2022 A/L Physics Past Paper · Question 38 (English Medium)

Answer: (5) $A$: $+q$, $B$: $+q$, $C$: $-q$

Given information

What we know:

Three identical metal spheres stand on insulating supports.

$A$ has $+3q$, $B$ has no net charge and $C$ has $-2q$.

First $B$ touches $C$ and is moved away. Then $B$ touches $A$ and is moved away.

The question has no numbers. So we keep $q$ and check with the sample value $q = 2\ \mu\text{C}$.

What the question asks

We want: the final charge on each of the spheres $A$, $B$ and $C$.

What each letter means:

$q$ ($\text{C}$): a unit of charge. The spheres hold $+3q$, $0$ and $-2q$.

$Q$ ($\text{C}$): the charge on a sphere.

$V$ ($\text{V}$): the potential of a sphere.

$C$ ($\text{F}$): the capacitance of a sphere.

Relevant theory

What is happening:

When $B$ touches $C$, electrons from the crowded negative sphere $C$ spread onto $B$. $B$ becomes negative and $C$ loses some of its negative charge.

When $B$ then touches $A$, the electrons of $B$ move onto the positive sphere $A$. They cancel some of its positive charge. What is left is shared by the two spheres.

The charge only moves from one sphere to another. So the three charges still add up to $+3q - 2q = +q$.

In a neutral body both types of charges exist equally.

When negative (-) charges are removed from a body it becomes a positively charged body due to the excess of positive charges remaining in it while the body receiving the negative charges becomes a negatively (-) charged body.

When the two spheres are connected by a conductor they reach the same potential V

Hence the capacitance of a capacitor is the ratio its charge subtending to the potential difference between its plates.

Two spheres that touch are joined by a conductor, so they reach the same potential $V$.

The charge on a sphere is $Q = CV$. Identical spheres have the same capacitance $C$. With the same $C$ and the same $V$, each sphere holds the same charge.

So two identical spheres that touch share their total charge equally.

$$Q_1 = Q_2 = \dfrac{Q_{total}}{2}$$

In words: each sphere ends with half of the total charge of the two.

For example, take two jugs of the same size, one full and one empty, and join them with a pipe at the bottom. The water settles at the same level in both. Each jug holds half of the water.

The two touching spheres are like those two jugs. The charge settles until both have the same share.

Required equations

$Q_1 = Q_2 = \dfrac{Q_{total}}{2}$

$Q_B = Q_C = \dfrac{0 + (-2q)}{2} = -q$

$Q_B = Q_A = \dfrac{(-q) + (+3q)}{2} = +q$

Applying it to this question

The charges add up before and after each touch, so the total does not change. Before: $+3q + 0 - 2q = +q$.

$-4$

$Q_{total} = 0 + (-4)$

$-2$

$Q_B = Q_C = \dfrac{-4}{2}$

$4$

$Q_{total} = (-2) + (+6)$

$2$

$Q_B = Q_A = \dfrac{4}{2}$

$Q_A = +2\ \mu\text{C} = +q$

$Q_B = +2\ \mu\text{C} = +q$

$Q_C = -2\ \mu\text{C} = -q$

Reasoning and calculation

The final charges are $A$: $+q$, $B$: $+q$ and $C$: $-q$.

That is option (5).

Checks

$(a + b) / 2$

Does it make sense?

Add the final charges: $+q + q - q = +q$. At the start we had $+3q + 0 - 2q = +q$. The total is the same, as it must be when charge only moves from one sphere to another.

$C$ is touched only once, by a neutral sphere, so it keeps half of its charge, $-q$. That matches.

$A$ lost positive charge by being touched to a negative sphere. It went from $+3q$ down to $+q$, and it is still positive.

Conclusion

$A: +q,\ B: +q,\ C: -q$

The final charges are $A$: $+q$, $B$: $+q$ and $C$: $-q$.

The answer is (5).

Take the touches one at a time. After each touch, add the two charges and share the sum equally between the two identical spheres.

Then check that the three charges still add up to what you started with.

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