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?