Charged particles, conductors, and electric and magnetic fields

A negatively charged particle is placed in an electric field, as shown in Figure 4.15. What will happen to the negatively charged particle in the field?

• It will move in the direction of the field with a decreasing acceleration.

• It will move in the opposite direction of the field with an increasing acceleration.

• It will move in the same direction of the field with an increasing acceleration.

• It will move in the opposite direction of the field with a decreasing acceleration.

Answer explanation

A negatively charged particle experiences a force in the opposite direction of the electric field and, as the field lines are more closely spaced in this direction, the force and hence the acceleration of the particle will increase as it moves.

If the initial velocity of the charged particle was doubled while the strength of the B field was halved, what would the new radius of curvature be?

• 2R

• 4R

• R/2

• R/4

Answer explanation

The radius of curvature is given by qvB = mv2/r and hence r = mv/qB. As r is proportional to v/B, doubling v and halving B will increase the radius by a factor of 4.

High-velocity ions and electrons are constantly emitted from the Sun towards the Earth. In what direction will a high-speed electron directed towards the equator from the Sun experience a force when it first encounters the Earth's magnetic field?

• south

• north

• east

• west

Answer explanation

The Earth's magnetic field is directed from the south to the north (unlike the field of a bar magnet). Applying the right-hand grip rule

gives a force towards the east for a positive particle and hence the force must be towards the west on a negative particle.

What is the most probable conclusion that can be drawn from these particle tracks?

• The magnetic field is directed into the page and the particles are slowing down.

• The magnetic field is directed into the page and the particles are speeding up.

• The magnetic field is directed out of the page and the particles are slowing down.

• The magnetic field is directed out of the page and the particles are speeding up.

Answer explanation

The force on the positively charged particle is upwards when it enters the field and this would only happen if the field was directed into the page (by the right-hand palm rule). The radius of curvature (r = mv/qB) is proportional to the velocity of the charged particle and, as the radius is decreasing, the charged particles must be slowing down.

Figure 4.18 shows a proton (q = +1.602 x 10^{-19} C) travelling horizontally about to enter an electric field midway between two charged plates. When answering you may ignore the gravitational force on the particle because it is negligible compared with the electric force on the particle. The proton is observed to follow a parabolic path and collides with the lower plate at point x. Which of the following changes could move the collision point from x to point y?

• The separation between the plates could be decreased

  or the initial velocity of the proton could be increased.

• The initial velocity of the proton could be increased or

  the plates could be moved further apart from each other.

• The voltage across the plates could be increased or the

  velocity of the particle increased.

• The strength of the electric field between the plates could be decreased or the initial velocity of the

  particle could be decreased.

Answer explanation

This is because it is like projectile motion. To increase the range of the projectile we could increase the initial speed, decrease the rate at which it accelerates downwards or increase the vertical distance it must fall before hitting the ground. Moving the plates further apart would decrease the strength of the electric field between the plates (as E = V/d), which would decrease the downwards acceleration and increase the range. It would also increase the vertical distance to the lower plate, which would increase the range.

Two horizontal charged plates separated from one another by 2 cm produce an upwards electric field of 200 NC⁻¹ between the plates. A proton (m = 1.67 x 10⁻²⁷ kg and q = +1.6 x 10⁻¹⁹ C) is released at the lower plate. You may neglect gravity in your calculations. Determine the electrical potential difference between the plates and state which of the plates is at the higher potential.

• 4 V, upper plate

• 4 V, lower plate

• 8 V, upper plate

• 8 V, lower plate

Answer explanation

E = V/d and hence V = Ed = 200 x 0.02 = 4 V. As the electric field is directed upwards the lower plate will be at a higher potential (as the field would do work on a positively charged particle, moving it from the lower to the upper plate).

Two horizontal charged plates separated from one another by 2 cm produce an upwards electric field of 200 NC⁻¹ between the plates. A proton (m = 1.67 x 10⁻²⁷ kg and q = +1.6 x 10⁻¹⁹ C) is released at the lower plate. You may neglect gravity in your calculations. Calculate the force on the proton.

• 3.20 x 10^-17 N upwards

• 3.20 x 10^-17 N downwards

• 1.25 x 10²¹ N upwards

• 1.25 x 10²¹ N downwards

Answer explanation

Two horizontal charged plates separated from one another by 2 cm produce an upwards electric field of 200 NC⁻¹ between the plates. A proton (m = 1.67 x 10⁻²⁷ kg and q = +1.6 x 10⁻¹⁹ C) is released at the lower plate. You may neglect gravity in your calculations. Determine the velocity of the proton just before it hits the top plate.

• 1.38 x 10^4 m/s

• 2.77 x 10^4 m/s

• 3.92 x 10^4 m/s

• 1.96 x 10^4 m/s

Answer explanation

Two parallel charged plates are separated by 10 cm and have a potential difference of 200 V across the plates. What is the electric field strength between the plates?

• 2000 N/C

• 20 N/C

• 200 N/C

• 20000 N/C

Figure 4.19 shows the path of two particles (X and Y) that entered a uniform magnetic field at 90° to the field. The magnetic flux density was 0.2 T, the charged particles were moving with an initial velocity of 120 m s⁻¹, both charged particles had a mass of 2 x 10⁻¹² kg and both moved in circular path of diameter d = 0.2 mm in the field. Find the charge on X and Y.

• X = + 1.2 x 10^-5 C

  Y = - 1.2 x 10^-5 C

• X = - 1.2 x 10^-5 C

  Y = + 1.2 x 10^-5 C

• X = + 2.4 x 10^-5 C

  Y = - 2.4 x 10^-5 C

• X = + 2.4 x 10^-5 C

  Y = - 2.4 x 10^-5 C

Answer explanation