Question

(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 (large-scale) charges?

Hint:

Quantisation: any charge is \(q=\pm n e\) with \(e=1.6\times10^{-19}\) C. It is ignored for macroscopic charges because \(e\) is so tiny that even 1 microcoulomb is about \(10^{13}\) electrons, making charge look continuous.

Answer 1

This is a conceptual question about the quantisation of charge.

Step 1 (part a): Meaning of quantisation. The statement "electric charge of a body is quantised" means that the total charge \(q\) on any body is always an integral (whole-number) multiple of a basic unit of charge, namely the magnitude of the electronic charge \(e\). In symbols,
\[q=\pm n\,e,\qquad n=0,1,2,3,\dots\]
where \(e=1.6\times10^{-19}\ \text{C}\). A body can carry charge \(0,\ \pm e,\ \pm 2e,\ \pm 3e,\dots\) but never a fractional value such as \(0.5\,e\) or \(1.7\,e\). The reason is that charge is transferred only by adding or removing whole electrons (or protons), each carrying charge of magnitude \(e\).

Step 2 (part b): Why quantisation is ignored on a large scale. The basic unit \(e=1.6\times10^{-19}\ \text{C}\) is extremely small. A macroscopic charge, for example \(1\ \mu\text{C}=10^{-6}\ \text{C}\), corresponds to
\[n=\frac{q}{e}=\frac{10^{-6}}{1.6\times10^{-19}}\approx 6\times10^{12}\]
electrons. Since \(n\) is a colossal number, adding or removing one electron changes the charge by only one part in about \(10^{12}\), which is far below any measurable resolution.

Step 3: Conclusion. At the macroscopic level the charge appears to vary continuously rather than in tiny discrete steps, just as a stream of sand appears continuous even though it is made of grains. Therefore quantisation can be safely ignored for large-scale charges.

Answer 2

Expert framing: granularity and the continuum limit.

Step 1 (part a): Charge is an intrinsic, conserved attribute carried in indivisible packets. Every free particle observed in the laboratory carries charge that is an integer multiple of \(e=1.6\times10^{-19}\ \text{C}\). Hence for any body \(q=n e\) with integer \(n\); intermediate values are forbidden because you cannot transfer a fraction of an electron. (Quarks carry \(\pm e/3,\ \pm 2e/3\), but they are permanently confined inside particles and never appear free, so the laboratory unit of free charge remains \(e\).)

Step 2 (part b): Treat \(e\) as the "step size" of a staircase. Quantisation matters only when the number of steps is small. For a macroscopic charge the step count is astronomically large.

Step 3: Quantify the smallness of one step relative to the whole. Even a modest charge of \(1\ \mu\text{C}\) needs about
\[n=\frac{1\times10^{-6}}{1.6\times10^{-19}}\approx 6.25\times10^{12}\ \text{electrons.}\]
The fractional change from gaining or losing a single electron is \(1/n\approx1.6\times10^{-13}\), utterly undetectable.

Step 4: This is the standard "continuum limit": when the quantum (the grain) is vanishingly small compared with the total, the discrete variable behaves as a smooth continuous one. So macroscopic electrostatics treats charge as continuously divisible, and quantisation is dropped.