Question 1
1
Question 1(a)
1(a)
Question 1(a)(ii)
1(a)(ii)
Define electric field.
Answer
force per unit positive charge B1
Question bank
A-Level CAIE Physics 18 1 Electric Fields And Field Lines Question Bank
Practice A-Level CAIE Physics 18 1 Electric Fields And Field Lines questions by syllabus topic with past-paper context, marks, difficulty and question previews on Eduninja.
Question 12
12
A tiny oil droplet with mass \(6.9 \times 10^{-13} \mathrm{~kg}\) is at rest in an electric field of electric field strength \(2.1 \times 10^{7} \mathrm{NC}^{-1}\), as shown. The weight of the droplet is exactly balanced by the electrical force on the droplet. What is the charge on the droplet? \(3.3 \times 10^{-20} \mathrm{C}\) \(-3.3 \times 10^{-20} \mathrm{C}\) \(3.2 \times 10^{-19} \mathrm{C}\) \(-3.2 \times 10^{-19} \mathrm{C}\)
Answer
D
Question 12
12
The diagrams show a negative electric charge situated in a uniform electric field and a mass situated in a uniform gravitational field. Which row shows the directions of the forces acting on the charge and on the mass?
Answer
B
Question 3
3
Question 3(b)
3(b)
Two point charges A and B are separated by a distance of 20 nm in a vacuum, as illustrated in Fig. 3.1. A point P is a distance x from A along the line A B. The variation with distance x of the electric potential \(V_{A}\) due to charge A alone is shown in Fig. 3.2. The variation with distance x of the electric potential \(V_{B}\) due to charge B alone is also shown in Fig. 3.2.
Question 3(b)(i)
3(b)(i)
State and explain whether the charges A and B are of the same, or opposite, sign.
Answer
either both potentials are positive/same sign or gradients are positive \& negative (so fields in opposite directions)
Question 13
13
A particle is in a uniform field. The particle experiences a force in the opposite direction to the field. Which field is the particle in, and on which property of the particle is the field acting?
Answer
A
Question 3
3
Question 3(a)
3(a)
State what is meant by a line of force in
Question 3(a)(ii)
3(a)(ii)
an electric field.
Answer
(tangent to line gives) direction of force on a (small test) charge charge is positive M1 A1 [2]
Question 4
4
Question 4(a)
4(a)
Define electric field strength.
Answer
electric field strength is the force per unit positive charge (acting on a stationary charge)
Question 4
4
Question 4(b)
4(b)
Two charged metal spheres A and B are situated in a vacuum, as illustrated in Fig. 4.1. The shortest distance between the surfaces of the spheres is 6.0 cm . A movable point P lies along the line joining the centres of the two spheres, a distance x from the surface of sphere A. The variation with distance x of the electric field E at point P is shown in Fig. 4.2.
Question 4(b)(i)
4(b)(i)
Use Fig. 4.2 to explain whether the two spheres have charges of the same, or opposite, sign.
Answer
(near to each sphere,) fields are in opposite directions or point (between spheres) where fields are equal and opposite or point (between spheres) where field strength is zero M1 so same (sign of charge) A1
Question 5
5
Question 5(a)
5(a)
Define electric field strength.
Answer
force per unit positive charge acting on a stationary charge B1
Question 3
3
Question 3(a)
3(a)
Define electric field strength.
Answer
electric field strength is force per unit positive charge B1 [1]
Question 3(c)
3(c)
Two horizontal metal plates are 14 mm apart in a vacuum. A potential difference (p.d.) of 1.9 kV is applied across the plates, as shown in Fig. 3.1. A uniform electric field is produced between the plates. The sphere S in (b) is charged and is held stationary between the plates by the electric field.
Question 3(c)(ii)
3(c)(ii)
Calculate the magnitude of the charge on S . charge = ..... C
Answer
F=Q E
Question 3(c)(iii)
3(c)(iii)
The magnitude of the p.d. applied to the plates is increased. Explain why S accelerates towards the top plate.
Answer
\(\begin{array}{lll}\text { (iii) } & \text { electric force increases/is greater (than weight) } & \text { B1 } \\ & \text { charge (on } \mathrm{S} \text { ) is negative to give resultant/net/sum/total force up } & \text { B1 }\end{array}\)