(N/A) The figure shows a set of electric field lines originating from a point charge $q$.
Consider two small area elements placed at points $R$ and $S$,oriented normal to the field lines.
The number of field lines passing through a given area is proportional to the magnitude of the electric field at that location. The diagram illustrates that the field at $R$ is stronger than at $S$ because the field lines are more densely packed at $R$.
In three dimensions,the solid angle $\Delta \Omega$ subtended by an area element $\Delta S$ at a distance $r$ from the charge is given by $\Delta \Omega = \frac{\Delta S}{r^2}$,which implies $\Delta S = r^2 \Delta \Omega$.
For a fixed solid angle $\Delta \Omega$,the number of radial field lines $n$ passing through the area element is constant.
At two points $P_1$ and $P_2$ at distances $r_1$ and $r_2$ from the charge,the area elements subtending the same solid angle $\Delta \Omega$ are $A_1 = r_1^2 \Delta \Omega$ and $A_2 = r_2^2 \Delta \Omega$,respectively.
The number of field lines $n$ cutting these area elements is the same. Therefore,the number of field lines per unit area (which represents the field strength $E$) is:
$E_1 = \frac{n}{A_1} = \frac{n}{r_1^2 \Delta \Omega}$
$E_2 = \frac{n}{A_2} = \frac{n}{r_2^2 \Delta \Omega}$
Since $n$ and $\Delta \Omega$ are constant,it follows that the strength of the electric field is inversely proportional to the square of the distance,i.e.,$E \propto \frac{1}{r^2}$.