2020 A/L Physics Past Paper · Question 50 (English Medium)

Answer: (2) $\dfrac{5\varepsilon_0 A}{3d}$

What the question asks

We must find: the equivalent capacitance between terminals $X$ (on plate $Q$) and $Y$ (on plate $R$), with wires joining $P$–$S$ and $R$–$T$.

Given information

Given: five plates $P, Q, R, S, T$, each of area $A$; equal gap $d$ between neighbouring plates; vacuum ($\varepsilon_r = 1$); $X$ is on plate $Q$ and $Y$ is on plate $R$.

Relevant theory

Consider a capacitor consisting of two parallel plates each of area. A placed at a distance d apart and having a charge Q spread uniformly over each one as shown.

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

Reasoning and calculation

Setup: each facing pair of plates is a parallel-plate capacitor — four gaps $PQ, QR, RS, ST$; in vacuum each one is $C_0 = \dfrac{\varepsilon_0 A}{d}$.

Symmetry: all four gaps share the same $A$ and the same $d$ — so all four equal $C_0$; what differs is only which two terminals each gap sits between.

Equal-potential points: plates joined by a wire sit at one potential. There are three nodes — $X = \{Q\}$, $Y = \{R, T\}$, and a floating node $F = \{P, S\}$. So $QR$: $X$–$Y$; $PQ$: $F$–$X$; $RS$: $Y$–$F$; $ST$: $F$–$Y$.

Reducing: $RS$ and $ST$ are both between $F$ and $Y$ — the same potential difference, so they are in parallel (charges add, $C = \dfrac{Q}{V}$). That combination is in series with $PQ$ through node $F$ — since $F$ is isolated its net charge is zero, so the same charge passes through both and the potential differences add. That whole branch is in parallel with $QR$ between $X$ and $Y$.

Required equations

$C_0 = \frac{\varepsilon_0 A}{d}$

$C = C_1 + C_2$

$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$

$2C_0$

$C_{RS} + C_{ST} = 2C_0$

$\frac{2}{3}C_0$

$\frac{C_0 \times 2C_0}{C_0 + 2C_0} = \frac{2C_0}{3}$

$C_{XY} = C_0 + \frac{2C_0}{3}$

$C_{XY} = \frac{5C_0}{3} = \frac{5\varepsilon_0 A}{3d}$

Circuit rule (charge conservation): if $V$ is applied between $X$ and $Y$, the floating node $F$ sits at $\dfrac{V}{3}$; the charge on $QR$ is $C_0V$, on $PQ$ it is $\dfrac{2}{3}C_0V$ — the total from $X$ is $\dfrac{5}{3}C_0V$, so $C = \dfrac{Q}{V} = \dfrac{5C_0}{3}$.

Reading the options

Comparing with the options: $\dfrac{5\varepsilon_0 A}{3d}$ — option (2).

Checks

$C0 + C0 * 2 * C0 / (C0 + 2 * C0)$

Second way (charge accounting): let the floating node be at $V_F$; its net charge is zero: $C_0(V - V_F) = 2C_0 V_F \Rightarrow V_F = \dfrac{V}{3}$. The charge from $X$ is $C_0V + C_0\left(V - \dfrac{V}{3}\right) = \dfrac{5}{3}C_0V$ ✓.

Boundary check: the answer is larger than $C_0$ (what $QR$ alone would give), and smaller than $4C_0$ (if all four were simply in parallel) ✓.

Conclusion

$C_{XY} = \frac{5\varepsilon_0 A}{3d}$

$C_{XY} = C_0 + \dfrac{C_0 \cdot 2C_0}{3C_0} = \dfrac{5}{3}\cdot\dfrac{\varepsilon_0 A}{d}$ — the answer is (2).

Looking only at the order of the plates and assuming all the gaps are in series (or all in parallel).

First mark the equal-potential nodes made by the wires; then check which two nodes each neighbouring pair actually sits between.