﻿ ﻿Practice Questions - Fluids - MCAT Physics and Math Review

## Chapter 5: Electrostatics and Magnetism

### Practice Questions

1.

1. Questions 1–3 refer to the figure below, in which F represents the electrostatic force exerted on charged particle S by charged particle R.

2. 3. In the figure, the magnitude of the electric force on R due to S is:

1. 2. F.

3. 2F.

4. 4F.

4. If the distance between the centers of the spheres is halved, the magnitude of the force on S due to R will be:

1. 2. 3. 2F.

4. 4F.

5. Assume the direction of F is the same direction as the electric field between R and S. If an electron were placed midway between R and S, the resultant electric force on the electron would be:

1. toward R.

2. toward S.

3. upward in the plane of the page.

4. downward in the plane of the page.

2. If the electric field at a distance r away from charge Q is what is the ratio of the electric fields at r, 2r, and 3r?

1. 9:3:1

2. 36:9:4

3. 36:18:9

4. 36:18:12

3. A positive charge of +Q is fixed at point R a distance d away from another positive charge of +2Q fixed at point S. Point A is located midway between the charges, and point B is a distance from +2Q, as shown below. In which direction will a positive charge move if placed at point A and point B, respectively? 1. Toward the +Q charge for both

2. Toward the +2Q charge for both

3. Toward the +Q charge at point A, and toward the right at point B

4. Toward the +2Q charge at point A, and toward the right at point B

4. Two parallel conducting plates are separated by a distance d. One plate carries a charge +Q and the other carries a charge −Q. The voltage between the plates is 12 V. If a +2 μC charge is released from rest at the positive plate, how much kinetic energy does it have when it reaches the negative plate?

1. 2.4 × 10–6 J

2. 4.8 × 10–6 J

3. 2.4 × 10–5 J

4. 4.8 × 10–5 J

5. The negative charge in the figure below (−1 μC) moves from y = −5 to y = +5 and is made to follow the dashed line. What is the work required to move the negative charge along this dashed line? 1. –10 J

2. −5 J

3. 0 J

4. 10 J

6. If the magnetic field a distance r away from a current-carrying wire is 10 T, what will be the net magnetic field at r if another wire is placed a distance 2r from the original wire and has a current twice as strong flowing in the opposite direction?

1. 0 T

2. 15 T

3. 20 T

4. 30 T

7. Given an electric dipole, the electrical potential is zero:

1. only at the midpoint of the dipole axis.

2. anywhere on any perpendicular bisector of the dipole axis and at infinity.

3. anywhere on the dipole axis.

4. only for points at infinity.

8. An electron is accelerated over a distance d by an electrical potential V. The electrical potential applied to this electron is then increased by a factor of 4 and the electron is accelerated over the same distance d. The speed of the electron at the end of the second trial will be larger than at the end of the first trial by a factor of:

1. 16.

2. 8.

3. 4.

4. 2.

9. Which of the following accurately depicts the field lines created by a proton that is moving toward the right on this page?

1. 2. 3. 4. 10.A certain 9 V battery is used as a power source to move a 2 C charge. How much work is done by the battery?

1. 4.5 J

2. 9 J

3. 18 J

4. 36 J

11.A proton and an alpha particle (a helium nucleus) repel each other with a force of F while they are 20 nm apart. If each particle combines with three electrons, what is the magnitude of the new force between them?

1. 9F

2. 3F

3. F

4. 12.A moving negative charge placed in an external magnetic field circulates counterclockwise in the plane of the paper as shown. In which direction is the magnetic field pointing?

1. Into the page

2. Out of the page

3. Toward the center of the circle

4. Tangent to the circle

13.A dipole is placed in an electric field and is allowed to come to equilibrium. How would the dipole react if the direction of the electric field is suddenly reversed?

1. It rotates to align with the new field.

2. It accelerates linearly along the field lines.

3. It experiences no rotational or linear movement.

4. It both rotates to align with the new field and accelerates linearly along the field lines.

PRACTICE QUESTIONS

1. BAccording to Newton’s third law, if R exerts a force on S, then S exerts a force with equal magnitude but opposite direction back on R. Therefore, the magnitude of the force on R due to S is F.

2. D

The force is inversely proportional to r2. Cutting the distance in half will therefore multiply the force by 22, making it four times its original value: 3. BAn electric field’s direction at a given point is defined as the direction of the force that would be exerted on a positive test charge in that position. Because electrons are negatively charged particles, they will therefore feel a force in the opposite direction of the electric field’s vector. In this case, because the force points to the left (toward R), an electron will feel a force pointing to the right (toward S) if E is in the same direction as F.

