# 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 superpositionstates 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.

#### Contents

## 1-Dimension

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?

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

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

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

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

A charge \(+16\) C is fixed at each of the points \(x=3, 9, 15, \ldots, \infty \) on the \(x\)-axis, and a charge \(-16\) C is fixed at each of the points \(x=6, 12, 18, \ldots, \infty \) on the \(x\)-axis. Then find the potential at the origin due to these charges.

If the potential is of the form \(\frac { A\ln B }{ C\pi { \varepsilon }_{ 0 } } \), find \(A+B+C\), where \(A\) and \(C\) are co-prime and \(B\) is prime.

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## 2-Dimensions

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?

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 \[\theta = \tan^{-1}\frac{y}{x} = \tan^{-1}\frac{4}{3} = 53.13^\circ.\ _\square\]

Three equal charges are placed in such a way that they form a right triangle with sides \(3\text{ cm}, 4\text{ cm}, 5\text{ cm}.\) Each charge has a magnitude of \(2\times 10^{-6}\text{C}\).

Find the net force acting on the charge at the right angle.

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**Details and assumptions:**

Take \(\dfrac 1{4\pi\epsilon_0} = 9\times 10^9\text{ Nm}^2/\text{C}^2\).

Round your answer to the nearest integer.

**Cite as:**Superposition of Electric Fields.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/superposition-electric-fields/