PhysicsICSE Class 9 · Current Electricity

Electric Charge, Current and Potential Difference: Q = ne, I = Q/t and V = W/q (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

Charge is quantised: Q = ne with e = 1.6 × 10⁻¹⁹ C. Current is the rate of flow of charge, I = Q/t (ampere = coulomb per second); conventional current flows from the positive terminal to the negative, opposite to electron flow. Potential difference is the work done per unit charge, V = W/q (volt = joule per coulomb).

Conventional current leaves the + terminal, while electrons move the opposite way; the ammeter sits in series to measure I = Q/t.

Illustration drawn for QuestionBuddy.

Understand Electric Charge, Current and Potential Difference: Q = ne, I = Q/t, V = W/q

Free electrons and conductors. In a metal the outer electrons are loosely held and move freely; a p.d. across the metal makes them drift, forming a current. Good conductors have many free electrons; insulators almost none; semiconductors are in between; a superconductor has zero resistance below a critical temperature.

Charge and its quantum nature. Electrons carry negative charge, protons positive. The unit is the coulomb (C). The smallest free charge is that of an electron or proton, $e = 1.6 \times 1 0^{- 19}$ C, and every charge on a body is a whole-number multiple of $e$ :

$Q = n e (n = 1, 2, 3, \dots)$

so $1 C = \frac{1}{1.6 \times 1 0 ^{- 19}} \approx 6.25 \times 1 0^{18}$ electrons. A charge that is not a whole multiple of $e$ cannot exist on a body.

Electric current. Rate of flow of charge:

$I = \frac{Q}{t}, 1 A = 1 C s^{- 1}$

with $1 mA = 1 0^{- 3}$ A and $1 μ A = 1 0^{- 6}$ A. If $n$ electrons pass in time $t$ , $I = \frac{n e}{t}$ .

*Direction:* conventional current flows from the positive terminal through the circuit to the negative terminal; the actual electron flow is the opposite way. An ammeter is connected in series.

Potential difference. Charges move only if there is a potential difference between two points (like water flowing from a higher to a lower level). The p.d. is the work done per unit charge in carrying a charge between the two points:

$V = \frac{W}{q}, 1 V = 1 J C^{- 1}$

so $W = q V$ . A voltmeter is connected in parallel. A cell maintains a p.d. between its terminals.

Typical chain of calculation. From current and time find the charge ( $Q = I t$ ), then the number of electrons ( $n = \frac{Q}{e}$ ), then the work done in moving the charge through a p.d. ( $W = Q V$ ).

Units to convert. Minutes to seconds, mA to A, μC to C.

Electric Charge, Current and Potential Difference: Q = ne, I = Q/t, V = W/q: formulas and key facts

Quantum nature of charge: $Q = n e, e = 1.6 \times 1 0^{- 19} C$; n is a whole number.
Current: $I = \frac{Q}{t} = \frac{n e}{t}$; A = C s⁻¹.
Potential difference: $V = \frac{W}{q} \Rightarrow W = q V$; V = J C⁻¹.
Electrons in one coulomb: $\frac{1}{e} = 6.25 \times 1 0^{18}$; Approximate.

How to solve Electric Charge, Current and Potential Difference: Q = ne, I = Q/t, V = W/q questions

Move step by step from current to charge to electrons to energy.

1Convert time to seconds and current to amperes.
2Find $Q = I t$ (or $Q = n e$ ).
3Divide by $e$ to get the number of electrons.
4Use $W = Q V$ for the work done by the source.
5State the direction of conventional current if asked.

Electric Charge, Current and Potential Difference: Q = ne, I = Q/t, V = W/q in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Charging a phone

Suppose a charger delivers a steady 2 A at 5 V for one hour.

Find the charge delivered, the number of electrons that cross a point and the electrical work done.

Try it first, then show the solutionSolution

$Q = I t = 2 \times 3600 = 7200$ C.
$n = \frac{7200}{1.6 \times 1 0 ^{- 19}} = 4.5 \times 1 0^{22}$ .
$W = Q V = 7200 \times 5 = 36000$ J.

