Question

How do you calculate the electric field between two charges?

Answer 1

Step 1: Calculate the electric field due to a point charge

1. The electric field intensity at any point is the strength of the electric field at the given point.
2. It is defined as the force exerted by a unit positive charge located at a given point.
3. The concept of the field was firstly given by Faraday.
4. If the force acting on this unit positive charge $+q_0$ at a point $r$, then electric field intensity is given by: $\overrightarrow{E}=\frac{\overrightarrow{F}}{q}$,
5. By Coulomb’s law, the force of attraction or repulsion between the charges $q_0$ and $Q$ at $P$ is $F=\frac{1}{4\pi {\epsilon }_0}\frac{Q{q}_0}{r^2}$, where $r$ is the distance between the charges.

Step 2: Calculate the general electric field intensity due to a system of charges

1. Each charge has its own electric field and two charge fields interact with each other. The net electric field between them will be the resultant vector of the two fields.
2. Therefore, the electric field intensity at any point due to a system or group of charges is equal to the vector sum of electric field intensities due to individual charges at the same point.

$\overrightarrow{E}=\overrightarrow{E_1}+{\overrightarrow{E}}_2+{\overrightarrow{E}}_3+.........\left(i\right)$

Step 3: Calculate the electric field between two charges $q_1$ and $q_2$ respectively

1. When the number of charges is two, then the equation (i), becomes $\overrightarrow{E}=\overrightarrow{E_1}+{\overrightarrow{E}}_2$
2. Therefore, the electric field between two charges is given by

$\begin{array}{rcl}\overrightarrow{E}& =& \overrightarrow{E_1}+{\overrightarrow{E}}_2\\ & =& \frac{1}{4\pi {\epsilon }_0}\left(\frac{{\displaystyle q_1}}{{\displaystyle {r_1}^2}}+\frac{{\displaystyle q_2}}{{\displaystyle {r_2}^2}}\right)\end{array}$.

The electric field intensity due to a system of charges is, E= $\begin{array}{l}\frac{1}{4\pi {\epsilon }_0}\left(\frac{{\displaystyle q_1}}{{\displaystyle {r_1}^2}}+\frac{{\displaystyle q_2}}{{\displaystyle {r_2}^2}}\right)\end{array}$

Hence, the answer is, $\begin{array}{l}\frac{1}{4\pi {\epsilon }_0}\left(\frac{{\displaystyle q_1}}{{\displaystyle {r_1}^2}}+\frac{{\displaystyle q_2}}{{\displaystyle {r_2}^2}}\right)\end{array}$.