Formula for the Electric Field Due to Continuous Charge Distribution

[prosteve037]

Homework Statement

I am having trouble understanding how

$\textit{Δ}\vec{E}\textit{ = k}_{e}\frac{Δq}{{r}^{2}}$

(where ΔE is the electric field of the small piece of charge Δq)

turns into

$\vec{E}\textit{ = k}_{e}\sum_{i}\frac{{Δq}_{i}}{{{r}_{i}}^{2}}$

then into

$\vec{E}\textit{ = k}_{e} \lim_{Δq→0}\sum_{i}{\frac{{Δq}_{i}}{{{r}_{i}}^{2}}}$

which finally takes the form

$\vec{E}\textit{ = k}_{e}\int{\frac{dq}{{r}^{2}}}$

Homework Equations

- Listed Above -

The Attempt at a Solution

I understand the summation part, which just takes the sum of all the electric fields of each individual part of the continuous charge distribution.

What I don't understand is the latter part, where the limit is taken of the summation and then turned into an integral. What's going on there and why is that done?

 

[SammyS]

Homework Statement

I am having trouble understanding how

$\textit{Δ}\vec{E}\textit{ = k}_{e}\frac{Δq}{{r}^{2}}$

(where ΔE is the electric field of the small piece of charge Δq)

turns into

$\vec{E}\textit{ = k}_{e}\sum_{i}\frac{{Δq}_{i}}{{{r}_{i}}^{2}}$

then into

$\vec{E}\textit{ = k}_{e} \lim_{Δq→0}\sum_{i}{\frac{{Δq}_{i}}{{{r}_{i}}^{2}}}$

which finally takes the form

$\vec{E}\textit{ = k}_{e}\int{\frac{dq}{{r}^{2}}}$

Homework Equations

- Listed Above -

The Attempt at a Solution

I understand the summation part, which just takes the sum of all the electric fields of each individual part of the continuous charge distribution.

What I don't understand is the latter part, where the limit is taken of the summation and then turned into an integral. What's going on there and why is that done?

Have you had a course in integral calculus?

 

[spaghetti3451]

Homework Statement

I am having trouble understanding how

$\textit{Δ}\vec{E}\textit{ = k}_{e}\frac{Δq}{{r}^{2}}$

(where ΔE is the electric field of the small piece of charge Δq)

turns into

$\vec{E}\textit{ = k}_{e}\sum_{i}\frac{{Δq}_{i}}{{{r}_{i}}^{2}}$

then into

$\vec{E}\textit{ = k}_{e} \lim_{Δq→0}\sum_{i}{\frac{{Δq}_{i}}{{{r}_{i}}^{2}}}$

which finally takes the form

$\vec{E}\textit{ = k}_{e}\int{\frac{dq}{{r}^{2}}}$

Homework Equations

- Listed Above -

The Attempt at a Solution

I understand the summation part, which just takes the sum of all the electric fields of each individual part of the continuous charge distribution.

What I don't understand is the latter part, where the limit is taken of the summation and then turned into an integral. What's going on there and why is that done?

Think of the first formula as giving the i'th electric field due to the i'th (discrete) charge at the i'th position. This equation assumes that the i'th charge has some spatial extent. The i'th position is approximately at the centre of this spatial extent. (This charge has to tend to 0 as the spatial extent goes to 0).

You need to sum all the i'th terms to obtain the total electic field. But alas! The charges in the equation are not infinitesimally small, so that the position of each charge in the equation is at best an approximation of the spatial extent of the charge. Therefore, to obtain the total electric field, we need to sum all the i'th electric fields in the limit of the i'th charge tending to 0. this condition constrains the spatial extent to tend to 0, so that the position becomes exact. In other words, you are converting the discrete second equation into the continuous third equation. The limit taken for the charge tending to zero converts the discrete charges into infinitesimally small charges at unique points that have no spatial extent.

The fourth equation is simply the commonly used shorthand notation for the more easily understood third equation.

 

[prosteve037]

Have you had a course in integral calculus?

I took some Calculus back in high school, which was 2 years ago. I can't seem to remember much/enough to understand the mathematical progression here :/

Think of the first formula as giving the i'th electric field due to the i'th (discrete) charge at the i'th position. This equation assumes that the i'th charge has some spatial extent. The i'th position is approximately at the centre of this spatial extent. (This charge has to tend to 0 as the spatial extent goes to 0).

You need to sum all the i'th terms to obtain the total electic field. But alas! The charges in the equation are not infinitesimally small, so that the position of each charge in the equation is at best an approximation of the spatial extent of the charge. Therefore, to obtain the total electric field, we need to sum all the i'th electric fields in the limit of the i'th charge tending to 0. this condition constrains the spatial extent to tend to 0, so that the position becomes exact. In other words, you are converting the discrete second equation into the continuous third equation. The limit taken for the charge tending to zero converts the discrete charges into infinitesimally small charges at unique points that have no spatial extent.

The fourth equation is simply the commonly used shorthand notation for the more easily understood third equation.

I'm a little confused about what you mean by "spatial extent". Are you saying that the discrete pieces of charge aren't taken to be at their exact locations in the first equation?

 

[asteriatic]

I can explain the reasoning behind the progression from the first equation to the final equation in terms of mathematical concepts and physical principles.

Firstly, the equation \vec{E}\textit{ = k}_{e}\frac{Δq}{{r}^{2}} represents the electric field at a point due to a small piece of charge Δq located at a distance r from the point. This is a simplification of the general equation for the electric field, which takes into account the contributions from all the charges in a continuous charge distribution.

To account for the contributions from all the charges, we use the concept of a summation. The second equation, \vec{E}\textit{ = k}_{e}\sum_{i}\frac{{Δq}_{i}}{{{r}_{i}}^{2}}, represents the electric field at a point due to the entire continuous charge distribution, where the sum is taken over all the small pieces of charge Δq in the distribution. This is a more accurate representation of the electric field at a point compared to the first equation.

Next, we take the limit of the summation as the size of each small piece of charge, Δq, approaches zero. This is because in a continuous charge distribution, the charges are infinitely small and therefore the summation can be seen as an approximation. As we take the limit, the summation becomes more accurate and approaches the true value of the electric field at a point. This is represented by the third equation, \vec{E}\textit{ = k}_{e} \lim_{Δq→0}\sum_{i}{\frac{{Δq}_{i}}{{{r}_{i}}^{2}}}.

Finally, we can use the mathematical concept of an integral to represent the limit of the summation. The integral allows us to sum up the contributions from all the infinitesimally small charges in the continuous charge distribution, giving us the final equation, \vec{E}\textit{ = k}_{e}\int{\frac{dq}{{r}^{2}}}. This is the most accurate representation of the electric field at a point due to a continuous charge distribution.

In conclusion, the progression from the first equation to the final equation is a result of using mathematical concepts and physical principles to accurately represent the electric field at a point due to a continuous charge distribution.