Mini Physics
Part of Mini EducationSwitch subject
SearchBookmarksMenu

Electric charge, current and Q = It

Key idea: Rubbing transfers electrons: gaining them makes an object negative and losing them makes it positive. Like charges repel, unlike charges attract, and current is charge passing a point each second.

Reading settings

On this page

Reading settings

Text size, line spacing and reading width for your lessons.

Text size
Line spacing
Reading width

Live preview

Take a moment to read at your own pace. Choose the text size and spacing that feel comfortable for you.

Pause after each idea, then explain it in your own words before moving on.

Narrower lines take effect when the screen has room.

Enlarged diagram

Pull a jumper over your head and your hair stands up; switch on a torch and the bulb lights. Both come down to electric charge. Rubbing two objects together moves charge from one to the other, and current measures how quickly charge flows through a circuit.

Charge and interactions

Charge is positive or negative

Electric charge, Q, is measured in coulombs (C). There are two kinds of charge: positive and negative. Protons carry positive charge; electrons carry negative charge.

A neutral object still contains charge. It has equal amounts of positive and negative charge, so its net charge is zero; the charged particles are still present.

When two solid objects are rubbed together, electrons can move from one to the other. The protons stay in the nuclei of the atoms. So:

  • an object that gains electrons becomes negative;
  • an object that loses electrons becomes positive.
Charge account for two objects rubbed togetherBefore rubbing, objects A and B are each neutral: each has six positive and six negative charges. During rubbing, two electrons move from A to B. After rubbing, A has six positive and four negative charges, a net charge of plus two units; B has six positive and eight negative charges, a net charge of minus two units. The total charge is zero before and after, and the oppositely charged objects attract.Before: both neutralAB+ + + + + +− − − − − −+ + + + + +− − − − − −A: 6 positive, 6 negative → 0B: 6 positive, 6 negative → 0Total charge: 0After: 2 electrons moved A → B2 electronsAB+ + + + + +− − − −+ + + + + +− − − − − − − −A: 6 positive, 4 negative → +2B: 6 positive, 8 negative → −2Total charge: (+2) + (−2) = 0Only electrons move between the objects; the positive charges stay in the atoms.After rubbing, A and B have unlike charges, so they attract.
Charge account for two objects rubbed togetherBefore rubbing, objects A and B are each neutral: each has six positive and six negative charges. During rubbing, two electrons move from A to B. After rubbing, A has six positive and four negative charges, a net charge of plus two units; B has six positive and eight negative charges, a net charge of minus two units. The total charge is zero before and after, and the oppositely charged objects attract.Before: both neutralA+ + + + + +− − − − − −B+ + + + + +− − − − − −A: 6 positive, 6 negative → 0B: 6 positive, 6 negative → 0Total charge: 0After: 2 electrons moved A → B2 electronsA+ + + + + +− − − −B+ + + + + +− − − − − − − −A: 6 positive, 4 negative → +2B: 6 positive, 8 negative → −2Total charge: (+2) + (−2) = 0Only electrons move between the objects; the positive charges stay in the atoms.After rubbing, A and B have unlike charges, so they attract.
Rubbing moves electrons from A to B. No charge is made or destroyed: the total stays zero, and A and B end with equal and opposite charges.

Worked example 1

Keep a charge account

Problem

Objects A and B start neutral. Rubbing moves 2 electrons from A to B. What charge does each object end with?

Show full solution

  1. Follow the electrons out of A

    Method

    Count what A loses.

    Reason

    A loses 2 negative charges but keeps all its positive charges.

    Working

    Answer

    A has a net positive charge

  2. Follow the electrons into B

    Method

    Count what B gains.

    Reason

    B gains the same 2 negative charges.

    Working

    Answer

    B has a net negative charge of the same size

  3. Check the total

    Working

    Answer

    The total was zero before and is still zero. Rubbing separates charge; it does not create it.

Check your understanding 1

A neutral plastic bead gains four electrons. What sign is its charge, and what happened to its protons?

Sign of the bead’s charge
What happened to its protons
Show answer

It is negative: it now has four more electrons than protons. Its protons did not move. Only electrons move between rubbed solids.

Check your understanding 2

A rod and cloth both start neutral, and charge passes only between them during rubbing. The rod becomes negatively charged. State what moved and in which direction, and state the sign of the charge on the cloth.

What moved, and which way
Charge on the cloth
Show answer

Electrons moved from the cloth to the rod. The cloth lost electrons, so it is positive, with a charge equal in size to the rod’s.

