2023 A/L Physics Past Paper · Question 33 (English Medium)

Answer: (3) $X$ and $Y$ rise up with equal accelerations.

What the question asks

What is changed: the plates are brought closer. The voltage across them stays the same, so only the distance $d$ between them gets smaller.

We want: what $X$ and $Y$ do after that.

What each letter means:

$V$ ($\text{V}$): the voltage across the plates.

$d$ ($\text{m}$): the distance between the plates.

$E$ ($\text{V m}^{-1}$): the electric field intensity between the plates.

$q$ ($\text{C}$): the charge of a droplet.

$m$ ($\text{kg}$): the mass of a droplet.

$g$ ($\text{m s}^{-2}$): the acceleration of gravity.

$a$ ($\text{m s}^{-2}$): the acceleration of a droplet.

Relevant theory

The electric field intensity at a certain point in an electric field is the force acting ona unit positive (+) charge placed at that point.

Hence, the force acting on a charge "q" placed at a point of intensity E is F = Eq.

Hence according to the above expression, the electric field intensity at any point in an electric field is equal to the potential gradient at that point.

The first quote says the electric field intensity at a point is the force on a unit positive charge there. The second quote gives the force on a charge $q$ as $F = Eq$. The third quote says the field intensity equals the potential gradient.

Between two parallel plates the potential changes evenly from one plate to the other, so the gradient is the voltage over the distance, $E = \dfrac{V}{d}$.

For example, think of two lifts hanging still at one floor, a big lift and a small one. The cable of each lift pulls up exactly as much as its weight, so the big lift has a harder pull.

Now make each cable pull a tenth more than before. For each lift the extra pull is a tenth of its weight, so both lifts speed up upwards by the same amount, however heavy they are.

Applying it to this question

$qE = mg$

$E = \dfrac{V}{d}$

$\dfrac{q}{m} = \dfrac{g}{E}$

Reasoning and calculation

What stays the same: the voltage $V$ (the battery is still connected), the charge $q$ and the mass $m$ of each droplet, and $g$.

What changes: the distance $d$ gets smaller. So the field $E = \dfrac{V}{d}$ gets bigger.

Which way is the net force?

Before: the upward force $qE$ was equal to the weight $mg$.

After: $E$ is bigger, so the upward force $qE'$ is bigger than the weight $mg$. The net force is up, so both droplets rise.

How much faster does each one go?

The net upward force is $qE' - mg$. Divide by the mass $m$.

$$a = \dfrac{qE'}{m} - g$$

Put in $\dfrac{q}{m} = \dfrac{g}{E}$.

$$a = g\left(\dfrac{E'}{E} - 1\right)$$

The mass is not in this. It is the same for $X$ and $Y$, and it is upward, so $a$ is positive.

Reading the options

Now match it with the five options.

(1) They stay at rest. The net force is not zero. It does not match.

(2) They fall. The net force is up. It does not match.

(3) They rise with equal accelerations. This matches.

(4) $X$ rises faster than $Y$. The accelerations are equal. It does not match.

(5) $X$ falls faster than $Y$. It does not match.

Why the other options fail

The upward force grows when $E$ grows, so the droplets are no longer in balance.

The net force is up, because $qE'$ is bigger than $mg$.

The mass cancels, so $X$ and $Y$ have the same acceleration.

The droplets rise, and the mass cancels in the acceleration.

Checks

Does it make sense?

If the plates were not moved, $E' = E$ and the formula gives $a = 0$, as it should for droplets at rest.

If the plates are very close, $E'$ is very big and $a$ is big and upward, as a strong field would give.

Conclusion

$a = g\left(\dfrac{E'}{E} - 1\right)$

Both droplets rise with the same acceleration $g\left(\dfrac{E'}{E} - 1\right)$. That is option (3).

The answer is (3).

The quotes give the field intensity, the force $F = Eq$ and the field as the potential gradient. That the field between two parallel plates is $V$ over $d$, and that $\dfrac{q}{m}$ is the same for the two droplets because they hang in balance, is our reasoning from them. We ignore the buoyancy of the air and the pull of the droplets on each other, as the question says.

Teacher's note

A droplet at rest in a field has $qE = mg$, so its $\dfrac{q}{m}$ is $\dfrac{g}{E}$. When the field changes, the acceleration is $g\left(\dfrac{E'}{E} - 1\right)$.

The mass is not in this, so droplets of different mass move together.

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