Calculation of electric field and electrostatic potential for a charge distribution

Definition

The electric field due to a charge distribution is the vector quantity obtained by adding the field contributions from all infinitesimal charge elements in the distribution.

For a continuous distribution, $$\mathbf{E} = \frac{1}{4\pi\varepsilon_0}\int \frac{dq\,\hat{\mathbf{r}}}{r^2}$$ where:

• $dq$ is a small element of charge,
• $r$ is the distance from $dq$ to the field point,
• $\hat{\mathbf{r}}$ is the unit vector from the charge element to the point.

The electrostatic potential due to a charge distribution is the scalar quantity obtained by adding the potential contributions from all infinitesimal charge elements: $$V = \frac{1}{4\pi\varepsilon_0}\int \frac{dq}{r}$$

For discrete charges: $$\mathbf{E} = \sum_i \frac{1}{4\pi\varepsilon_0}\frac{q_i \hat{\mathbf{r}}_i}{r_i^2}, \qquad V = \sum_i \frac{1}{4\pi\varepsilon_0}\frac{q_i}{r_i}$$

Main Content

1. Electric field from a charge distribution

• The electric field is a vector quantity, so both magnitude and direction must be considered while adding contributions from different parts of the distribution.
• For a continuous charge distribution, the total field is found by dividing the charged object into very small elements $dq$, calculating the field due to each element using Coulomb’s law, and integrating over the entire distribution.

A continuous charge distribution may be:

Linear charge distribution

• with charge density $\lambda$ in coulombs per metre: $$dq = \lambda\,dl$$

Surface charge distribution

• with charge density $\sigma$ in coulombs per square metre: $$dq = \sigma\,dA$$

Volume charge distribution

• with charge density $\rho$ in coulombs per cubic metre: $$dq = \rho\,dV$$

The electric field due to an element $dq$ at a distance $r$ is: $$d\mathbf{E} = \frac{1}{4\pi\varepsilon_0}\frac{dq}{r^2}\hat{\mathbf{r}}$$ and the net field is obtained by integration: $$\mathbf{E} = \int d\mathbf{E}$$

Example idea: For a uniformly charged ring, symmetry helps show that horizontal components of field from opposite elements cancel, leaving only the field along the axis. This makes the problem much simpler than direct vector addition of every element individually.

2. Electrostatic potential from a charge distribution

• Electrostatic potential is a scalar quantity, so contributions from all charge elements are added algebraically without considering direction.
• This makes potential calculation often easier than direct field calculation, especially for symmetric charge distributions.

For a continuous charge distribution, the potential at a point is: $$V = \frac{1}{4\pi\varepsilon_0}\int \frac{dq}{r}$$ where $r$ is the distance from the charge element to the observation point.

Because potential is scalar:

• it can be calculated by direct integration more easily,
• then the electric field can be found using $$\mathbf{E} = -\nabla V$$ or in one dimension, $$E = -\frac{dV}{dr}$$

The potential is usually defined with respect to infinity: $$V(\infty)=0$$ Hence, the potential at a point is the work done per unit positive charge in bringing the test charge from infinity to that point slowly, without changing its kinetic energy.

Example idea: For a uniformly charged spherical shell, the potential outside the shell is the same as that of a point charge at the center, while inside the shell it remains constant. This is easier to derive using potential than by direct field calculation.

3. Relationship between electric field and potential

• The electric field and electrostatic potential are closely related, and one can often be derived from the other.
• The electric field points in the direction of the greatest decrease of potential.

Mathematically, $$\mathbf{E} = -\nabla V$$ This means:

• if the potential changes rapidly with position, the field is strong,
• if the potential is constant in a region, the electric field there is zero.

For one-dimensional radial symmetry: $$E_r = -\frac{dV}{dr}$$

This relationship is extremely useful:

• If the potential is easy to find, the field can be obtained by differentiation.
• If the field is known, the potential difference can be obtained by integration: $$V_B - V_A = -\int_A^B \mathbf{E}\cdot d\mathbf{l}$$

In electrostatics, the field is conservative, so the integral depends only on the end points and not on the path taken.

Example idea: If the potential between two points increases along a certain direction, the electric field must point in the opposite direction.

Working / Process

1. Identify the type of charge distribution

• Determine whether the charge is discrete or continuous.
• For continuous distributions, identify whether it is linear, surface, or volume charge.
• Write the correct charge element: $$dq=\lambda dl,\quad dq=\sigma dA,\quad dq=\rho dV$$

2. Choose the correct observation point and use symmetry

• Locate the point where the electric field or potential is required.
• Use symmetry to simplify the problem:
  • opposite components may cancel,
  • only one directional component may survive,
  • distances may become constant over the distribution.
• Symmetry is often the key to making the integration manageable.

3. Set up and evaluate the integral

• For field: $$d\mathbf{E}=\frac{1}{4\pi\varepsilon_0}\frac{dq}{r^2}\hat{\mathbf{r}}$$ and integrate vector components carefully.

• For potential: $$dV=\frac{1}{4\pi\varepsilon_0}\frac{dq}{r}$$ and integrate over the full distribution.

• If the electric field is required after finding potential, use: $$\mathbf{E}=-\nabla V$$

• Check the result for consistency in units, direction, and limiting cases.

Advantages / Applications

• It helps in calculating the electric field and potential for rods, rings, disks, spheres, and other charged bodies commonly used in electrostatics problems.
• It is essential in determining work done in moving charges, potential energy of systems, and the behavior of charges in external electric fields.
• It is widely used in the analysis and design of capacitors, electrostatic shielding, charged conductors, and many practical devices in physics and engineering.

Summary

• Electric field and electrostatic potential due to a charge distribution are found by adding contributions from all small charge elements.
• Field is a vector and usually requires careful handling of direction; potential is a scalar and is often easier to calculate.
• Symmetry greatly simplifies the calculation, and the relationship $\mathbf{E}=-\nabla V$ connects the two quantities.