Answer: 7200 C; 4.5 × 10²² electrons; 36 000 J.

Memory hook: It, then ÷e, then × V.

Tricks to remember Electric Charge, Current and Potential Difference: Q = ne, I = Q/t, V = W/q

Short memory aids. Say each one aloud once, then use it on a question the same day.

Whole numbers of e
Q = ne: divide by 1.6 × 10⁻¹⁹; a whole number is allowed, a fraction is not.
Use it when: Quantisation questions.
Ammeter in series, voltmeter in parallel
Ammeter: A in a line. Voltmeter: V across.
Use it when: Circuit diagrams.
Positive to negative outside
Conventional current leaves the + terminal; electrons go the other way.
Use it when: Direction questions.

Electric Charge, Current and Potential Difference: Q = ne, I = Q/t, V = W/q: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Ammeter connection · 1 mark · MCQBoard level

An ammeter is connected in a circuit:
(a) in parallel with the resistor
(b) in series with the resistor
(c) across the cell
(d) in parallel with the key

Try it first, then show the solutionSolution

An ammeter must carry the whole current, so it goes in series.

Answer: (b)

Example 2: Number of electrons · 2 marks · Short answerBoard level

A body carries a charge of $3.2 \times 1 0^{- 18}$ C. How many electrons does it have in excess ( $e = 1.6 \times 1 0^{- 19}$ C)? What is the sign of the charge if the excess particles are electrons?

Try it first, then show the solutionSolution

$n = \frac{Q}{e} = \frac{3.2 \times 1 0 ^{- 18}}{1.6 \times 1 0 ^{- 19}} = 20$ . (1½)
The charge is negative. (½)

Answer: 20 electrons; negative

Example 3: Current, charge and work · 3 marks · Short answerHarder board level

A current of 0.4 A flows through a lamp for 2 minutes at 6 V. Find the charge, the number of electrons passing and the work done.

Try it first, then show the solutionSolution

$Q = I t = 0.4 \times 120 = 48$ C. (1)
$n = \frac{48}{1.6 \times 1 0 ^{- 19}} = 3.0 \times 1 0^{20}$ . (1)
$W = Q V = 48 \times 6 = 288$ J. (1)

Answer: 48 C; 3.0 × 10²⁰ electrons; 288 J

Example 4: Time for electrons to pass · 3 marks · Short answerHarder board level

How long does it take for $1.2 \times 1 0^{20}$ electrons to pass a point if the current is 0.8 A? If the work done is 115.2 J, find the p.d.

Try it first, then show the solutionSolution

$Q = n e = 1.2 \times 1 0^{20} \times 1.6 \times 1 0^{- 19} = 19.2$ C. (1)
$t = \frac{Q}{I} = \frac{19.2}{0.8} = 24$ s. (1)
$V = \frac{W}{Q} = \frac{115.2}{19.2} = 6$ V. (1)

Answer: 24 s; 6 V

Example 5: A power bank · 4 marks · Case-basedHOTS / competency

A power bank supplies 2 A at 5 V for one hour to charge a phone.
(a) Find the charge that flows.
(b) Find the number of electrons.
(c) Find the work done by the supply.
(d) A student says electrons flow from the + terminal to the − terminal outside the supply. Correct her and name the conventional direction.

Try it first, then show the solutionSolution

(a) $Q = 2 \times 3600 = 7200$ C. (1)
(b) $\frac{7200}{1.6 \times 1 0 ^{- 19}} = 4.5 \times 1 0^{22}$ . (1)
(c) $W = 7200 \times 5 = 36000$ J. (1)
(d) Electrons flow from the negative to the positive terminal; the conventional current runs from + to − outside the supply. (1)

Answer: (a) 7200 C (b) 4.5 × 10²² (c) 36 000 J (d) electrons from − to +; conventional current + to −

Where students lose marks in Electric Charge, Current and Potential Difference: Q = ne, I = Q/t, V = W/q

The slips examiners see most often, and what to write instead.

