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A-Level CAIE Physics 18 2 Uniform Electric Fields Question Bank

Practice A-Level CAIE Physics 18 2 Uniform Electric Fields questions by syllabus topic with past-paper context, marks, difficulty and question previews on Eduninja.

10 matching questions · Open interactive library

Question 2

2

A sphere of mass \(1.6 \times 10^{-10} \mathrm{~kg}\) has a charge of +0.27 nC . The sphere is in a uniform electric field that acts vertically upwards, as shown in the side view in Fig. 2.1. SIDE VIEW The force exerted on the sphere by the electric field causes the sphere to remain at a fixed vertical height in a horizontal plane. There is a uniform magnetic field in the region of the electric field. The sphere moves at a speed of \(0.78 \mathrm{~m} \mathrm{~s}^{-1}\) in the horizontal plane. The magnetic field causes the sphere to move in a circular path of radius 3.4 m , as shown in the view from above in Fig. 2.2.

structured2 marks

Question 2(b)

2(b)

Calculate the strength of the uniform electric field.

Mediumstructured2 marks

Answer

m g=E q C1 \[ \begin{aligned} E =\left(1.6 \times 10^{-10} \times 9.81\right) /\left(0.27 \times 10^{-9}\right) =5.8 \mathrm{~N} \mathrm{C}^{-1} \end{aligned} \] A1

Question 13

13

A small water droplet of mass \(3.0 \mu \mathrm{~g}\) carries a charge of \(-6.0 \times 10^{-11} \mathrm{C}\). The droplet is situated in the Earth's gravitational field between two horizontal metal plates. The potential of the upper plate is +500 V and the potential of the lower plate is -500 V . What is the motion of the droplet? It accelerates downwards. It remains stationary. It accelerates upwards. It moves upwards at a constant velocity.

Hardmcq1 marks

Answer

Question 3

3

4 marks

Question 3(c)

3(c)

Two horizontal metal plates are separated by a distance of 1.8 cm in a vacuum. A potential difference of 270 V is maintained between the plates, as shown in Fig. 3.1. A proton is in the space between the plates. Explain quantitatively why, when predicting the motion of the proton between the plates, the gravitational field is not taken into consideration.

Mediumstructured4 marks

Answer

\[ \begin{aligned \text { gravitational force =1.67 \times 10^{-27} \times 9.81 =1.6 \times 10^{-26} \mathrm{~N} \end{aligned} \] \[ \begin{aligned} \text { electric force } =1.6 \times 10^{-19} \times 270 /\left(1.8 \times 10^{-2}\right) =2.4 \times 10^{-15} \mathrm{~N} \end{aligned} \]}} C1 A1 electric force very much greater than gravitational force B1 [4]

Question 4

4

2 marks

Question 4(b)

4(b)

A uniform electric field is produced by applying a potential difference of 1200 V across two parallel metal plates in a vacuum, as shown in Fig. 4.1. The separation of the plates is 14 mm . A particle P with charge \(3.2 \times 10^{-19} \mathrm{C}\) and mass \(6.6 \times 10^{-27} \mathrm{~kg}\) starts from rest at the lower plate and is moved vertically to the top plate by the electric field. Calculate

structured2 marks

Question 4(b)(i)

4(b)(i)

the electric field strength between the plates, electric field strength = \(\mathrm{Vm}^{-1}\)

Easystructured2 marks

Answer

E=V / d \(=1200 / 14 \times 10^{-3}\)

Question 4

4

3 marks

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.

structured3 marks

Question 4(b)(ii)

4(b)(ii)

A proton is at point P where \(x=5.0 \mathrm{~cm}\). Use data from Fig. 4.2 to determine the magnitude of the acceleration of the proton. acceleration = \(\mathrm{m} \mathrm{s}^{-2}[3]\)

Mediumstructured3 marks

Answer

(at \(x=5.0 \mathrm{~cm}\),) \(E=3.0 \times 10^{3} \mathrm{~V} \mathrm{~m}^{-1}\) C1 a=q E / m C1 \[ \begin{aligned} a=\left(1.60 \times 10^{-19} \times 3.0 \times 10^{3}\right) /\left(1.67 \times 10^{-27}\right) a=2.9 \times 10^{11} \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} \] A1

Question 3

3

2 marks

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.

structured2 marks

Question 3(c)(i)

3(c)(i)

Calculate the electric field strength between the plates. electric field strength = \(\mathrm{Vm}^{-1}\)

Mediumstructured2 marks

Answer

\(E=1.9 \times 10^{3} / 14 \times 10^{-3}\)

