2018 A/L Physics Past Paper · Question 50 (English Medium)

An electron between four plates — the voltages for the v–t curve

A,B,CA, B, C and DD represent vertical cross sections of four identical parallel rectangular metal plates placed normal to the plane of the paper. Each of the plates B,C,B, C, and DD has a small hole at its centre. The three plates are arranged so that their holes are coaxially placed as shown in figure (a). Plate AA is grounded and the entire system is in a vacuum. A stationary electron is created at time t=0t = 0, at the position OO on the axis through the holes as shown. Which of the voltages VB,VCV_B, V_C and VDV_D should be applied to the plates to obtain the velocity (v)(v) – time (t)(t) curve shown in figure (b) for the electron? (Assume that the given voltages are suitable for practical use and that the edge effects and gravitational effects can be neglected.)

  1. (1) VB=−3V_B = -3 kV, VC=+2.6V_C = +2.6 kV, VD=0V_D = 0 V
  2. (2) VB=+2.5V_B = +2.5 kV, VC=−2.6V_C = -2.6 kV, VD=+3V_D = +3 kV
  3. (3) VB=+2.5V_B = +2.5 kV, VC=+2.4V_C = +2.4 kV, VD=+200V_D = +200 V
  4. (4) VB=+3V_B = +3 kV, VC=+2.6V_C = +2.6 kV, VD=−2.8V_D = -2.8 kV
  5. (5) VB=+3V_B = +3 kV, VC=+3.2V_C = +3.2 kV, VD=−2.2V_D = -2.2 kV
Answer and explanation

Answer: (4) VB=+3V_B = +3 kV, VC=+2.6V_C = +2.6 kV, VD=−2.8V_D = -2.8 kV

What the question asks

We want: the voltages VBV_B, VCV_C, VDV_D that give the vv–tt curve of figure (b).

Relevant theory

The potential difference between two points in an electric field is the work done in taking a unit positive charge from one point to the other

The electron starts at rest at AA, where V=0V = 0.

Where the potential is VV, its kinetic energy is

12mv2=eV\tfrac12 mv^2 = eV

So it speeds up towards higher potential.

It slows down where the potential falls.

Where V=0V = 0, it stops and turns back.

Reading the options

Read figure (b) in three parts.

AA to BB: the speed rises. So VB>0V_B > 0.

BB to CC: the speed falls slowly. So VCV_C is a little less than VBV_B, but still positive.

CC to DD: the speed falls fast and the electron turns back. So V=0V = 0 somewhere between CC and DD. That needs VD<0V_D < 0.

Reasoning and calculation

Test each option.

Option (1): VB<0V_B < 0. The electron cannot leave AA.

Option (2): VC<0V_C < 0. The electron turns back between BB and CC. That is too early.

Option (3): VD>0V_D > 0. The potential is never 00. The electron never turns back.

Option (4): +3>+2.6>0>−2.8+3 > +2.6 > 0 > -2.8. All three parts fit.

Option (5): VC>VBV_C > V_B. The electron speeds up between BB and CC.

So options (1), (2), (3) and (5) each break one part of figure (b).

Only option (4) fits all three parts.

Why the other options fail

This option has VB=−3V_B = -3 kV.

This option has VC=−2.6V_C = -2.6 kV.

This option has VD=+200V_D = +200 V.

This option has VC=+3.2V_C = +3.2 kV, more than VBV_B.

Conclusion

Only option (4) is left: VB=+3V_B = +3 kV, VC=+2.6V_C = +2.6 kV, VD=−2.8V_D = -2.8 kV.

Check option (4).

At CC the kinetic energy is 2.62.6 keV.

From CC to DD the potential falls from +2.6+2.6 kV to −2.8-2.8 kV. So it passes 00 on the way.

There the electron stops. Then it goes back the same way to AA.

VB=+3V_B = +3 kV, VC=+2.6V_C = +2.6 kV, VD=−2.8V_D = -2.8 kV.

The answer is option (4).

Explore

A stationary electron is made at OO. Plate AA is earthed (00 V).

BB is at +3+3 kV. The electron speeds up towards BB and goes through its hole.

From BB to CC the potential falls a little, to +2.6+2.6 kV. The electron slows down gently.

After CC the potential falls fast, to −2.8-2.8 kV at DD. Where the potential is 00, the electron stops and turns back.

It goes back the same way and stops again at AA. That is figure (b). So the answer is option (4).

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