Like charges repel, unlike charges attract

Charges on the two objectsKindWhat happens
positive and negativeunlikeattract
positive and positivelikerepel
negative and negativelikerepel

Work out each object’s charge first, then apply the rule. After rubbing, A is positive and B is negative, so A and B attract. Two objects that have both gained electrons are both negative, so they repel.

Check your understanding 3

A student says, “A positive sphere has gained protons, and a neutral sphere has no charged particles.” Correct both claims. Then predict what happens between two initially neutral spheres that have each lost electrons.

Show answer

A sphere becomes positive by losing electrons, not by gaining protons. A neutral sphere has equal amounts of positive and negative charge. Two spheres that have both lost electrons are both positive, so they repel.

Current, charge flow and direction

Current is the rate of flow of charge

Current, I, is the rate of flow of charge: the charge that passes a point in a circuit each second. It is measured in amperes (A).

Picture standing at one point in a wire and counting the charge that goes past.

Counting charge at one point in a wireA section of wire carrying a steady current, with a dashed counting point marked across it. Six dots, each standing for 1.0 coulomb of charge, lie before the counting point and cross it during the next 2.0 seconds, in the direction of the conventional current arrow. Below, 6.0 coulombs divided by 2.0 seconds gives 3.0 amperes, meaning 3.0 coulombs cross the point every second.wirecounting pointconventional current →6 dots cross in the next 2.0 sEach dot stands for 1.0 C of charge.I = Q ÷ t = 6.0 C ÷ 2.0 s = 3.0 A3.0 A means 3.0 C crosses the point every second.
Counting charge at one point in a wireA section of wire carrying a steady current, with a dashed counting point marked across it. Six dots, each standing for 1.0 coulomb of charge, lie before the counting point and cross it during the next 2.0 seconds, in the direction of the conventional current arrow. Below, 6.0 coulombs divided by 2.0 seconds gives 3.0 amperes, meaning 3.0 coulombs cross the point every second.wirecounting pointconventional current →6 dots cross in the next 2.0 sEach dot stands for 1.0 C of charge.I = Q ÷ t = 6.0 C ÷ 2.0 s = 3.0 A3.0 A means 3.0 C crosses the point every second.
For the steady current shown, 6.0 C crosses the point in 2.0 s: 3.0 C crosses every second, a current of 3.0 A.
I = Q/t

For a steady current, equal amounts of charge pass in equal time intervals. If current varies, Q/t gives its average over the measured interval. One ampere means one coulomb of charge passes a point each second: 1 A = 1 C/s.

A lamp does not use up current: the same charge per second enters and leaves it. If less current left a lamp than entered it, charge would pile up inside the lamp every second. The lamp transfers energy, not charge.

Conventional current and electron flow

Opposite directions in one metal wireThe cell has its negative short plate on the left and positive long plate on the right. Electrons drift up the left wire, right across the top and down through the lamp to the positive terminal. Conventional current is opposite, left along the top wire. Negative dots represent electrons; the current arrow is only a direction convention.Conventional current ←Electrons →lamp−+cell
Follow the same top wire: electron drift is right; conventional current is left.
  1. Find the cell’s + long plate and − short plate. The cell and lamp form one closed loop.
  2. Watch the − dots drift right in the top wire, from the negative terminal towards the positive terminal. The current arrow points left.
  3. The arrow is a convention, not a second stream of particles. Both directions reverse if the cell is reversed.

This schematic slows the drift and leaves out random electron motion. The arrow shows a convention, not moving positive ions; the circuit is not to scale.

Conventional current is the direction in which positive charge would flow: out of the positive terminal of the cell, round the circuit and back to the negative terminal.

In a metal wire, the charges that actually move are electrons. They are negative, so they drift the other way: from the negative terminal towards the positive terminal.

FlowDirection round the circuitWhat it describes
Conventional currentpositive terminal → negative terminalthe agreed current direction
Electron flow in a metalnegative terminal → positive terminalthe motion of the electrons

Current arrows on circuit diagrams show conventional current.

Check your understanding 4

12 C passes a point in a wire in 4.0 s. Find the current. The electrons move to the left. Which way is the conventional current?

Unit: A
Conventional current direction
Show answer

I = Q/t = 12/4.0 = 3.0 A. The conventional current is to the right, opposite to the electron motion.

Use Q = It with time in seconds

Rearranging I = Q/t gives

Q = It

For constant current, use Q in coulombs, I in amperes and t in seconds. If current varies, I in this equation is the average over that interval.

