Electric Field Formula

Electric Field Formula

Electric fields are a fundamental concept in electromagnetism, a branch of physics. It denotes the area around a charged object where the object exerts an electric force on nearby charged objects. The electric field is a vector, which means it has magnitude and direction. Michael Faraday introduced the concept of an electric field. It is useful for visualizing and calculating the force that a charge might experience in the presence of other charges without direct contact. This field propagates away from a positive charge and inward toward a negative charge. Learn more about the electric field, its definition, formula, and examples.

What is Electric Field?

An electric field is a region around a charged particle or object within which a force would be exerted on other charged particles or objects. It is a vector field, meaning it has both magnitude and direction, and it represents how a charge influences the space around it.

Electric Field Definition

The electric field \(\mathbf{E}\) at a point in space is defined as the force \(\mathbf{F}\) that a positive test charge \(q\) would experience at that point, divided by the magnitude of the test charge:

\[ \mathbf{E} = \frac{\mathbf{F}}{q} \]

Unit of Eelctric Field

The electric field is measured in newtons per coulomb (N/C) or volts per meter (V/m).

What is Electric Field Formula?

An electric field surrounds an electric charge while simultaneously exerting force on other charges in the field. It either attracts or repels them.  The Electric Field Formula is formally defined as a vector field that correlates the (electrostatic or Coulomb) force or unit of charge exerted on an infinitesimal positive test charge at rest at each location in space. The Electric Field formula is E = F/q

For a point charge \(Q\), the electric field \(\mathbf{E}\) at a distance \(r\) from the charge is given by:

\[ \mathbf{E} = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2} \hat{\mathbf{r}} \]

where:

• \(\epsilon_0\) is the permittivity of free space (\(\approx 8.854 \times 10^{-12} \,\text{F/m}\)),
• \(Q\) is the charge creating the electric field,
• \(r\) is the distance from the charge,
• \(\hat{\mathbf{r}}\) is a unit vector pointing from the charge to the point of interest.

The electric charge or time-varying magnetic fields create the electric field. At the atomic level, the electric field is responsible for the attractive forces that hold the atomic nucleus and electrons together.

Formula and Derivation of Electric Field

The Electric Field Formula “field” is the area around an electric charge that may be observed to be influenced by it. When the additional charge is introduced into the field, the presence of an electric field may be felt. Depending on the composition of the incoming charge, the electric field will either attract or repel it.Any electric charge may be thought of as having an electric field. The charge and electrical force operating in the field define the electric field’s strength or intensity.

The Electric Field Formula for the electric field is

The Electric Field Formula E = F/q

Where

• E represents the electric field
• F is a force acting on the charge
• The charge q is surrounded by its electric field.

When two charges, Q and q, are separated by a distance r, the electrical force may be defined as

F= k Qq/r²

• F is the electrical force.
• Q and q are two charges
• R is the distance between the two charges.
• k is Coulomb’s constant

As a result, the electric field E may be defined as

E = F/q

This is the electric field felt by charge Q as a result of charge q.

This electric field intensity formula is defined by Coulomb’s law.

If there is a voltage V across a distance r, the electric field is defined as

E= V/r

The SI unit for an electric field is N/c, which stands for Newton/Coulomb.

If there are many electric fields in a location, the electric fields add up vectorially, taking the field’s direction into account.

Properties of Electric Field

Electric field lines have varying qualities. Some of the properties are listed below.

• Field lines never cross each other.
• They’re perpendicular to the surface charge.
• The intensity of an electric field grows as the field lines become closer together, indicating a greater force. In contrast, when field lines move apart, the field diminishes.
• The number of field lines is exactly proportional to the amount of the charge.
• These lines often begin with positive charges and conclude with negative charges, indicating the direction of the electric field.

Solved Examples on Electric Field Formula

Example 1: A force of 10 N is acting on the charge of 20 μC at any point. Find the electric field intensity at that point.

Solution:

Given

Force F = 10 N

Charge q = 20 μC

Electric field formula is given by

E = F / q

= 10 N / 20×10⁻⁶C

E = 0.5 × 10⁵ N/C

Example 2: Calculate the electric field at a distance of 2 meters from a point charge of \(5 \mu C\).

