A thin rod of length \(L\) possesses a charge \(\lambda\) per unit length distributed uniformly alongit. Find the force on a charge \(Q\) lying in the same line as the rod a distance \(h\) from the nearest end. Comment on the behaviour of the force as \(L \rightarrow 0\) and as \(L \rightarrow \infty\), in each case keeping \(h\) constant.

Coulomb's law is a fundamental principle in electrostatics that describes the force between two point charges. The law states that the electrostatic force (\f\(F\f\)) between two point charges is directly proportional to the product of the charges (\f\(q_1\f\) and \f\(q_2\f\)), and inversely proportional to the square of the distance (\f\(r\f\)) between them. It's given by the formula:
\f[F = k_e \frac{|q_1 \times q_2|}{r^2}\f]
where \f\(k_e\f\) is Coulomb's constant, which is approximately equal to \f\(8.988 \times 10^9 N \times m^2/C^2\f\). This constant is derived from the permittivity of free space, \f\(\fepsilon_0\f\), as \f\(k_e = 1/(4\xf \fepsilon_0)\f\).

Coulomb's Law is crucial when calculating the electric field of a uniformly charged rod. By taking an infinitesimally small charge element \f\(dq\f\), we can apply Coulomb's law to this element and then integrate along the length of the rod to obtain the total electric field or force affecting another charge in proximity to the rod.

The electric field intensity is a vector quantity that represents the electric force per unit charge experienced by a small positive test charge placed in the field. Suppose there is a charge distribution in space; the electric field at a point is defined as:
\f[E = \frac{F}{Q}\f]
where \f\(E\f\) is the electric field intensity, \f\(F\f\) is the force experienced by the test charge \f\(Q\f\).

In the context of our uniformly charged rod, the electric field intensity due to a small charge element \f\(dq\f\) at a distance \f\(r\f\) from a point charge \f\(Q\f\) can be calculated using the differential form of Coulomb's Law. To find the total electric field intensity at the location of \f\(Q\f\), you take the sum of the electric fields by all infinitesimal charges that make up the rod. This continuous sum is done through the process of integration, which is a fundamental concept in calculating electric fields in electrostatics.

Integration is a mathematical tool used to sum up small quantities over a continuous range to find the total amount. In electrostatics, particularly when dealing with distributed charges like that of a rod or a sphere, integration becomes essential. Instead of looking at the whole charge, we consider infinitesimally small elements of charge \f\(dq\f\) that contribute to the electric field at a point.

The process of integration enables us to calculate the influence of a continuously distributed charge by integrating the effect of each small charge element across the whole volume, surface, or length of the distribution. In our example of a charged rod, to find the total electric field or force at a point, we integrate the electric field contribution of each elemental charge \f\(dq\f\) over the length of the rod. This approach provides an exact solution to a problem that would otherwise be impossible to solve using standard algebraic methods.