Q.What important fact did Millikan's experiment establish?
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Quantization of Charge
Imagine you are at a water fountain. You can fill your bottle with any amount of water — a little, a lot, or anything in between. Water is continuous. Now imagine instead that you are buying marbles. You can only buy marbles in whole numbers: 1 marble, 2 marbles, 15 marbles. You cannot buy half a marble or 2.7 marbles. Electric charge behaves like marbles, not like water.
That is the core intuition: charge comes in discrete packets. You cannot have an arbitrary amount of charge. You can only have whole-number multiples of a smallest possible chunk.
The smallest chunk: the elementary charge
That smallest chunk is called the elementary charge, denoted by the symbol e. Its value is:
e=1.602×10−19 coulombs
This is the magnitude of charge carried by a single proton (positive) or a single electron (negative). A proton has charge +e, an electron has charge −e.
Every charged object in the universe — from a rubbed balloon to a lightning bolt — carries a total charge that is an integer multiple of e. No exception has ever been observed among free, isolated charges. (Quarks carry fractional charges of ±e/3,±2e/3 but are always confined inside composite particles such as protons and neutrons, whose own net charge is still an integer multiple of e.)
The precise statement
If q is the total charge on any object, then:
q=ne
where n is an integer (n=0,±1,±2,±3,…).
The sign of n tells you whether the charge is positive or negative. The magnitude ∣n∣ tells you how many elementary charges are present (in excess or deficit).
q=ne,n∈Z
Why this matters
This is not a mathematical trick. It is a fundamental law of nature. It means:
- You cannot have a charge of 0.5e or 1.7e.
- If you measure the charge on any object, you will always find it to be 0, ±e, ±2e, ±3e, and so on.
- All charge transfer — rubbing, conduction, induction — happens by moving whole electrons or protons. You cannot transfer a fraction of an electron.
A common mistake is to think that because charge values like 3.2×10−19 C look like decimals, they are not multiples of e. But 3.2×10−19 C is exactly 2e (since 2×1.6×10−19=3.2×10−19). Always check by dividing by e — the result must be an integer.
A concrete example
A glass rod rubbed with silk acquires a charge of +4.8×10−19 C. How many electrons were transferred?
Divide the total charge by e:
n=1.6×10−19+4.8×10−19=+3
So the rod lost exactly 3 electrons. It could not have lost 2.5 or 3.7 electrons. The charge is +3e.
Why this is called "quantization" …
By balancing tiny charged oil drops in an electric field, Millikan measured the charge on each drop and found every value was a whole-number multiple of a fixed smallest charge. …
Millikan's oil-drop experiment proved that electric charge is quantised, occurring only in integral multiples of the elementary charge e = 1.6 x 10^-19 C.
In the experiment, charged oil droplets were introduced between two horizontal plates. By adjusting the electric field, a droplet could be held stationary (electric force balancing gravity) or made to move at known speeds. Measuring the field and the drop's motion gave the charge q on each drop.
Millikan found that the charge on every drop was always a whole-number multiple of a definite smallest value:
q = n e, where n = 1, 2, 3, ... and e = 1.6 x 10^-19 C.
…
Showing the 12 most recent of 27 on this concept.
- CBSE 2026Set V11 markMCQQ.Which one of the following charge cannot exist on a body?(a) 2e(b) 3e(c) 3.5e(d) −4e
›Reveal solutionSolution
- CBSE 2026Set ANNUAL1 markMCQQ.Number of electrons in one coulomb charge are -(i) 6.25×1018(ii) 6.25×1012(iii) 6.25×1010(iv) 6.25×1014
›Reveal solutionSolution
1 C of charge contains 1/e electrons.
The charge on one electron is e=1.6×10−19 C. The number of electrons whose total charge is 1 coulomb is n=e1=1.6×10−191=6.25×1018.
…
- CBSE 2026Set DS1 markMCQQ.Number of electrons emitted from a piece of metal for giving 1×10−7 coulomb charge will be:i) 107ii) 1.6×1019iii) 6.25×1011iv) 9×1012
›Reveal solutionSolution
Charge is quantised, so n=Q/e=6.25×1011 electrons.
Concept. Charge exists only in whole multiples of the electronic charge e=1.6×10−19 C (quantisation of charge). If a metal piece loses a charge Q, the number of electrons removed is n=eQ.
Calculation. …
- CBSE 2026Set ANNUAL1 markMCQQ.An object has a negative charge of 1 coulomb. The number of excess electrons on it is(a) 6.25 x 10^-18(b) 1.6 x 10^19(c) 1.6 x 10^-19(d) 6.25 x 10^18
›Reveal solutionSolution
Charge is quantised: any charge Q is made up of a whole number of electron charges e, so n = Q/e.