Solution:
Given:
Charge \(Q = 5 \mu C = 5 \times 10^{-6} \, C\)
Distance \(r = 2 \, m\)
Permittivity of free space \(\epsilon_0 \approx 8.854 \times 10^{-12} \, F/m\)

The electric field \( \mathbf{E} \) due to a point charge is given by:

\[ \mathbf{E} = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2} \]

Plugging in the values:

\[ \mathbf{E} = \frac{1}{4 \pi \times 8.854 \times 10^{-12}} \frac{5 \times 10^{-6}}{2^2} \]

\[ \mathbf{E} = \frac{1}{4 \pi \times 8.854 \times 10^{-12}} \frac{5 \times 10^{-6}}{4} \]

\[ \mathbf{E} = \frac{1}{4 \pi \times 8.854 \times 10^{-12}} \times 1.25 \times 10^{-6} \]

\[ \mathbf{E} = \frac{1.25 \times 10^{-6}}{4 \pi \times 8.854 \times 10^{-12}} \]

\[ \mathbf{E} \approx \frac{1.25 \times 10^{-6}}{1.11265 \times 10^{-10}} \]

\[ \mathbf{E} \approx 1.12 \times 10^4 \, \text{N/C} \]

So, the electric field at 2 meters from the charge is approximately \(1.12 \times 10^4 \, \text{N/C}\).

Example 3: Two point charges, \(Q_1 = 3 \mu C\) and \(Q_2 = -4 \mu C\), are placed 0.5 meters apart. Find the electric field at the midpoint between the charges.

Solution:
Given:
Charge \(Q_1 = 3 \mu C = 3 \times 10^{-6} \, C\)
Charge \(Q_2 = -4 \mu C = -4 \times 10^{-6} \, C\)
Distance \(d = 0.5 \, m\)
Midpoint distance from each charge \(r = 0.25 \, m\)
Permittivity of free space \(\epsilon_0 \approx 8.854 \times 10^{-12} \, F/m\)

The electric field due to \(Q_1\) at the midpoint is:

\[ \mathbf{E}_1 = \frac{1}{4 \pi \epsilon_0} \frac{Q_1}{r^2} \]

\[ \mathbf{E}_1 = \frac{1}{4 \pi \times 8.854 \times 10^{-12}} \frac{3 \times 10^{-6}}{(0.25)^2} \]

\[ \mathbf{E}_1 = \frac{3 \times 10^{-6}}{4 \pi \times 8.854 \times 10^{-12} \times 0.0625} \]

\[ \mathbf{E}_1 = \frac{3 \times 10^{-6}}{1.11265 \times 10^{-12} \times 0.0625} \]

\[ \mathbf{E}_1 \approx \frac{3 \times 10^{-6}}{6.953125 \times 10^{-14}} \]

\[ \mathbf{E}_1 \approx 4.32 \times 10^7 \, \text{N/C} \]

The electric field due to \(Q_2\) at the midpoint is:

\[ \mathbf{E}_2 = \frac{1}{4 \pi \epsilon_0} \frac{Q_2}{r^2} \]

\[ \mathbf{E}_2 = \frac{1}{4 \pi \times 8.854 \times 10^{-12}} \frac{-4 \times 10^{-6}}{(0.25)^2} \]

\[ \mathbf{E}_2 = \frac{-4 \times 10^{-6}}{4 \pi \times 8.854 \times 10^{-12} \times 0.0625} \]

\[ \mathbf{E}_2 = \frac{-4 \times 10^{-6}}{1.11265 \times 10^{-12} \times 0.0625} \]

\[ \mathbf{E}_2 \approx \frac{-4 \times 10^{-6}}{6.953125 \times 10^{-14}} \]

\[ \mathbf{E}_2 \approx -5.76 \times 10^7 \, \text{N/C} \]

The total electric field at the midpoint is the vector sum of \(\mathbf{E}_1\) and \(\mathbf{E}_2\):

\[ \mathbf{E} = \mathbf{E}_1 + \mathbf{E}_2 \]

\[ \mathbf{E} = 4.32 \times 10^7 + (-5.76 \times 10^7) \]

\[ \mathbf{E} = -1.44 \times 10^7 \, \text{N/C} \]

So, the electric field at the midpoint is approximately \(-1.44 \times 10^7 \, \text{N/C}\) directed towards \(Q_2\).