Why can we say that charge of any body is always an integral multiple of $'e'$ ?

Vedclass pdf generator app on play store
Vedclass iOS app on app store

If the protons and electrons are the only basic charges in the universe, all the observable charges have to be integral multiples of $e$.

If a body contains $n_{1}$ electrons and $n_{2}$ protons, the total amount of charge on the body is $n_{2} e+$ $n_{1}(-e)=\left(n_{2}-n_{1}\right) e$

where $n_{1}$ and $n_{2}$ is integral multiples.

and their difference $=n_{2} e-n_{1}(-e)$

$=\left(n_{2}+n_{1}\right) e$ also an integer

Thus, the charge on any body is always an integral multiple of $e$ and can be increased or decreased also in steps of $e$.

Similar Questions

What is called earthing ? Give importance of earthing in wiring.

Two objects are rubbed against each other, the nature of electric force, when they are placed at some distance is

From what the name electricity is coined ? Explain its meaning.

Mention types of methods of charging the body.

A charged metallic sphere $A$ is suspended by a riylon thread. Another charged metallic sphere $B$ held by an insulating handle is brought close to $A$ such that the distance between their centres is $10 \,cm ,$ as shown in Figure $(a) .$ The resulting repulsion of $A$ is noted (for example, by shining a beam of light and measuring the deflection of its shadow on a screen). Spheres $A$ and $B$ are touched by uncharged spheres $C$ and $D$ respectively. as shown in Figure $(b)$ $C$ and $D$ are then removed and $B$ is brought closer to $A$ to a distance of $5.0 \,cm$ between their centres, as shown in Figure $(c)$ What is the expected repulsion of A on the basis of Coulomb's law? Spheres $A$ and $C$ and spheres $B$ and $D$ have identical sizes. Ignore the sizes of $A$ and $B$ in comparison to the separation between their centres.