2022 A/L Physics Past Paper · Question 32 (English Medium)

Answer: (4) $2\dfrac{\varepsilon_0 A}{d}$

Given information

What we know:

There are four plates, each at distance $d$ from the next. Where two plates face each other, the overlapping area is $A$.

The wire from $+$ splits and joins the top plate and the bottom plate.

The wire to $-$ joins the two middle plates to each other.

The question has no numbers. So we keep the letters and check with sample values $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F m}^{-1}$, $A = 0.01\ \text{m}^2$ and $d = 0.001\ \text{m}$.

What the question asks

We want: the equivalent capacitance of the four plates between $+$ and $-$.

What each letter means:

$\varepsilon_0$ ($\text{F m}^{-1}$): the permittivity of free space.

$A$ ($\text{m}^2$): the overlapping area of two neighbouring plates.

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

$C$ ($\text{F}$): the capacitance of one pair of plates.

$C_{eq}$ ($\text{F}$): the equivalent capacitance of the whole system.

Relevant theory

What is happening:

Join the plates to the battery. The battery pulls electrons off the top and bottom plates, so they become positive. It pushes electrons onto the two middle plates, so they become negative.

Top pair: a positive plate faces a negative plate. An electric field forms between them and charge is stored.

Bottom pair: the same, a negative plate faces a positive plate.

Middle pair: both plates are negative and a wire joins them. They are at the same potential, so no field forms between them and no charge collects on the faces that look at each other.

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

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

maintain the same potential difference across them, the capacitors are said to be connected parallel to each other.

So the four plates work as two capacitors: the top pair and the bottom pair. The middle pair does nothing.

One pair of plates of area $A$ and distance $d$ apart has the capacitance $C = \dfrac{\varepsilon_0 A}{d}$.

Both working pairs are joined between $+$ and $-$. So the potential difference across each is the same. Capacitors like this are in parallel, and their capacitances add.

$$C_{eq} = C + C$$

In words: the equivalent capacitance is the sum of the two capacitances.

For example, two equal buckets stand side by side and are filled from one tap to the same level. Together they hold twice the water of one bucket.

The two working pairs are like those buckets. They have the same potential difference, so together they hold twice the charge of one pair.

Required equations

$C = \dfrac{\varepsilon_0 A}{d}$

$C_{eq} = C + C$

$C_{eq} = 2\dfrac{\varepsilon_0 A}{d}$

Applying it to this question

$\dfrac{\text{F m}^{-1} \times \text{m}^2}{\text{m}} = \text{F}$, which is a capacitance. Adding two capacitances gives a capacitance.

$8.85e-14$

$\varepsilon_0 A = 8.85 \times 10^{-12} \times 0.01 = 8.85 \times 10^{-14}$

$88.5$

$C = \dfrac{8.85 \times 10^{-14}}{0.001}$

$177$

$C_{eq} = 88.5 + 88.5$

Reasoning and calculation

The equivalent capacitance is twice the capacitance of one pair, $C_{eq} = 2\dfrac{\varepsilon_0 A}{d}$.

That is option (4).

Checks

$2 * e * a / d$

Does it make sense?

The two working pairs both sit between $+$ and $-$. Charge can collect on both, so together they hold more charge for the same potential difference than one pair does. A larger capacitance than one pair is what we expect.

The middle pair has the same potential on both plates. It cannot hold any extra charge in this circuit. So the answer is twice one pair, not three times.

Conclusion

$C_{eq} = 2\dfrac{\varepsilon_0 A}{d}$

Two equal capacitors in parallel give $C_{eq} = 2\dfrac{\varepsilon_0 A}{d}$.

The answer is (4).

When plates are joined by a wire, they are at one potential. Redraw the plates with the same-potential ones marked and count only the pairs with different potentials on the two plates.

Capacitors that sit between the same two terminals are in parallel, and their capacitances add.

Exam tip