4. B

The first step in answering this question is to remember that the magnitude of the electric field is inversely proportional to the square of the distance: Therefore, if the electric field at radius r, Er, is then the electric field at radius 2r will be Similarly, the electric field at radius 3r is equal to Therefore, the ratio of Er:E2r:E3r is 36:9:4.

5. CA positive charge placed at A will experience two forces: a force to the left due to +2Q and a force to the right due to +Q. Because point A is the same distance from +Q and +2Q, the force due to +2Q will be larger than that due to +Q, and there will be a net force to the left (toward +Q). At point B, the forces from both +Q and +2Q will point to the right, so there will be a net force to the right.

6. CRecall that the change in potential energy, ΔU, and the change in potential, ΔV, are related by W = ΔU = qΔV. Therefore, ΔU = (2 × 10–6 C) × (–12 V) = –2.4 × 10–5 J. The positive charge is moving from the positive to the negative plate, and is therefore decreasing in potential energy; this is reflected by the fact that the voltage is –12 V rather than +12 V. The potential energy that is lost is converted into kinetic energy, so the charge must gain 2.4 × 10–5 J of kinetic energy.

7. CThere will be work done in moving the negative charge from its initial position to y = 0. However, in moving the negative charge from y = 0 to the final position, the same amount of work is done but with the opposite sign. This is because the force changes direction as the electron crosses y = 0. Therefore, the two quantities of work cancel each other out. This argument depends crucially on the symmetry of the initial and final positions.

8. D

The safest way to answer this question is to quickly draw a diagram: Notice right away that between the two wires, the direction of the magnetic field is the same: into the page. Therefore, because the vector direction is the same, we can just focus on the magnitudes of the two magnetic fields. We know that B1 = 10 T at a distance r. Consider the relationships in the equation Magnetic field and current are directly proportional, whereas magnetic field and distance are inversely proportional. Therefore, doubling the current will result in double the magnetic field as the first wire, or 20 T. The overall magnitude of the magnetic field is 10 T + 20 T = 30 T into the page.

9. B

Potential is a scalar quantity. The total potential is the sum of the potentials of the positive and negative charges: where r+ and r represent the distances from the positive and negative charge, respectively. The sum of these terms will be zero at any point where r+ = r. This will be at any point along the perpendicular bisector of the dipole axis, as well as at infinity.

10.DThe electrical potential (V) is equal to the amount of work done (W) divided by the test charge (q). This means that the potential is directly proportional to the amount of work done, which is equal to the amount of energy gained by the particle; therefore, the overall amount of energy increases by a factor of 4. Because energy is directly proportional to the square of the speed (according to ), the speed must increase by a factor of 2.

11.DYou should know that the field lines for a positively charged particle will always point away from the particle in a radial pattern, regardless of the direction in which the particle is moving. This is because field lines point in the direction a positive test charge would move in that field (that is, the direction that a force would be exerted on a positive test charge in that field).

12.CVoltage (ΔV) is equal to the quotient of the amount of work done (W) divided by the charge of the particle on which the work is done (q), according to the equation Because the voltage equals 9 V and the charge equals 2 C, the work done must equal 9 V × 2 C = 18 J.

13.CThe magnitude of the electrostatic force is given by the equation Because the distance does not change during the interaction in the question, the value of r is irrelevant to the answer. Currently, q1 and q2 are equal to +1e and +2e, respectively;the addition of three electrons (each of which carries a charge of −e) will change the charges to −2e and −1e. Therefore, the product q1q2 before the interaction is equal to the product q1q2 after the interaction (+2e2). Because k and r also remain constant in this system, the magnitude of the force does not change.

14.BThis problem is an application of the right-hand rule. The velocity vector v is always tangent to the circle. The magnetic force must always point radially toward the center of the circle. Use the point labeled in the diagram to set up the right-hand rule. Your thumb points up the page, tangent to the circle at this point. The back of your hand, which represents the force on a negative charge, points to the left, radially toward the center of the circle. Your fingers must point out of the page to get your hand into this position. Therefore, the direction of the magnetic field must be out of the page.

15.CTorque is a function of both force applied and the angle at which it is applied. A dipole placed in an electric field will experience a torque until it comes to rest oriented within the field, at which point the angle between the plane of the dipole and the electric field is 0°. Once this point is reached, inverting the electric field has no impact on the dipole because it will now have an angle of 180°, the sine of which is still 0.

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