Q.In Milikan's experiment, the charge found on a charged droplet was -6.4 x 10^-19 C, then write the number of electrons in that charged droplet.
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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" …
Millikan's oil-drop experiment showed that electric charge is always quantised in integer multiples of the elementary charge e, so the number of excess electrons on the droplet is found by dividing its total charge magnitude by e. …
Charge is quantised in integer multiples of the electronic charge e; dividing the droplet's charge by e gives the number of excess electrons on it.
Millikan's oil-drop experiment showed that electric charge is always an integral multiple of the elementary charge e=1.6×10−19 C:
q=ne
…
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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