2021 A/L Physics Past Paper · Question 32 (English Medium)

Answer: (1) not changed.

Given information

We know:

Three capacitors, each $C$. One is on its own. The other two are side by side.

The key $K$ is joined across the two side-by-side capacitors.

The battery gives $V$. The key $K$ is closed first, then opened.

What the question asks

We want: how the total charge of the three capacitors changes when $K$ is opened.

The plan: find the charges with $K$ closed. Then find them with $K$ open, and compare.

Relevant theory

The capacitance of a capacitor can be considered as a measure of its ability to store electric charges.

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

Required equations

$Q = CV$

$Q_2 = 2C\,V_2$

$V_1 + V_2 = V$

Applying it to this question

$0$

$Q_2 = 2C \times 0$

$CV$

$Q_1 = CV$

$Q_1 + Q_2 = CV + 0$

$-Q_1 + Q_2 = -CV$

$\dfrac{Q_1}{C} + \dfrac{Q_2}{2C} = V$

$3Q_1 = 3CV$

$2Q_1 + (Q_1 - CV) = 2CV$

$Q_2 = 0$

$Q_1 = CV, \quad Q_2 = CV - CV$

Every term is a charge. $CV$ has units $\text{F} \times \text{V} = \text{C}$, and the equations compare charge with charge.

Reasoning and calculation

With $K$ open, $Q_1 = CV$ and $Q_2 = 0$. The total charge is $CV + 0 = CV$.

With $K$ closed it was also $CV$. The total charge is not changed. This is option (1).

Checks

$(2 * c * v + c * v) / 3$

Does it make sense?

After the key is opened the pair has $V_2 = \dfrac{Q_2}{2C} = 0$. So the single capacitor still has the whole $V$.

Nothing in the circuit needs to change, so no charge moves.

Conclusion

$Q_1 + Q_2 = CV \text{ (not changed)}$

With $K$ open, $Q_1 = CV$ and $Q_2 = 0$. With $K$ closed, $Q_1 = CV$ and $Q_2 = 0$.

The total charge is $CV$ both times. The answer is (1).

A closed key across a capacitor leaves that capacitor with no voltage and no charge.

When the key opens, check what charge is cut off. A cut-off part keeps its charge. Then use the loop equation.

Exam tip