separate trials, a device is used to accelerate each particle in the −y-direction from rest through a potential difference of absolute value ΔV . The polarity of the potential difference can be adjusted so that a particle with either positive charge or negative charge can be accelerated in the −y-direction by the device. Gravitational effects are negligible.

After moving through the potential difference, particles 1 and 2 exit the device with kinetic energies \(K_{1}\) and \(K_{1}\), respectively

(a) Calculate the ratio \(\frac{K_{2}}{K_{1}}\)

After exiting the device, the particles enter a large region of constant uniform magnetic field of magnitude \(b_{0}\) that is directed in the +z-direction (out of the page), as shown in Figure 1. Each particle is moving in
the −y-direction when entering the region, and each particle is moving in the +y-direction when exiting the region

(b) i. Determine an expression for the speed of Particle 2 in the region. Express your answer in terms of M , \(K_{2}\) , and physical constants, as appropriate.
ii. Derive an expression for the horizontal distance Δx between the locations where Particle 2 enters and leaves the region. Express your answer in terms of M , Q, \(K_{2}\) , \(B_{0}\) , and physical constants, as appropriate

(c) On the following diagram in Figure 2, sketch and clearly label the paths of both particles 1 and 2 in the region.

(d) A uniform electric field is added to the region such that Particle 1 of negative charge −Q travels with constant speed in a straight line through the region. Determine the direction of the electric field.

▶️Answer/Explanation

1(a)Example Response

\(E_{0} =E_{f}\)

\(\Delta U + \Delta V = 0\)

\(-\Delta U_{E} = \Delta K\)

\(\left|q\Delta V \right| = K\)

\(K_{1}=\left|-Q\Delta V \right| = Q\Delta V\)

\(K_{2}=\left|+2Q\Delta V \right| = 2Q\Delta V\)

\(\frac{K_{2}}{K_{1}} = \frac{2Q\Delta V}{Q\Delta V}\)

\(\frac{K_{2}}{K_{1}} = 2\)

1(b)Example Response 

\( \vec{a} = \frac{\sum \vec{f}}{m}\)

\(\vec{a_{c}} = \frac{q\vec{v} * \vec{B}}{m}\)

\(\frac{^{v2}}{r} =\frac{qvB}{m}\)

\(r =\frac{mv}{qB}\)

\(r =\frac{\left ( \frac{M}{2} \right )\left ( \sqrt[2]{\frac{k_{2}}{m}} \right )}{2QB_{0}}\)

\(r =\frac{\sqrt{K_{2}}M}{2QB_{0}}\)

\(\Delta x=2r\)

\(\Delta x =\frac{\sqrt{K_{2}}M}{2QB_{0}}\)

1(c)Example Response 

1(d)Example Response

+x -direction

Question

Three point charges produce the electric equipotential lines shown on the diagram above.
a. Draw arrows at points L, N. and U on the diagram to indicate the direction of the electric field at these points.
b. At which of the lettered points is the electric field E greatest in magnitude? Explain your reasoning.
c. Compute an approximate value for the magnitude of the electric field E at point P.
d. Compute an approximate value for the potential difference, VM – VS, between points M and S.
e. Determine the work done by the field if a charge of +5 × 10–12 coulomb is moved from point M to point R.
f. If the charge of +5 × 10–12 coulomb were moved from point M first to point S, and then to point R, would the answer to e. be different, and if so, how?

Answer/Explanation

Ans:

a. 

     The field lines point perpendicular to the equipotential lines from high to low potential.
b. The magnitude of the field is greatest at point T because the equipotential lines are closest together, meaning ΔV has the largest gradient, which is related to the strength of the electric field.
c. E = ΔV/d = (10 V)/(0.02 m) = 500 V/m
d. VM – VS = 40 V – 5 V = 35 V
e. W = –qΔV and ΔV = – 10 V which gives W = 5 ×10–11 J
f. The work done is independent of the path so the answer would be the same.

Question

Consider the electric field diagram above.
a. Points A, B, and C are all located at y = 0.06 m .
i. At which of these three points is the magnitude of the electric field the greatest? Justify your answer.
ii. At which of these three points is the electric potential the greatest? Justify your answer.
b. An electron is released from rest at point B.
   i. Qualitatively describe the electron’s motion in terms of direction, speed, and acceleration.
   ii. Calculate the electron’s speed after it has moved through a potential difference of 10 V.
c. Points B and C are separated by a potential difference of 20 V. Estimate the magnitude of the electric field midway between them and state any assumptions that you make.
d. On the diagram, draw an equipotential line that passes through point D and intersects at least three electric field lines.

Answer/Explanation

Ans:

a. i. The magnitude of the field is greatest at point C because this is where the field lines are closest together.
ii. The potential is greatest at point A. Electric field lines point from high to low potential.
b. i. The electron moves to the left, against the field lines. As the field gets weaker the electron’s acceleration to the left decreases in magnitude, all the while gaining speed to the left.
ii. W = qΔV = ½ mv2 gives v = 1.9 × 106 m/s
c. If we assume the field is nearly uniform between B and C we can use E = ΔV/d where the distance between B and C d = 0.01 m giving E = 20 V/0.01 m = 2000 V/m
d.

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