Mistake: Time in minutes used in I = Q/t
Time must be in seconds.
Wrong working
Q = 0.4 × 2 = 0.8 C
Right working
Q = 0.4 × 120 = 48 C
Do this: Convert first.
Usually costs: The answer.
Mistake: Direction of electron flow reversed
Electrons flow from negative to positive.
Wrong working
Electrons leave the positive terminal.
Right working
Conventional current + to −; electrons − to +.
Do this: Write both directions.
Usually costs: 1 mark.
Mistake: Voltmeter in series
A voltmeter is in parallel.
Wrong working
Voltmeter in line with the resistor.
Right working
Voltmeter across the resistor.
Do this: V across.
Usually costs: Circuit marks.

Exam tips

Convert time to seconds and current to amperes before substituting.
Show Q = It, then n = Q/e as separate lines.
State the sign convention when asked about direction.
Include units (C, A, V, J) in every line of the working.

Practice questions on Electric Charge, Current and Potential Difference: Q = ne, I = Q/t, V = W/q

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Electric current is the rate of flow of:
(a) mass
(b) energy
(c) charge
(d) potential
Hint
Definition.
Answer
(c) charge
Q2 · 1 mark · Very short answerBoard level
State the SI unit of potential difference.
Hint
Joule per coulomb.
Answer
Volt (V).
Q3 · 2 marks · Short answerHarder board level
How many electrons are there in a charge of 0.5 C? ( $e = 1.6 \times 1 0^{- 19}$ C)
Hint
n = Q/e.
Answer
$n = \frac{0.5}{1.6 \times 1 0 ^{- 19}} = 3.1 \times 1 0^{18}$ (3.125 × 10¹⁸).
Q4 · 3 marks · Short answerHarder board level
A charge of 0.6 C passes through a bulb in 20 s while the p.d. across it is 3 V. Find the current and the work done.
Hint
I = Q/t, W = QV.
Answer
$I = \frac{0.6}{20} = 0.03$ A $= 30$ mA; $W = 0.6 \times 3 = 1.8$ J.
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): A body cannot carry a charge of $2.4 \times 1 0^{- 19}$ C.
Reason (R): Charge is quantised in whole-number multiples of $1.6 \times 1 0^{- 19}$ C.
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Hint
$2.4/1.6$ is not a whole number.
Answer
(a). $\frac{2.4}{1.6} = 1.5$ , not an integer.
Q6 · 3 marks · Short answerHOTS / competency
A student measures the charges on three tiny drops as $6.4 \times 1 0^{- 19}$ C, $9.6 \times 1 0^{- 19}$ C and $12.8 \times 1 0^{- 19}$ C. (a) Find the largest quantity of which all three are whole multiples. (b) Can she conclude the elementary charge is that quantity? (c) Express each charge as $n e$ using $e = 1.6 \times 1 0^{- 19}$ C.
Hint
Find the common factor, then think about the smallest charge seen.
Answer
(a) $3.2 \times 1 0^{- 19}$ C. (b) No: all drops could carry 2, 3 and 4 units of it, but a drop with $1.6 \times 1 0^{- 19}$ C might exist; the elementary charge is the smallest charge found, which needs more measurements. (c) $n = 4$ , $6$ and $8$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Value of the elementary charge?Show answer
1.6 × 10⁻¹⁹ C.
2.Definition of current?Show answer
Rate of flow of charge, I = Q/t.
3.1 volt equals?Show answer
1 joule per coulomb.
4.Which instrument is connected in parallel with a component?Show answer
The voltmeter.
5.Direction of conventional current?Show answer
From the positive to the negative terminal.

Electric Charge, Current and Potential Difference: Q = ne, I = Q/t, V = W/q: frequently asked questions

What is the quantum nature of charge?

Charge exists only in whole-number multiples of the electron charge, Q = ne with e = 1.6 × 10⁻¹⁹ C.

What is the difference between conventional current and electronic current?

Conventional current is taken from the positive to the negative terminal; electrons actually flow the opposite way.

What is potential difference?

The work done per unit charge in moving charge between two points, V = W/q, measured in volts.