Electric Field for Distributed Charge systems (Line, Surface and Volume)

# Standard procedure for finding the Electric Field due to distributed charge.

Electric field arises because of the charge. We are well aware of its standard formulas. This article explains the standard procedure for deriving the expression for Electric Field Intensity due to distributed charge. The standard steps discussed over here could be applied to get the E due to distributed charges i.e. line charge, surface charge and volume charge.

## What is Electric Field Intensity?

According to Coulomb’s Law, every charge is surrounded by its Electric Field. Any other charge will experience a force if it enters into the Electric field. The strength of the force experienced by the charge may be different at different points of the field. In other words, the strength of the field varies from point to point. And this strength of the field at any point is termed as the Electric Field Intensity of that field.

Electric Field Intensity | Definition and Physical Significance.

#### Electric Field Intensity because of Point Charge.

Theoretically, we assume that all the charge is concentrated at a point known as a point charge. The electric field due to this point charge say Q, at a distance r is given as,

$E=\frac1{4\pi\varepsilon}\frac Q{r^2}$

This is the standard formula for E assuming that all the given charge concentrated at a point. Now the point is like a dot in Mathematics having no physical dimensions.

## What is distributed Charge?

As stated above, theoretically any charge is shown at a point assuming that all the charge is concentrated at that point. But, practically the charge is deposited on a physical body. Hence rather than being concentrated at a point, the charge is distributed all over the hosting body depending on many factors like dimensions, nature etc of the hosting body. So practically the charge is distributed charge.

The charge can be deposited along the length, over the surface or within the volume of the hosting body. And hence the distributed charges are line charge, surface charge and volume charge.

For each type of distribution, charge density is defined. In the case of line charge, linear charge density is charge per unit length. For surface charge, surface charge density is charge per unit area. Similarly, volume charge density which is charge per unit volume is defined for volume charge distribution.

Types of Charge Densities: Line, Surface & Volume

## Steps for getting Electric Field due to distributed Charge

So we may wish to derive the expression for Electric field intensity due to line charge or surface charge or the volume charge. In all the cases the basic procedure is the same. Yes but mathematical simplification in each case is different.

#### Step 1: Take infinitesimal charge from the given charge distribution and consider it as a point charge

Every time for deriving the formula for E, we are going to use the standard formula i.e.

$E=\frac1{4\pi\varepsilon}\frac Q{r^2}$

But, this formula is for the point charge, isn’t it? It is assumed that the charge you are considering is concentrated at a point.

So basically we consider a very small portion of the given charge distribution, say an infinitesimal charge or differential charge denoted by dq. Note that the total charge is q but we are considering a very small part of it viz. dq.

For example, say the given charge distribution is a line charge with linear charge density λ. Then to get the small portion of charge i.e. the infinitesimal charge, you have to consider a very small length, say dl, of that line charge. Ideally, dl should be approaching zero. And the charge contained by this dl, i.e. dq can easily be considered as a point charge.

So from the above figure, we can write dq for line charge as

$dq=\lambda dl$

Similarly, the infinitesimal charge (dq) from the total charge q can be obtained for the Surface and Volume charge. For the surface charge, we need to consider the infinitesimal surface while for the volume charge, the infinitesimal volume. And dq is obtained by multiplying respective densities.

$\begin{array}{l}dq=\sigma ds\;\;(for\;surface\;charge)\\dq=\rho dv\;\;(for\;volume\;charge)\end{array}$

#### Step 2: Find the E at the given point considering the dq from step 1. Say that field the dE

Once you have obtained the infinitesimal charge dq from the given charge distribution, forget rest for some time. And find the Electric field at the required point considering the only dq. It is our normal case of a point charge. Use the normal formula.

We are supposed to find the Electric field because of complete charge q. But in this step, we are writing only small E because of small q i.e. dq from the original charge. So let us call this infinitesimal field be dE.

$dE=\frac1{4\pi\varepsilon}\frac{dq}{r^2}$

#### Step 3: Properly note down and decompose the dE vector

We know that the electric field intensity is a vector quantity. Hence dE calculated in the above step is also a vector one. Now according to your position of ‘dq’ this dE vector can be pointing anywhere in the space. We should decompose this dE vector into its constituent components say vertical and horizontal according to the geometry of your diagram. For example, if there is an infinite line charge as shown in the figure and we are finding the E at point P. In this case, we decompose dE as shown in the figure. Similarly, if we have the surface charge as shown in figure

#### Step 4: Final E by integrating dE covering all the ‘dq’s with proper limits of integration

This might be the most crucial step. We are supposed to find out the Electric Field of the Distributed Charge at the given point because of the charge q. But, as you can realize in the previous steps, we have found out the infinitesimal field, dE considering the infinitesimal charge ‘dq’ from the given charge q.

Now it is the time to aggregate all the ‘dE’s of all the ‘dq’s. And in Mathematics the aggregation of the infinitesimal elements is the integration. So by using proper integration, the final answer for the E can be calculated.

1. For line charges, the proper form of line integration is used according to the coordinate system. Note that by applying the line integration, we are collecting all the ‘dq’s present over the line. So limits of the integration should be properly written according to the problem.
All the concepts of the line integration are properly applied over here e.g. the form of dl, the variable coordinate variable, fixed coordinates etc. Go through the concepts of Line Integration here.
2. For the surface charges, the proper form of surface integration should be applied. Go through the Surface Integration here.
3. For the volume charges volume integration with proper limits should be applied.

Note that separate integration should be used for each component of dE i.e. for horizontal and vertical in our illustration. Most of the time one component is zero because of symmetry of the charges.

Then final E at that point will be the vector sum of all the components obtained by the integration.

#### Alias for Step 3

If your problem doesn’t support symmetry and you are finding it difficult to decompose the dE into constituent components then instead of forming components you should write dE in vector notation.

$dE={\left(dE\right)}_x{\widehat a}_x+{\left(dE\right)}_y{\widehat a}_y+{\left(dE\right)}_z{\widehat a}_z$

Then as discussed in step 4, you should integrate for each term to get x, y and z components of the final E. Then vector addition of these terms will be our final Electric Field of the Distributed Charge.

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