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Test: Electrostatics - MCAT MCQ


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10 Questions MCQ Test Physics for MCAT - Test: Electrostatics

Test: Electrostatics for MCAT 2024 is part of Physics for MCAT preparation. The Test: Electrostatics questions and answers have been prepared according to the MCAT exam syllabus.The Test: Electrostatics MCQs are made for MCAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Electrostatics below.
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Test: Electrostatics - Question 1
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After MgCl2 dissolves in a neutrally charged solvent, what is the net charge of the solution?

Detailed Solution for Test: Electrostatics - Question 1
  • The law of conservation of charge states that the net charge of an isolated system remains constant.
  • When the ions are added to a neutrally charged solvent like water, the overall solution remains the same.
  • MgCl2 dissolves into one +2 cation and two -1 anions.
  • The only way to change the net charge of a system is to introduce a charge from elsewhere, or to remove a charge from the system. In this closed system, the resulting Mg2+ cation and two Cl− anions will yield an overall neutral charge.
Test: Electrostatics - Question 2
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If one sphere has a mass of 2 kg and an electric charge of +4 μC while another sphere, 3 meters away, has a mass of 1 kg and an electric charge of +5 μC, what is the acceleration of the more massive sphere?

Detailed Solution for Test: Electrostatics - Question 2
  • In order to determine the acceleration of an object, we must first determine the magnitude and direction of forces acting on the object.
  • Since like charges repel, the more massive sphere experiences an electric force in the direction away from the less massive sphere. It also experiences a gravitational force attracting it to the less massive sphere.
  • The gravitational force is much weaker than electric force by many orders of magnitude, thus it’s effect on the more massive sphere is negligible.
  • The electric force felt by the massive sphere can be determined by Coulomb’s Law:
  • In order to determine the acceleration, we must divide the net force by the mass of the object:
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Test: Electrostatics - Question 3
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In the figure below, spheres q1 ,q2 , and q3 have the same positive charge magnitude. Which of the following is a possible direction for the net electric field vector on q4?

Detailed Solution for Test: Electrostatics - Question 3
  • The superposition principle means that the net electric field vector is the sum of all individual electric field vector arrows made by each charge’s interaction with the charge being affected.
  • If q1 and q2 have the same positive charge magnitude, then their electric field vector arrows should cancel out in the horizontal direction. This will occur whether q4 has a positive or negative charge.
  • Since there is no other charge to influence the direction of the net electric field vector in the horizontal direction, the correct vector direction should not be angled toward the left or the right. An arrow that points straight up or straight down are possible directions for the net electric field vector.
Test: Electrostatics - Question 4
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The magnitude of electric field experienced by a charge at a certain distance from a source charge is equal to 64 N/C. What will be the magnitude of the electric field at four times that distance and with a source charge half as strong?
Detailed Solution for Test: Electrostatics - Question 4
  • The equation for determining the electric field is  Where r is the distance between the charges, Q is the source charge, and k0 is Coulomb's constant.
  • If we use the equation for determining the electric field and change the value of r to 4r and change the value of  we obtain:
  • Simplifying our formula, we obtain:

