NCEA Level 2, 2020 Physics exam

91173 Demonstrate understanding of electricity and electromagnetism

QUESTION TWO: LIGHTNING

The question is a straight forward parallel plates question, with the context of a charged cloud to add a bit of interest. How realistic (or otherwise) the situation is can be ignored.

Use the equation `E = V/d` and rearrange, giving answer to 3 significant figures (all the given values are 3sf).

`d = V/E = (1.75 \times 10^8) / (8.57 \times 10^4) = 2040 m`

Use the equation `E = V/d = F/q`. This sort of question is quite common; you need to recognise the two different forms of the equation – and what they're used for – and that you already have most of it. These equations give two equivalent units for electric field; volts per metre, and newtons per coulomb. Some questions require you to directly change between those units.

Between the cloud and the ground, the electric field and the charge on the balloon are both constant, so the force is also constant, meaning the balloon's exact position is irrelevant for this question.

`F = Eq = (8.57\times10^4) * (3.70\times10^-6) = 0.317 N`

Between the cloud and the ground, the electric field and the charge on the electron are both constant, so because of `F = Eq` the force on the electron is also constant.

Note that the question ignores the matter of whether it would be possible for an electron to actually make the journey – low energy electrons are unlikely to travel more than a few centimetres in air before ionising something. High energy electrons can be generated in thunderstorms, but large distances (>1 km) prefer higher altitudes with less air to get in the way.

I do not believe the given situation would allow/facilitate an electron to travel from the cloud to the ground. A better wording would be "from the cloud towards the ground".

With initial speed, final speed, and wanting to find the distance travelled, `v_f^2 = v_i^2 + 2ad` is appropriate.

To replace `a` in the equation, rearrange `F = ma` to get `a = F / m`.

The force on an electron in the electric field is given by `F = Eq`. Don't fall into the trap of using the amount of force we calculated earlier for the balloon.

We can ignore force due to gravity because it is considerably less than the force due to the electric field (see the box to the right).

The base of the cloud is negatively charged so the electron will be repelled downward, so we can ignore the sign of the electron's charge – we know which way it's going to move.

Gravity:

`F = ma`
`= 9.11\times10^-31 · 9.81`
`= 8.94\times10^-30 N`

Electric field:

`F = Eq`
`= (8.57\times10^4) · (1.60\times10^-19)`
`= 1.37\times10^-14 N`

`v_f^2 = v_i^2 + 2ad`

`v_f^2 = v_i^2 + (2Fd)/m`

`v_f^2 = v_i^2 + (2Eqd)/m`

`v_f^2 - v_i^2 = (2Eqd)/m`

`d = ((v_f^2 - v_i^2)m)/(2Eq)`

`d = (((4.20\times10^5)^2 - (1.20\times10^5)^2) · (9.11\times10^-31))/(2 · (8.57\times10^4) · (1.60\times10^-19)) = 5.38\times10^-6 m`