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Exercises · 1.4

Q.(a) Explain the meaning of the statement 'electric charge of a body is quantised'.

(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?
Uttarakhand UbseTextbookSubjective· 2mImportance★★★★★
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Electric charge is quantised, meaning it exists only in integer multiples of the elementary charge e=1.6×10−19 Ce = 1.6 \times 10^{-19} \, \text{C}. At macroscopic scales, the number of charge carriers is so enormous that the discrete jumps become negligible, and charge behaves as if it were continuous.

(a) What "quantised" means for electric charge

The statement "electric charge of a body is quantised" means that any observable charge QQ on an object is always an integer multiple of a fundamental, smallest unit of charge. That unit is the magnitude of the charge on a single electron or proton, denoted by ee:

e=1.602×10−19 Ce = 1.602 \times 10^{-19} \, \text{C}

So if a body has a net charge QQ, it must satisfy:

Q=±ne,where n=0,1,2,3,…Q = \pm n e, \quad \text{where } n = 0, 1, 2, 3, \dots

You cannot have, say, 0.5e0.5e or 1.7e1.7e on a body in isolation. Charge comes in discrete packets — it is not a continuous fluid that can be divided arbitrarily. This was first demonstrated convincingly by Millikan's oil drop experiment, which showed that every measured charge was a multiple of ee.

Important

The quantisation of charge is a fundamental law of nature. It arises because matter is made of electrons and protons, each carrying exactly ±e\pm e. Any transfer of charge involves moving whole electrons or protons — you cannot transfer a fraction of an electron.

(b) Why we ignore quantisation at macroscopic scales

When dealing with macroscopic (large-scale) charges — say, a charged metal sphere carrying 1 μC1 \, \mu\text{C} — the number of excess electrons (or protons) is enormous. Let's calculate:

n=Qe=1×10−6 C1.6×10−19 C≈6.25×1012n = \frac{Q}{e} = \frac{1 \times 10^{-6} \, \text{C}}{1.6 \times 10^{-19} \, \text{C}} \approx 6.25 \times 10^{12}

That's over six trillion elementary charges. Now, if you add or remove even a few billion electrons, the change in charge is:

ΔQ=(109)×(1.6×10−19)=1.6×10−10 C\Delta Q = (10^9) \times (1.6 \times 10^{-19}) = 1.6 \times 10^{-10} \, \text{C}

This is 0.160.16 nanocoloumbs — far below the sensitivity of most macroscopic measuring instruments. The relative jump between allowed charge values is:

eQ≈1.6×10−1910−6=1.6×10−13\frac{e}{Q} \approx \frac{1.6 \times 10^{-19}}{10^{-6}} = 1.6 \times 10^{-13} …

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