When we think about electric fields, we often imagine them being produced by point charges. However, in many practical situations, charges are not concentrated at a single point but are distributed over a region of space. Understanding how these continuous charge distributions produce electric fields is crucial in fields like electromagnetism, material science, and even biology. In this post, we’ll explore how to calculate the electric field produced by a continuous distribution of charge
The Concept of Electric Field
The electric field at a point in space is defined as the force per unit charge experienced by a small positive test charge placed at that point:
[math]\mathbf{E} = \frac{\mathbf{F}}{q}[/math]
This vector field describes how a charge would interact with its surroundings. The direction of the field is the direction of the force that a positive test charge would experience.
Continuous Charge Distributions
In real-world scenarios, charges are often spread out over a line, surface, or volume. Depending on how the charge is distributed, we describe the distribution using one of the following:
- Linear Charge Density (λ\lambda): Charge per unit length (C/m)
- Surface Charge Density (σ\sigma): Charge per unit area (C/m²)
- Volume Charge Density (ρ\rho): Charge per unit volume (C/m³)
For each of these distributions, the electric field at a point in space can be calculated by summing the contributions of all the infinitesimal charge elements that make up the distribution.
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Calculating the Electric Field
The general approach to finding the electric field [math]\mathbf{E}[/math] due to a continuous charge distribution involves integrating the contributions of all infinitesimal charge elements dqdq in the distribution.
Electric field due to line charge distribution
Consider a line of charge with linear charge density [math]\mathbf{. To calculate the electric field at a point due to this distribution, consider a small element of the line dx carrying a charge dq.
The electric field [math]\mathbf{E}[/math] due to this element at a distance r from P is given by Coulomb’s law:
[math]d\mathbf{E} = \frac{1}{4\pi\epsilon_0} \frac{dq}{r^2} \hat{\mathbf{r}}[/math]
Here, [math]\hat{\mathbf{r}}[/math] is the unit vector pointing from the charge element to the point P. The total electric field is then found by integrating this expression along the entire line of charge.
Electric Field Due to a Surface Charge Distribution
For a surface charge distribution with surface charge density [math]\sigma[/math], the approach is similar. An infinitesimal element of area dA carries a charge dq=σdAdq
The electric field due to this element is:
[math]d\mathbf{E} = \frac{1}{4\pi\epsilon_0} \frac{\sigma dA}{r^2} \hat{\mathbf{r}}[/math]
To find the total electric field, integrate this expression over the entire surface.
Electric Field Due to a Volume Charge Distribution
For a volume charge distribution with volume charge density ρ\rho, the infinitesimal charge element is dq=ρdVdq , where dV is an infinitesimal volume element.
The electric field due to this element is:
[math]d\mathbf{E} = \frac{1}{4\pi\epsilon_0} \frac{\rho dV}{r^2} \hat{\mathbf{r}}[/math]
The total electric field is obtained by integrating over the entire volume:
[math]\mathbf{E} = \frac{1}{4\pi\epsilon_0} \int_V \frac{\rho(\mathbf{r’}) dV’}{|\mathbf{r} – \mathbf{r’}|^2} \hat{\mathbf{r}}[/math]
Applying the Concept: Examples
Let’s see how these principles apply to some classic examples
Electric Field of a Charged Rod
Consider a uniformly charged rod of length with total charge Q. The linear charge density is [math]\lambda = \frac{Q}{L}. To find the electric field at a point on the axis of the rod, we integrate the contributions of each infinitesimal charge element along the length of the rod.
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Electric Field of a Charged Rod
Consider a uniformly charged rod of length LL with total charge Q. The linear charge density is [math]\lambda = \frac{Q}{L}[/math]. To find the electric field at a point on the axis of the rod, we integrate the contributions of each infinitesimal charge element along the length of the rod.
for detail pf this toplic click here