Worked example 2

Charge from current and time

Problem

A current of 0.40 A flows for 3.0 minutes. Find the charge that passes.

Show full solution

  1. Convert the time

    Method

    Change minutes to seconds.

    Reason

    Q = It needs amperes and seconds.

    Working

    Answer

    t = 3.0 × 60 = 180 s

  2. Substitute

    Method

    Write Q = It before substituting.

    Reason

    Charge is current multiplied by time.

    Working

    Answer

    Q = 0.40 × 180 = 72 C

  3. Check

    Working

    Answer

    0.40 C passes every second for 180 seconds, so 72 C is sensible.

Milliamperes and milliseconds. Small currents and short times use the prefix milli-: 1 mA = 0.001 A and 1 ms = 0.001 s. Convert both before you use Q = It.

GivenConvert toWorking
250 mAamperes250 × 0.001 = 0.250 A
40 msseconds40 × 0.001 = 0.040 s
2.5 minutesseconds2.5 × 60 = 150 s

Worked example 3

Charge with prefixed units

Problem

A current of 250 mA flows for 40 ms. Find the charge that passes.

Show full solution

  1. Convert to base units

    Method

    Change milliamperes to amperes and milliseconds to seconds.

    Reason

    Q = It gives coulombs only from amperes and seconds.

    Working

    Answer

    I = 0.250 A, t = 0.040 s

  2. Substitute

    Working

    Answer

    Q = It = 0.250 × 0.040 = 0.010 C, which is 10 mC

Check your understanding 5

A camera flash receives 5.0 mC of charge in 2.0 ms. Write the charge in coulombs and the time in seconds, then find the current.

Unit: C
Unit: s
Show the steps Optional working
Unit: C
Unit: s
Unit: A
Show answer

Q = 5.0 mC = 0.0050 C and t = 2.0 ms = 0.0020 s, so I = Q/t = 0.0050/0.0020 = 2.5 A.

Check your understanding 6

A steady 0.35 A flows through a lamp for 4.0 minutes. How much charge enters the lamp? A student says only half of it leaves because the lamp uses it up. How much charge actually leaves the lamp in that time?

Show the steps Optional working
Unit: s
Unit: C
Unit: C
Show answer

t = 4.0 × 60 = 240 s, so Q = It = 0.35 × 240 = 84 C. The same 84 C leaves the lamp in that time. The lamp transfers energy; it does not use up charge.

Check your understanding 7 Use changes in charge readings

A counter records the total charge that has passed one point since timing began: 0 C at 0 s, 0.80 C at 2.0 s and 2.0 C at 6.0 s. Find the average current from 0 to 2.0 s and from 2.0 to 6.0 s. A learner divides the final 2.0 C by 6.0 s and says the current was steady throughout. Review that claim, and decide whether the interval averages give the current at every instant.

Unit: A
Show the steps Optional working
Unit: C
Unit: s
Unit: A
Was the current steady throughout?
Do interval averages give the current at every instant?
Show answer

From 0 to 2.0 s: current = (0.80 − 0)/2.0 = 0.40 A. From 2.0 to 6.0 s: charge change = 2.0 − 0.80 = 1.2 C, time change = 4.0 s, so average current = 0.30 A. The whole-run average is 2.0/6.0 ≈ 0.33 A, but the two different interval averages show that the current was not steady throughout. Each interval average describes charge per second over that interval; endpoint readings alone do not show whether current was constant within it or what its value was at a particular instant.

Challenge yourself

A current of 0.30 A flows for 4 minutes. Calculate the charge transferred. Then find the current needed to transfer the same charge in 90 s.

Check your thinking

First convert 4 minutes to 240 s. Q = It = 0.30 × 240 = 72 C. For the second case, I = Q/t = 72/90 = 0.80 A. Transferring the same charge in less time needs a larger current.

Common mistakes

  • Saying a positive rod gained protons. Protons stay in the nuclei; the rod lost electrons.
  • Saying electrons flow the same way as conventional current. Conventional current goes from + to −; electrons in a metal drift from − to +.
  • Saying a lamp uses up some of the current. The current into a lamp equals the current out of it.
  • Using minutes in Q = It. For 0.25 A for 2.0 minutes, Q = 0.25 × 120 = 30 C, not 0.25 × 2.0 = 0.50 C.

Spotted an error? Report a correction

Syllabus and review details

Last reviewed:

Back to top