MDCAT Physics chapter 8: Physics · Electrostatics MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
18 exam-style questions on Physics · Electrostatics, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.
Ten Physics · Electrostatics questions with answers
Try each one, then open the answer. These ten stay the same; the test draws new ones every time.
1. An oil drop of mass 3.2 × 10−15 kg is held stationary between two horizontal plates 1.0 cm apart with a potential difference of 500 V across them. Taking g = 10 m s−2 and e = 1.6 × 10−19 C, how many excess electrons does the drop carry?
A
40
B
4
C
400
D
2
Show answer
Answer: B. For balance qE = mg, with E = V/d = 500/0.010 = 5.0 × 104 V m−1. So q = (3.2 × 10−15 × 10) ÷ (5.0 × 104) = 6.4 × 10−19 C = 4e. Leaving d in centimetres gives 400.
2. Three identical capacitors connected in parallel have an equivalent capacitance of 27 μF. If the same three are connected in series, the equivalent capacitance is
A
9 μF
B
3 μF
C
81 μF
D
1 μF
Show answer
Answer: B. In parallel, capacitances add, so each capacitor is 27/3 = 9 μF. In series 1/C = 3/9, so C = 3 μF, smaller than any one of them.
3. An electron (charge −1.6 × 10−19 C) is in a uniform electric field of 5.0 × 104 N C−1. The electric force on it is
A
8.0 × 10−15 N, along the field
B
3.2 × 10−24 N, opposite to the field
C
8.0 × 10−14 N, opposite to the field
D
8.0 × 10−15 N, opposite to the field
Show answer
Answer: D. F = qE = 1.6 × 10−19 × 5.0 × 104 = 8.0 × 10−15 N. The field direction is the direction of the force on a positive charge, so the force on the negative electron is opposite to the field.
4. The separation between the plates of an air-filled parallel-plate capacitor of capacitance C is halved, and the gap is then completely filled with a dielectric of relative permittivity 3. The new capacitance is
A
3C/2
B
2C/3
C
6C
D
C/6
Show answer
Answer: C. C = εrε0A/d: halving d doubles C and the dielectric multiplies it by 3, giving 6C. 3C/2 wrongly treats C as proportional to d.
5. Two point charges of +2.0 μC and +8.0 μC are 0.20 m apart in a vacuum. Taking k = 9 × 109 N m2 C−2, the force between them is
A
3.6 N
B
0.72 N
C
36 N
D
0.36 N
Show answer
Answer: A. F = kq1q2/r2 = 9 × 109 × 2.0 × 10−6 × 8.0 × 10−6 ÷ 0.040 = 0.144 ÷ 0.040 = 3.6 N (a repulsion). Dividing by r instead of r2 gives 0.72 N.
6. Two small identical spheres each carry a charge +q and repel each other with a force F. Half of the charge on one sphere is moved to the other, and their separation is unchanged. The force between them becomes
A
3F/4
B
F
C
F/4
D
3F/2
Show answer
Answer: A. The charges become q/2 and 3q/2, so their product is 3q2/4 and the force is 3F/4. The total charge is unchanged, but the force depends on the product of the charges, not on their sum.
7. An alpha particle (charge +2e), starting from rest, is accelerated through a potential difference of 1000 V. The kinetic energy it gains is
A
1000 eV
B
500 eV
C
4000 eV
D
2000 eV
Show answer
Answer: D. Energy gained = qV = 2e × 1000 V = 2000 eV (3.2 × 10−16 J). 1000 eV is what a singly charged particle such as a proton would gain.
8. Point charges +q and −q are fixed a distance d apart. A third charge +q is placed at the mid-point between them. The net force on the third charge is
A
zero, as the two forces cancel
B
4kq2/d2, towards the −q charge
C
8kq2/d2, towards the +q charge
D
8kq2/d2, towards the −q charge
Show answer
Answer: D. Each fixed charge is d/2 away, so each force is kq2/(d/2)2 = 4kq2/d2. The +q pushes the third charge towards −q and the −q pulls it the same way, so the forces add to 8kq2/d2. They would cancel only if the two fixed charges were alike.
9. A 2 μF capacitor is charged to 100 V and disconnected from the supply. It is then connected across an uncharged 2 μF capacitor. The final potential difference across each capacitor is
A
100 V
B
50 V
C
200 V
D
25 V
Show answer
Answer: B. The charge Q = CV = 2 × 100 = 200 μC is conserved and is now spread over 4 μF in parallel, so V = 200/4 = 50 V. Each capacitor cannot keep 100 V, because the charge is shared.
10. Taking k = 9 × 109 N m2 C−2, the electric field strength at a distance of 0.20 m from a point charge of +8.0 nC is
A
1800 N C−1
B
360 N C−1
C
180 N C−1
D
3600 N C−1
Show answer
Answer: A. E = kq/r2 = 9 × 109 × 8.0 × 10−9 ÷ 0.040 = 72 ÷ 0.040 = 1800 N C−1, directed away from the charge. Dividing by r instead of r2 gives 360 N C−1.
18, each with the answer and a short explanation, across 2 topics: Coulomb's law and electric field, Potential and capacitors. Every chapter test draws ten of them, so you can take it many times.
Is the MDCAT Physics chapter 8 test free?
Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.
How is the chapter test marked?
The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.
