Superposition of Electric Fields

Every charged particle in the universe creates an electric field in the space surrounding it. This field can be calculated with the help of Coulomb's law. The principle of superposition allows for the combination of two or more electric fields.

The principle of superposition states that every charge in space creates an electric field at point independent of the presence of other charges in that medium. The resultant electric field is a vector sum of the electric field due to individual charges.

In 1-dimension, electric fields can be added according to the relationship between the directions of the electric field vectors.

• Same direction: Add the magnitudes together to find the net field.

• Opposite directions: Subtract the smaller magnitude from the larger magnitude to find the net field. The net field will point in the direction of the greater field.

Two charges $q_{1}$ and $q_{2}$ are kept at the endpoints of a rod $AB$ of length $L = 2\text{ m}$ in vacuum. What is the magnitude of electric field at the center of the rod due to these 2 charges?

• $q_{1} = +10^{-4} C$
• $q_{2} = +10^{-5} C$
• $\dfrac{1}{4\pi\epsilon_{0}} = 9 \times 10^{9} \dfrac{\text{Nm}^{2}}{\text{C}^{2}}$

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According to the principle of superposition, each charge creates its own electric field independent of the other charge. Let the electric fields by $q_{1}$ and $q_{2}$ be $E_{1}$ and $E_{2},$ respectively. Using Coulomb's law, we have

$$\begin{aligned} |E_{1}| &= \dfrac{q_{1}}{4\pi\epsilon_{0}(1)^{2}} = 9 \times 10^{9} \times 10^{-4} = 9 \times 10^{5} \left(\dfrac{\text{N}}{\text{C}}\right)\\ |E_{2}| &= \dfrac{q_{2}}{4\pi\epsilon_{0}(1)^{2}} = 9 \times 10^{9} \times 10^{-5} = 9 \times 10^{4} \left(\dfrac{\text{N}}{\text{C}}\right). \end{aligned}$$

Now, since $E_{1}$ and $E_{2}$ are oppositely directed and the angle between them is $\theta = \pi$ radians, we have

$$\begin{aligned} |E_{net}| &= \sqrt{E_{1}^{2} + E_{2}^{2} + 2E_{1}E_{2}\cos\theta} \\ &= \sqrt{E_{1}^{2} + E_{2}^{2} - 2E_{1}E_{2}} \\ & = \Big| |E_{1}| - |E_{2}| \Big|\\ \Rightarrow E_{net} &= 9 \times 10^{5} - 9 \times 10^{4} \\ &= 86 \times 10^{4} \left(\frac{\text{N}}{\text{C}}\right).\ _\square \end{aligned}$$

In 2-dimensions, considerations need to be made for the relative directions of the electric fields.

The two electric fields at a point in space are $\vec{E}_1 = 3\hat{x} \frac{\text{N}}{\text{C}}$ and $\vec{E}_2 = 4\hat{y} \frac{\text{N}}{\text{C}}.$ What is the net electric field?

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The net electric field is $$\vec{E}_{\text{net}} = \vec{E}_1 + \vec{E}_2 = (3\hat{x} + 4\hat{y})\frac{\text{N}}{\text{C}}.$$

The magnitude of this electric field is $$E = \sqrt{3^2 + 4^2} = 5 \left(\frac{\text{N}}{\text{C}}\right).$$

The direction is $$\theta = \tan^{-1}\frac{y}{x} = \tan^{-1}\frac{4}{3} = 53.13^\circ.\ _\square$$