    Indicating that the adjusted electric field will be 1/32 the magnitude of the original electric field.
  • The original electric field was 64 N/C, which means that the new adjusted electric field will be 64/32 = 2N/C.
Test: Electrostatics - Question 5
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In five seconds, a point charge moves from position A to position B as shown in the diagram below. Which one of the following two paths require the most energy? 
Detailed Solution for Test: Electrostatics - Question 5
  • The amount of time is not relevant to this problem.
  • In electrostatics, energy is determined by product of the point charge magnitude, the displacement of the point charge, and the value of the electric field, for a constant field in the direction of displacement:
    W = Fd
    W = qEd
  • When the displacement is the same for both paths of a point charge along a constant electric field, the energy expended is equivalent.
Test: Electrostatics - Question 6
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A constant electric field moves a positive point charge from one position to another position so that the change in electric potential is −450 V. Which of the following must increase during this movement?
Detailed Solution for Test: Electrostatics - Question 6
  • According to the question stem, there is an electric potential difference between the two positions.
  • The negative change in electric potential indicates that the point charge moves from a position of higher electric potential to a position of lower electric potential.
  • The point charge displacement as a result of a constant electric field indicates that the point charge moves to a more favorable, downstream position. This indicates that the charge loses potential energy and gains kinetic energy during the movement.
  • Just like a ball dropped from a height due to the force of gravity, the point charge moves to its new location due to the force generated by the electric field on the charge. In both situations, the object being moved will increase in velocity during its movement.
Test: Electrostatics - Question 7
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The figure below shows a source charge Q. Which of the following can best describe the electric potential difference as a positive point charge is moved from position A to position B through the indicated path?
Detailed Solution for Test: Electrostatics - Question 7
  • The path does not matter when determining the change in electric potential difference of a point charge.
  • The electric potential difference with respect to infinity of a charge Q, or voltage, at a point is determined by the formula  Where Q is the value of a source charge and r is  the distance between the source charge and the point charge.
  • At both points, the k0 constant and value of source charge Q is the same. The only difference is the distance, r, between the test charge and the source charge.
  • Since the distance r is larger at point B, the electric potential difference with respect to inifinity is smaller.
  • The electric potential difference from A to B is this smaller V amount, at position B, subtracted by the original larger 
  • V at position A. The resulting electric potential difference is negative.
Test: Electrostatics - Question 8
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The magnitude of the electric force experienced by a point charge a distance d away from a source charge is affected by which of the following?
Detailed Solution for Test: Electrostatics - Question 8
  • The sign of the source charge affects only the direction of the electric force (either towards or away from a source charge), it does not affect the magnitude.
  • Similarly, the sign of the point charge affects only the direction of the electric force and not its magnitude.
  • The mass of an object is important for determining gravitational force, not electrostatic force. The electric force of a single point charge q at a distance d away from a source charge Q is given by the equation  The signs of both charges will only affect the overall sign on the force value. The distance d is the only answer choice that affects the magnitude of the electric force.
Test: Electrostatics - Question 9
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Sphere A and B have the same mass and charge magnitudes. They are placed a distance r away from each other so that gravitational forces and electrostatic forces allow both spheres to be in equilibrium. What would occur if sphere A doubled in charge?
Detailed Solution for Test: Electrostatics - Question 9
  • In order for the two spheres to be in equilibrium, the forces on each sphere must cancel out. Drawing a free-body diagram will visually demonstrate the forces on each sphere.
  • The gravitational force can only act to attract both spheres toward each other. Therefore the electrostatic force in this situation must repel the spheres away from each other.
  • The formula for electrostatic force is  Doubling the charge of sphere A will double the repulsive electrostatic force but not change the attractive gravitational force measurably since the added charge would have such a small mass.
  • The electrostatic force experienced by each sphere is the same because the electrostatic force equally accounts for the source charge as well as the point charge in the determination of its magnitude. Therefore, the spheres will move away from each other with the same force.
Test: Electrostatics - Question 10
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In the diagram below, which of the following statements is true about a point charge in position A, which is midway between two positive source charges of equal magnitude?
Detailed Solution for Test: Electrostatics - Question 10
  • The electric potential at a point charge is the sum of each individual source charge’s contribution. Since both source charges are positive, they contribute a positive electric potential difference at point A.
  • The electric field arrows of each source charge’s contribution to the point charge is in the direction away from the source charge. Since the two source charges are in opposing directions, are equidistant from the point charge, and are equal in magnitude, the electric field vectors cancel out, thus electric field at position A is 0.
  • If the electric field is 0, then there is no electrostatic force moving the point charge. If an object is not displaced, no work is being done.
  • Since the point charge is equally repelled by both source charges, the net electrostatic force experienced by the charge at that point must be 0.
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