A negative charge means the object has more electrons than protons. Each electron carries a charge of magnitude e = 1.6 x 10^-19 C. If the object carries a total (excess) charge of magnitude Q, the number of excess electrons is
…
- CBSE 2025Set 55/4/11 markMCQQ.A body acquires charge 8.0×10−12 C. The mass of the body: (A) increases by 4.5×10−7 kg (B) decreases by 1.0×10−6 kg (C) decreases by 4.55×10−23 kg (D) increases by 9.1×10−23 kg
›Reveal solutionSolution
When a body gains positive charge, it loses electrons; the mass change equals the number of electrons lost times the electron mass. For 8.0×10−12 C, the body decreases by 4.55×10−23 kg.
Why charging changes mass
Charging a body means adding or removing electrons. Since electrons have mass (me=9.1×10−31 kg), any change in the number of electrons changes the body's total mass.
The sign of the charge tells us what happened:
- Positive charge: electrons were removed → mass decreases
- Negative charge: electrons were added → mass increases
The magnitude of charge tells us how many electrons moved, since each electron carries charge e=1.6×10−19 C.
Finding the mass change
-
Determine the number of electrons involved
The charge acquired is Q=8.0×10−12 C (positive). The number of electrons that must have been removed is:
n=eQ=1.6×10−198.0×10−12
n=5.0×107 electrons
-
Calculate the total mass of these electrons
Each electron has mass me=9.1×10−31 kg, so the total mass lost is:
Δm=n×me=5.0×107×9.1×10−31
Δm=45.5×10−24=4.55×10−23 kg
- Determine the direction of change …
- CBSE 2025Set JS1 markQ.A conductor has positive charge of 2.4×10−18 coulomb. Find how much electrons are in deficit/excess on the conductor.
›Reveal solutionSolution
A positive charge of 2.4×10−18 C corresponds to a shortage of n=q/e=15 electrons.
Concept — quantisation of charge. Charge exists only in integer multiples of the electronic charge e=1.6×10−19 C: q=ne. A positive conductor has lost electrons, so it has a deficit of electrons.
Solution. …
- CBSE 2025Set A1 markQ.Fill in the blank with appropriate word: One coulomb charge has ______ electrons.
›Reveal solutionSolution
One coulomb of charge corresponds to about 6.25 × 10¹⁸ electrons.
Charge is quantised: total charge q=ne, where n is the number of electrons (or elementary charges) and e=1.6×10−19 C is the magnitude of charge on a single electron. For q=1 C:
…
- CBSE 2025Set A1 markQ.Write answer in one sentence: Write mathematical form of quantisation of electric charge.
›Reveal solutionSolution
Quantisation of charge is expressed mathematically as q = ne.
Experiments (starting with Millikan's oil-drop experiment) show that electric charge is never continuous but always occurs in discrete, integral multiples of a smallest indivisible unit of charge, called the elementary charge e=1.6×10−19 C (the magnitude of the charge on an electron or proton). This is expressed mathematically as:
q=ne
…
- CBSE 2025Set ANNUAL1 markMCQQ.How many electrons will have a charge of one Coulomb?(a) 6.25 x 10^18(b) 6.25 x 10^19(c) 5.25 x 10^18(d) 5.25 x 10^19
›Reveal solutionSolution
Charge is quantized in units of the electronic charge e = 1.6 x 10^-19 C, so the number of electrons needed to make up 1 C is n = Q/e.
Any charge Q is an integer multiple of the elementary charge: Q = n e.
Here Q = 1 C and e = 1.6 x 10^-19 C, so
n = Q/e = 1 / (1.6 x 10^-19) = 6.25 x 10^18
…
- CBSE 2024Set A1 markMCQQ.Number of electrons present in 8 coulomb negative charge is (A) 5 × 10^19 (B) 2.5 × 10^19 (C) 12.8 × 10^19 (D) 1.6 × 10^19
›Reveal solutionSolution
n = Q/e = 8 ÷ (1.6×10⁻¹⁹) = 5×10¹⁹ electrons.
Charge is quantised: total charge Q=ne, where e=1.6×10−19 C.
…
- CBSE 2024Set ANNUAL1 markMCQQ.The minimum amount of charge observed so far is(a) 1 C(b) 4.8 x 10^-13 C(c) 1.6 x 10^-19 C(d) 1.6 x 10^19 C
›Reveal solutionSolution
Charge is quantised: every observable free charge is an integer multiple of the elementary charge e = 1.6 x 10^-19 C, the smallest charge ever measured on a free particle.
Millikan's oil-drop experiment established that electric charge does not take arbitrary values but always occurs as an integral multiple of a smallest unit, the electronic charge
e=1.6×10−19 C
…
- CBSE 2024Set ANNUAL1 markQ.What does q1+q2=0 signify in electrostatics?
›Reveal solutionSolution
q1+q2=0 means the two charges are equal in magnitude and opposite in sign, giving zero net charge for the pair — the defining condition of a system like an electric dipole.
If q1+q2=0, then q2=−q1: the two charges have exactly equal magnitude but opposite polarity (one is +q, the other is −q). This does not mean there is no electric field around them — a pair of equal and opposite charges separated by some distance still produces a field (this is precisely the definition of an electric dipole, with dipole moment p=q×(separation), and it is only the algebraic (net) charge of the pair that vanishes, not the field. It also means that if this pair is enclosed inside a Gaussian …
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