42 practice MCQs on Coulomb's law, electric field and potential, Gauss's law and capacitors. Every question has the correct answer and an explanation.
Start free — 20 questionsElectrostatics is part of the MDCAT Physics syllabus. This page collects Electrostatics MCQs in the style of past MDCAT papers — below are 12 of them with answers and explanations, and the full set of 42 questions is in the HighYield MDCAT QBank.
This chapter covers: Coulomb's law, electric field and potential, Gauss's law and capacitors.
Correct answer: B. 5 x 10-6 ampere second
A charge of 5 microcoulombs, expressed in base SI units, is 5×10⁻⁶ ampere-seconds, since one coulomb is defined as one ampere flowing for one second.
Correct answer: C. Capacitance becomes four times
Since capacitance C = ε0A/d, doubling the area (A) and halving the separation (d) both increase capacitance, together making it four times the original value.
Correct answer: A. Double
Since capacitance C = ε0A/d, doubling both the plate dimensions (length and width, so area quadruples) and the separation (d doubles) gives an overall factor of 4/2 = 2, so capacitance doubles.
Correct answer: C. 10
A capacitor connected to an AC source charges and discharges (effectively 'saturating' and reversing) twice per cycle, so over 5 complete cycles it does so 10 times.
Correct answer: B. Faraday
Michael Faraday, a British scientist, introduced the concept of an electric field in the early 19th century. He used the concept to describe the influence that a charged object exerts on its surroundings, even in the absence of other charges.
Correct answer: C. 6 sec
The time constant of an RC circuit is τ = RC = 3000 Ω × 2×10⁻³ F = 6 seconds.
Correct answer: D. nature of medium and system of units
Coulomb's constant, denoted by k, is a fundamental constant in electrostatics that determine the strength of the electrostatic force between two charged particles. It is expressed as: k = 1 / (4πεo) Where εo is the permittivity of free space. The value of k depends on both the nature of the medium and the system of units being used. The nature of the medium affects the permittivity, which is a measure of how easily electric fields can propagate through a material. Similarly, the system of units used can also affect the value of Coulomb's constant.
Correct answer: B. Double layer
To store electric charge, ultra-capacitors use the double-layer effect.
Correct answer: C. Product of the magnitude of charges
Option C is correct. Coulomb's law states that the force is directly proportional to the product of the magnitude of charges (q1 * q2).
Correct answer: D. 0 V
At the midpoint between two equal and opposite charges, the electric potential contributions from each charge are equal in magnitude but opposite in sign. Therefore, they cancel each other out, resulting in a total electric potential of 0 V at that point. This is why the correct answer is 0 V. The other options suggest a non-zero potential, which does not consider the crucial fact that the charges are equal and opposite, leading to a complete cancellation of their effects at the midpoint.
Correct answer: C. Coulomb's law
Coulomb's law states that the magnitude of the electrostatic force (F) between two point charges (q1 and q2) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (r) between them:F ∝ (q1 × q2) / r²orF = k × (q1 × q2) / r²where k is Coulomb's constant.
Correct answer: A. Volt
The SI unit of electric potential, also known as voltage, is the volt (V). It is defined as the potential difference required to produce a current of one ampere through a resistance of one ohm.
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