Let us investigate the phenomenon of electricity. The theoretical basis for this phenomenon is Coulomb’s Law which was originally discovered by Charles Coulomb. This law states that the electric force exerted by any charged particle \(q_1\) on any another charged particle \(q_2\) is
$$\vec{F}_{1,2}=k_e\frac{q_1q_2}{r_{1,2}^2}\hat{r}_{12}.\tag{1}$$
where \(\hat{r}_{12}\) is a unit vector drawn from \(q_1\) towards \(q_2\), \(r_{1,2}\) is the separation distance between the two charged particles, and \(k_e\) is Coulomb’s constant which has been measured to be
$$k_e=8.99×10^9\frac{N∙m^2}{C^2}.\tag{2}$$
This equation tells us that the electric force manifests whenever there are two particles containing charge separated by a distance. The electric field \(E\) produced by any point charge \(q_1\) is defined as
$$\vec{E}≡\frac{\vec{F}_{12}}{q_2}.\tag{3}$$
If we substitute Equation (1) into Equation (3), then Equation (3) becomes
$$\vec{E}_1=\frac{k_e\frac{q_1q_2}{r_{1,2}^2}\hat{r}_{12}}{q_2}=k_e\frac{q_1}{r_{1,2}^2}\hat{r}_{1,2}.\tag{4}$$
We see from Equation (3) that the electric field \(\vec{E}\) produced by any point charge depends only on its own charge \(q_1\) and the distance \(r_{1,2}\) away from itself. In Equation (3), since the dependence on \(q_2\) cancels out, the electric field \(\vec{E}_1\) (produced by \(q_1\)) does not depend on the other charge \(q_2\). Even if the charge \(q_2\) disappeared, the electric field \(\vec{E}_1\) produced by \(q_1\) would still be present.
We can also use Coulomb’s law to calculate the electric force exerted on any point charge \(q_t\) by a collection of three point charges. Using Coulomb’s law we already know how to calculate the electric force exerted on one point charge by another using Equation (1). In other words we already know that the force exerted by \(q_1\) on \(q_t\) is \(\vec{F}_{1,t}=k_e\frac{q_1q_t}{r^2}\hat{r}_{1,t}\), \(q_2\) on \(q_t\) is \(\vec{F}_{2,t}=k_e\frac{q_2 q_t}{r^2}\hat{r}_{2,t}\), and \(q_3\) on \(q_t\) is \(\vec{F}_{3,t}=k_e\frac{q_3q_t}{r^2}\hat{r}_{3,t}\). According to Newtonian mechanics, the total force acting on \(q_t\) is just the sum of all the individual forces exerted on \(q_t\):
$$\vec{F}_{total}=k_e\sum_{i=1}^3\frac{q_iq_t}{r^2}\hat{r}_{i,t}.$$
The result can be generalized to an \(n\) number of charged particles acting on \(q_t\). The total force exerted by \(n\) point charges on \(q_t\) is given by
$$\vec{F}_{total}=k_e\sum_{i=1}^n\frac{q_iq_t}{r^2}\hat{r}_{i,t}.\tag{5}$$
Lastly we’ll be interested in the electric force produced by an infinite number of charged particles. An infinite number of charged particles is called a continuous charge distribution. The net charge on a charged rod or plate is due to an enormous number of charged electrons and protons. Although we could, in principle, account for all of these charged particles, there are so many of them that it would make calculating the force and electric field that they produce impractical. Instead, we can just model these kinds of charge distributions as being continuous and made of infinitely many charged particles. To calculate the force exerted by an infinite number of charged particles, we need to take the infinite sum of the infinite number of individual forces exerted by each charged particle:
$$\vec{F}=\lim_{n→∞}k_e\sum_{i=1)}^n\vec{F}_{i,t}=\lim_{n→∞}k_e\sum_(i=1)^n\frac{q_iq_t}{r^2}\hat{r}_{i,t}.$$
As the number of charges become infinite, the sum \(\sum\) becomes an infinite sum \(\int\) and the charges become infinitesimally small:
$$\vec{F}=k_e\int\frac{q}{r^2}dq.\tag{6}$$
Like force vectors, electric field vectors also add. We can calculate the electric field produced by three charged particles at any point in space by taking the sum of each electric field vector:
$$\vec{E}_T=\vec{E}_1+\vec{E}_2+\vec{E}_3=k_e\frac{q_1}{r_{1,t}^2}\hat{r}_{1,t}+k_e\frac{q_2}{r_{2,t}^2}\hat{r}_{2,t}+k_e\frac{q_3}{r_{3,t}^2}\hat{r}_{3,t}=k_e\sum_{i=1}^3\frac{q_i q_t}{r_{i,t}^2}\hat{r}_{i,t}.$$
This result can be generalized to the electric field produced by n charged particles:
$$\vec{E}=k_e\sum_{i=1}^n\frac{q_iq_t}{r_{i,t}^2}\hat{r}_{i,t}.\tag{7}$$
If we take the limit \(\lim_{n→∞}k_e\sum_{i=1}^n\frac{q_iq_t}{r_{i,t}^2}\hat{r}_{i,t}\) , the summation \(\sum\) becomes infinite since we would be taking the sum of an infinite number of terms. Thus we must replace the finite sum \(\sum\) with the infinite sum \(\int\). Since the index “\(i\)” can take on an infinite number of possible values, the variables \(r_{i,t}\), \(q_i\) and \(\hat{r}_{i,t}\) become continuous; we can replace there variables with the labels \(r\), \(q\) and \(\hat{r}\). Lastly, since the number of point charges go to infinity, the charges will become infinitesimally small. Recall from your days in calculus that we can take the infinite sum of infinitesimally small quantities (i.e. \(∫dx\)) to build a finite quantity (i.e. \(∫dx\) could just be a finite length along the x-axis). The electric field produced by an infinite number of charged particles is given by
$$\vec{E}=k_e\int\frac{1}{r^2}\hat{r}dq.\tag{8}$$
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