What is a Capacitor?

A capacitor is a component that stores electrical charge and, by extension, electrical energy.

We help you move from easy GCSE ideas to tough A-level math. You will learn how these parts store energy like tiny buckets.

We show you how to avoid simple mistakes that lose marks. Our steps make hard exam questions easy to solve.

Read on to master your physics exam and get a top grade!

What is a Capacitor?

A capacitor is a component that stores electrical charge and, by extension, electrical energy. It consists of two conducting plates separated by an insulator known as a dielectric.

Capacitors Formula

Every student must master the fundamental relationship:

Q = CV

• Q: Charge (Coulombs, C)
• C: Capacitance (Farads, F)
• V: Potential Difference (Volts, V)

How does a capacitor actually store energy?

A capacitor doesn’t store energy chemically like a battery. It stores it in an electric field. When you connect it to a power source, electrons are forced onto one plate and pulled off the other.

This creates a “tension” or potential difference between the plates. The energy is stored in the space between the plates—the electric field.

What is the difference between a capacitor and a battery?

Speed: A capacitor can discharge its entire energy in a fraction of a second (like a camera flash). A battery releases energy slowly through a chemical reaction.

Lifespan: Capacitors can be charged and discharged millions of times without wearing out. Batteries degrade over time.

Storage Capacity: Batteries can store massive amounts of energy (energy density); capacitors store very little by comparison.

What happens if you exceed the voltage rating on a capacitor?

If you go over this limit, the insulating layer (dielectric) between the plates breaks down. This causes a “dielectric breakdown,” leading to a short circuit.

In many cases, the capacitor will overheat, leak fluid, or even pop/explode. This is why Unit Checking and Estimation are so important in A-level practicals.

How do you calculate the time constant?”

The time constant ($\tau$) tells you how long a capacitor takes to charge or discharge. It is calculated using the formula:

$\tau$ = $R \times C$

• R is the resistance in the circuit ($\Omega$).
• C is the capacitance (F). It represents the time taken for the voltage to fall to approximately 37% of its original value.

Why does a capacitor block DC but pass AC?

• DC (Direct Current): When connected to DC, the capacitor charges up until its voltage equals the source voltage. Once full, no more charge can flow. The circuit is “blocked.”
• AC (Alternating Current): Because AC constantly changes direction, the capacitor is constantly charging and discharging. This creates a flow of electrons back and forth in the wires, which looks like the current is “passing through” even though no electrons actually cross the gap between the plates.

What is a dielectric and what does it do?”

A dielectric is an insulating material (like ceramic, plastic, or paper). When placed between the plates, it becomes polarized.

This means the molecules inside it line up to oppose the electric field of the plates. This reduces the overall voltage for the same amount of charge, which — according to C = Q/V — effectively increases the capacitance.

Why Capacitors Feel Harder Than GCSE

The transition from GCSE to A-level involves several “level-up” moments in your cognitive thinking.

Maths Level Jumps

At GCSE, you mostly used basic algebra. At A-level, you have to work with exponential growth and decay.

This means using the natural logarithm (ln) and the constant e to find out how fast a capacitor is losing charge.

Abstract Ideas

You cannot see an electric field. Unlike a circuit with a glowing bulb, capacitors involve “invisible” storage of energy.

Visualizing how electrons build up on one plate and repel them from the other takes a higher level of thinking.

Multi-Step Questions

A single exam question might require you to find the capacitance, then the time constant, and finally the voltage after a certain number of seconds. If you fail the first part, the rest becomes very difficult.

Less Memorisation

In lower grades, you could pass by memorizing facts. Here, you must understand the mechanisms.

You need to explain why adding a dielectric increases capacitance, not just state that it does.

Where Marks are Lost in Exams

Even students who understand the topic lose marks due to simple errors.

• Units: Forgetting that $1 \mu F$ is $1 \times 10^{-6}$ F.
• Rearranging Equations: Making mistakes when moving terms from one side of an equals sign to the other, especially with fractions.
• Not Explaining Reasoning: Examiners want to see your logic. If you show a final answer without the steps, you might get zero marks.
• Sign Errors: In discharge equations, a missing minus sign in the exponent will lead to a nonsensical answer.

Applying Equations: The “We Teach” Method

To ensure you get every mark, we recommend a four-step approach to any calculation.

1. Symbol Meaning

Before touching your calculator, list your variables: $C=10 \mu F$, $V=12V$, $R=2k\Omega$.

2. Unit Checking

Convert everything to base units immediately. $10 \mu F$ becomes $10 \times 10^{-6}$ F. $2k\Omega$ becomes $2000 \Omega$.

3. Step-by-Step Substitution

Don't skip steps. Write the formula, then the formula with numbers, then the answer. This secures "method marks" even if you have a "fat-finger" error on your calculator.

4. Estimation

Does your answer make sense? A capacitor in a circuit shouldn’t take 400 years to discharge. If the number looks weird, check your powers of ten.

Exam-Style Questions and Answers

Question 1 (Calculation Focus)

A $470 \mu F$ capacitor is charged to a potential difference of 12V. Calculate the energy stored.

Answer:

1. Identify formula: W = $\frac{1}{2}CV^2$
2. Convert units: C = $470 \times 10^{-6} F$
3. Substitute: W = $0.5 \times (470 \times 10^{-6}) \times 12^2$
4. Calculate: W = $0.5 \times 470 \times 10^{-6} \times 144 = 0.03384 J$
5. Final Answer: $3.38 \times 10^{-2} J$ (to 3 significant figures).

Question 2

Explain what happens to the capacitance of a parallel-plate capacitor when a dielectric is inserted.

Answer:

The dielectric material becomes polarized in the electric field between the plates. This creates an opposite internal electric field, which reduces the overall potential difference (V) between the plates for a given amount of charge (Q). Since C = Q/V, a decrease in V for the same Q results in an increase in capacitance.

Handling Multi-Step Calculations

If you are “Maths-strong but Physics-weak,” you might find the “why” harder than the “how.” We focus on:

• Concept Visualisation: Imagine the capacitor as a tank of water being filled through a narrow pipe (the resistor).
• Real-world examples: Think of a camera flash or a heart defibrillator.
• Diagrams: Always draw the circuit. It helps you see if capacitors are in series or parallel.

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Capacitor Components

When moving through multi-step questions, use this table as a quick reference for your units and symbols.

Quantity	Symbol	Standard Unit	Common Sub-units
Capacitance	C	Farad (F)	Microfarad ($\mu F$), Nanofarad ($nF$)
Charge	Q	Coulomb (C)	Millicoulomb ($mC$), Microcoulomb ($\mu C$)
Time Constant	$\tau$	Second (S)	Milliseconds ($ms$)
Energy Stored	W or E	Joule (J)	Microjoules ($\mu J$)
Resistance	R	Ohm ($\Omega$)	Kilohm ($k\Omega$), Megohm ($M\Omega$)

Read More What is Biomass Energy?

FAQs

What is a dielectric?

It is an electrical insulator that increases capacitance by reducing the internal electric field.

Why is the energy formula $1/2 QV$ and not just QV?

Because as the capacitor charges, the voltage increases from 0 to V. The work done is the average voltage multiplied by charge, which is $1/2 V$.

How do you add capacitors in parallel?

You simply add them: $C_{total} = C_1 + C_2 + C_3$$.

What is the time constant?

It is the time taken for the charge to drop to $37$% of its initial value.

Can a capacitor replace a battery?

Only for short bursts. They store much less energy than batteries but release it much faster.