MCQs / Physics / Electrostatics
Electrostatics MCQs — Physics 120 multiple-choice questions on Electrostatics from Physics, each with the correct answer and, where available, a teacher's explanation. Read through them to revise the chapter, then practise the same questions one by one to test yourself.
1. The Earth's field at its surface can be calculated as if all its charge (or mass) were: concentrated at its centre on the North Pole spread through space on the equator Answer: A) concentrated at its centre
Explanation: Same principle as for a sphere. SLO P-11-E-05 · Knowledge
Discuss this question
2. An electron is accelerated from rest through a p.d. of 50 V. The kinetic energy it gains is: 8 10^-18 J 8 10^-17 J 8 10^-19 J 1.6 10^-17 J Answer: A) 8 10^-18 J
Explanation: KE = eV = 1.6 10^-19 50 = 8 10^-18 J (= 50 eV). SLO P-11-E-04 · Understanding
Discuss this question
3. Workers on live high-voltage lines wear conductive suits that act as: Faraday cages insulators capacitors magnets Answer: A) Faraday cages
Explanation: Protect them from the fields. SLO P-11-E-06 · Understanding
Discuss this question
4. Charges +Q and -Q are a distance d apart. The field at the midpoint is: 2kQd^2 8kQd^2 4kQd^2 0 Answer: B) 8kQd^2
Explanation: Each gives kQ(d/2)^2 = 4kQd^2 in the same direction. SLO P-11-E-08 · Application
Discuss this question
5. Inside a closed hollow conductor placed in an external electric field, the field is: zero reversed and doubled equal to outside stronger than outside Answer: A) zero
Explanation: Electrostatic shielding. SLO P-11-E-06 · Understanding
Discuss this question
6. An electron entering a uniform electric field at right angles to the field follows: a straight line a spiral a parabolic path a circular path Answer: C) a parabolic path
Explanation: Like a projectile: constant force perpendicular to initial velocity. SLO P-11-E-04 · Understanding
Discuss this question
7. The electric field strength due to a point charge Q at distance r is: E = kQ^2r^2 E = kQr^2 E = kQr^2 E = kQr Answer: C) E = kQr^2
Explanation: Inverse square law. SLO P-11-E-08 · Application
Discuss this question
8. Coaxial cables have an outer braided conductor to: carry extra current make the cable stronger cool the wire shield the inner wire from external interference Answer: D) shield the inner wire from external interference
Explanation: Faraday shielding. SLO P-11-E-06 · Understanding
Discuss this question
9. The electric field strength 0.05 m from a point charge of 4 10^-8 C in free space is: 1.44 10^5 N/C 7190 N/C 71,900 N/C 2.88 10^5 N/C Answer: A) 1.44 10^5 N/C
Explanation: E = kQr^2 = 8.99 10^9 4 10^-8(0.05)^2 = 1.44 10^5 N/C. SLO P-11-E-08 · Application
Discuss this question
10. A Faraday cage works because conductors contain: free electrons that can move no charges fixed ions only only protons Answer: A) free electrons that can move
Explanation: Electrons redistribute instantly. SLO P-11-E-06 · Understanding
Discuss this question
11. For a point outside a charged spherical conductor, the charge can be treated as: located on the nearest surface point spread uniformly in space a point charge at the centre of the sphere zero Answer: C) a point charge at the centre of the sphere
Explanation: The field outside is the same as a point charge's. SLO P-11-E-05 · Knowledge
Discuss this question
12. The field 1 m from a point charge is 9000 N/C. The charge is: 1 10^-5 C 2 10^-6 C 1 10^-6 C 5.01 10^-7 C Answer: C) 1 10^-6 C
Explanation: Q = Er^2k = 1 10^-6 C. SLO P-11-E-08 · Application
Discuss this question
13. Coulomb's law is valid for: point charges (or spherically symmetric charges) any shape at any distance exactly only moving charges only magnets Answer: A) point charges (or spherically symmetric charges)
Explanation: Point charges. SLO P-11-E-07 · Application
Discuss this question
14. Near a sharp point on a charged conductor, the field lines are: parallel to the surface absent far apart very close together (strong field) Answer: D) very close together (strong field)
Explanation: Charge concentrates on sharp points. SLO P-11-E-03 · Understanding
Discuss this question
15. A charge of 1.6 10^-19 C is placed in an electric field of 10,000 N/C. The force on it is: 1.6 10^-15 N 3.2 10^-15 N 1.6 10^-16 N 1.6 10^-14 N Answer: A) 1.6 10^-15 N
Explanation: F = qE = 1.6 10^-19 10,000 = 1.6 10^-15 N. SLO P-11-E-02 · Application
Discuss this question
16. A Faraday cage is made of: a conducting material (e.g. metal mesh) wood an insulator such as plastic glass Answer: A) a conducting material (e.g. metal mesh)
Explanation: Free charges must move. SLO P-11-E-06 · Understanding
Discuss this question
17. The SI unit of electric field strength is: N J C N/C (or V/m) Answer: D) N/C (or V/m)
Explanation: Force per unit charge. SLO P-11-E-01 · Knowledge
Discuss this question
18. Why is the 'point charge' model useful for spheres? it makes the field uniform it removes all charges it changes the sphere's radius it allows Coulomb's law to be used for charged spheres Answer: D) it allows Coulomb's law to be used for charged spheres
Explanation: Simplifies calculation. SLO P-11-E-05 · Knowledge
Discuss this question
19. The charge on an isolated conducting sphere resides: at its centre on its outer surface nowhere uniformly through its volume Answer: B) on its outer surface
Explanation: Charges repel to the surface. SLO P-11-E-05 · Knowledge
Discuss this question
20. The electric field around an isolated negative charge points: radially inward nowhere radially outward in circles Answer: A) radially inward
Explanation: Towards negative charges. SLO P-11-E-01 · Knowledge
Discuss this question
21. Electric field strength is defined as: charge per unit force force × charge work per unit charge force per unit positive charge Answer: D) force per unit positive charge
Explanation: E = F/q. SLO P-11-E-02 · Application
Discuss this question
22. Two parallel plates 5 cm apart have a p.d. of 1000 V between them. The field strength between them is: 20,000 V/m 2 10^5 V/m 2000 V/m 200 V/m Answer: A) 20,000 V/m
Explanation: E = Delta VDelta d = 10000.05 = 20,000 V/m. SLO P-11-E-02 · Application
Discuss this question
23. A charge of 2 10^-6 C is placed in an electric field of 5000 N/C. The force on it is: 0.01 N 0.001 N 0.02 N 0.1 N Answer: A) 0.01 N
Explanation: F = qE = 2 10^-6 5000 = 0.01 N. SLO P-11-E-02 · Application
Discuss this question
24. A positive charge released in a uniform electric field moves: at constant velocity in a circle against the field in the direction of the field, with uniform acceleration Answer: D) in the direction of the field, with uniform acceleration
Explanation: Constant force qE. SLO P-11-E-04 · Understanding
Discuss this question
25. Cathode-ray oscilloscopes use deflecting plates to: heat the screen make magnetic fields move the electron beam by electric fields slow the electrons to rest Answer: C) move the electron beam by electric fields
Explanation: Electrostatic deflection. SLO P-11-E-04 · Understanding
Discuss this question
Page 1 of 5 Next
Practise Electrostatics MCQs
More Physics chapters