(N/A) The characteristics of an electric field are as follows:
$(i)$ The charge $Q$,which produces the electric field,is called a source charge,and the charge $q$,which tests the effect of the source charge,is called a test charge.
However,if a charge $q$ is brought to any point around $Q$,it is bound to experience an electrical force due to $Q$ and will tend to move. $A$ way out of this difficulty is to make $q$ negligibly small. The force $\vec{F}$ is then negligibly small,but the ratio $\frac{F}{q}$ is finite and defines the electric field: $\overrightarrow{E} = \lim_{q \rightarrow 0} \frac{\overrightarrow{F}}{q}$.
$(ii)$ Note that the electric field $\overrightarrow{E}$ due to $Q$,though defined operationally in terms of some test charge $q$,is independent of $q$. This is because $\vec{F}$ is proportional to $q$,so the ratio $F/q$ does not depend on $q$. The field exists at every point in three-dimensional space.
$(iii)$ For a positive charge,the electric field is directed radially outwards from the charge,as shown in figure $(a)$. For a negative charge,the electric field vector at each point points radially inwards,as shown in figure $(b)$.
$(iv)$ Since the magnitude of the force $F$ on charge $q$ due to charge $Q$ depends only on the distance $r$ of the charge $q$ from charge $Q$,the magnitude of the electric field $\vec{E}$ will also depend only on the distance $r$. Therefore,$E \propto \frac{1}{r^2}$.