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When we calculate electric field intensity for a point charge at any point inside electric field the field intensity is $E = F/q$ where $F$ is the force acting on charge $q$. In this case, the charge $q$ should be very small. The practical value of $q$ cannot be so small as needed. In defining electric field by measuring value $F/q$ is smaller than the actual value.

My question: How can the accurate value of electric field intensity be calculated?
I'm quite confused. Can someone point me out?

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up vote 3 down vote accepted

The electric field intensity is usually defined in introductory textbooks as the limit $$\lim_{q\rightarrow0}\frac{\mathbf F}{q},$$ i.e. the force on a test charge, per unit charge, when that test charge goes to zero. However, you are right in noting that the limit $q\rightarrow 0 $ is impossible to take because charge is quantized, and you can never have charges below one electron charge $q=e$.

Because of that, it is easier to define the electric field produced by a point charge $q_1$ at a relative separation $\mathbf r$ as the vector $$\mathbf E:=\frac{1}{4\pi\epsilon_0} \frac{q_1 \hat{\mathbf r}}{r^2},$$ and then stipulate as a physical law that the force such a field (or any electric field) exerts on a second charge $q_2$ is given by $\mathbf F=q_2\mathbf{E}$, with due stipulation that self-interactions are not allowed. This procedure is both mathematically consistent and concurs with physical, experimental observations.

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