Question 4

4

7 marks

Question 4(b)

4(b)

Fig. 4.1 shows a pair of parallel metal plates with a potential difference (p.d.) of 2400 V between them. The plates are separated by a distance of 4.6 cm . The plates are in a vacuum.

structured2 marks

Question 4(b)(ii)

4(b)(ii)

Calculate the strength of the electric field between the plates. electric field strength = \(\mathrm{NC}^{-1}\)

Mediumstructured2 marks

Answer

E=V / d C1 \[ \begin{aligned} =2400 / 0.046 =5.2 \times 10^{4} \mathrm{NC}^{-1} \end{aligned} \] A1

Question 4(c)

4(c)

A moving proton enters the region between the plates from the left, as shown in Fig. 4.2.

structured5 marks

Question 4(c)(i)

4(c)(i)

The proton is deflected by the electric field. On Fig. 4.2, draw a line to show the path of the proton as it moves through and out of the region of the electric field.

Mediumstructured2 marks

Answer

smooth curve in region of field and straight line outside field B1 direction of deflection shown as downwards in region of field B1

Question 4(c)(ii)

4(c)(ii)

A helium nucleus \(\left({ }_{2}^{4} \mathrm{He}\right)\) now enters the region of the electric field along the same initial path as the proton and travelling at the same initial speed. State and explain how the final speed of the helium nucleus compares with the final speed of the proton after leaving the region of the electric field.

Hardstructured3 marks

Answer

helium nucleus has double the charge but four times the mass B1 velocity parallel to plates same and acceleration perpendicular to plates smaller (for helium) B1 final speed is lower (for helium) B1

Question 4

4

2 marks

Question 4(b)

4(b)

Fig. 4.1 shows a pair of parallel metal plates with a potential difference (p.d.) of 2400 V between them. The plates are separated by a distance of 4.6 cm . The plates are in a vacuum.

structured2 marks

Question 4(b)(ii)

4(b)(ii)

Calculate the strength of the electric field between the plates. electric field strength = \(\mathrm{NC}^{-1}\)

Easystructured2 marks

Answer

E=V / d C1 \[ \text { = } 2400 \text { / } 0.046 \] \(=5.2 \times 10^{4} \mathrm{~N} \mathrm{C}^{-1}\) A1

Question 4

4

7 marks

Question 4(b)

4(b)

Two horizontal metal plates are 20 mm apart in a vacuum. A potential difference of 1.5 kV is applied across the plates, as shown in Fig. 4.1. A charged oil drop of mass \(5.0 \times 10^{-15} \mathrm{~kg}\) is held stationary by the electric field.

structured1 marks

Question 4(b)(ii)

4(b)(ii)

Calculate the electric field strength between the plates. electric field strength = \(\mathrm{Vm}^{-1}\)

Mediumstructured1 marks

Answer

\(E=1500 / 20 \times 10^{-3}=75000 \mathrm{Vm}^{-1}\)

Question 4(b)(iii)

4(b)(iii)

Calculate the charge on the drop. charge = C

Mediumstructured4 marks

Answer

F=q E ( W=m g and) q E=m g \(q=m g / E=5 \times 10^{-15} \times 9.81 / 75000\) negative charge

Question 4(b)(iv)

4(b)(iv)

The potential of the upper plate is increased. Describe and explain the subsequent motion of the drop.

Mediumstructured2 marks

Answer

F>m g or F now greater B1 drop will move upwards

Question 7

7

Two oppositely-charged parallel metal plates are situated in a vacuum, as shown in Fig. 7.1. particle, mass m charge +q speed v The plates have length L. The uniform electric field between the plates has magnitude E. The electric field outside the plates is zero. A positively-charged particle has mass m and charge +q. Before the particle reaches the region between the plates, it is travelling with speed v parallel to the plates. The particle passes between the plates and into the region beyond them.

structured3 marks

Question 7(a)

7(a)

3 marks

Question 7(a)(i)

7(a)(i)

On Fig. 7.1, draw the path of the particle between the plates and beyond them.

Mediumstructured2 marks

Answer

path: reasonable curve upwards between plates B1 straight and at a tangent to the curve beyond the plates B1 [2]

Question 7(a)(ii)

7(a)(ii)

For the particle in the region between the plates, state expressions, in terms of E, m, q, v and L, as appropriate, for 1. the force F on the particle, 2. the time t for the particle to cross the region between the plates.

Mediumstructured1 marks

Answer

1. (F=) E . g B1 [1] 2. (t=) L / v B1 [1]