- #1

LLT

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Electric Field and Electric Potential Please?

How they are related and stuff...

Gauss' Law and Coulomb's Law...

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- Thread starter LLT
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- #1

LLT

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Electric Field and Electric Potential Please?

How they are related and stuff...

Gauss' Law and Coulomb's Law...

- #2

LLT

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I mean how to use Gauss' Law to derive Coulomb's Law from the point charge

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- #4

Oxymoron

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Take a flat surface area (A) which a UNIFORM electric field (**E**) passes through. The amount of electric flux (the number of electric field lines if you like) that pass through this area is simply E*A. This is only for UNIFORM electric fields!!! For non-UNIFORM electric fields you have to integrate! And this is called Gauss' Law. Here is how it works...

Take any surface area (A) and any electric field (**E**). We divide up the area into tiny little areas dA which are almost infinitely small. This way we can treat each one like little flat pieces. The electric field, which does not have to be uniform, can now be considered uniform over our tiny area. Now the electric flux equals the summation of all the E's and the A's. So Electric flux = integral of E dot dA. Where the dot is the dot product. I assume you know this and how to integrate.

With Gauss' Law we tend to integrate over a closed surface. Let me explain a little more about this.

Take a point charge anywhere in space. We want to know the electric flux around that point (ie. the intensity of the electric field caused by that charge). Gauss' Law says that we can surround that charge with an imaginary surface (A Gaussian Surface as it is often called). This surface MUST have symmetry if you want easy calculations (...like a sphere, cylinder or flat plane), of course you could take any surface. Now take the point charge and surround it with an imaginary sphere shell with zero thickness. Because of the symmetry we know that the electric flux coming out of the point charge must be equal to the electric flux passing through our shell. If you don't believe this draw a dot surrounded by a circle and draw lines directed radially outward from the dot. See how every line passes at some point through the circle! This is exactly what Gauss' Law states!

If we center our sphere on the point,**E** must have the same intensity everywhere on the surface and we know that **E** is parallel to d**A**.

Notice here that I have written the area as a vector. This is because in algebra, every flat plane can be defined by its normal vector (ie. a vector that is perpendicular to its surface). With our Gaussian sphere, we can divide it up into tiny little areas which are all flat. Take any one of these and its normal vector points radially outward just like the electric field lines emmanating from the point charge. This is why they are parallel.

So...

integral(**E .**d**A**) = integral(EdA) = E integral(dA)

We took the E outside of the integral because it becomes constant because it is the same at all points on the surface.

The integral of dA is simply the sum of all the tiny areas which is equal to the surface area of a sphere.

= E(4*pi*r^2)

Gauss' Law states that the if you perform a closed integral on**E .**d**A** then it equals the charge enclosed by the Gaussian surface. (we just did that with a sphere and point charge). However electric flow through space does not go unimpeded hence we introduce a new constant which depends on the medium which the electric field penetrates. Epsilon-naught (e0) which is the permittivity of free space.

So E(4*pi*r^2) = Q/epsilon-naught. Solve for E...

E = Q/(4*pi*epsilon0*r^2).

Notice the 1/r^2 factor?? This makes sense because the electric field obeys the inverse proportionality rule.

Plus this is where Coulomb's Law comes from!!!!!

Remember F = k(q1q2/r^2) (Coulombs Law)

Well E = F/q right!?

So E = [k(q1q2/r^2)]/q = k(Q/r^2) (cancel one of the q's)

We know k is a proportionality constant and is equal to 1/(4*pi*epsilon0)

So E = Q/(4*pi*epsilon0*r^2). Just like above.

This is equal because Coulomb's law ONLY deals with point charges.

However we could easily perform Gauss' Law on any surface like a metal conducting ball! How could we have done this using Coulomb's Law!?!?! Impossible. But with Gauss it is simple! Watch...

Take an conducting sphere which has charge Q distributed evenly throughout. It has a radius of r. We can determine the electric field by performing a closed integral around the ball!

integral(**E . **d**A**) = E(4*pi*r^2) = Q/epsilon0 (just like above!!)

This result is the same as a point charge because Gauss' Law thinks the sphere is just an enlarged version of the point charge.

Lets do a long conducting wire!

We can calculate the electric field somewhere near the long wire (make sure it is far from the ends as possible). This time we are going to construct a cylindrical Gaussian surface so that it symmetrically encloses the wire!

Again...

integral(**E . **d**A**) = E(2*pi*r*l) (No need to include the surface areas of the ends of the cylinder because no electric flux exists through it (they are perpendicular!)).

Well that is all I can say about Gauss and Coulomb. But what about Potential?

Take any surface area (A) and any electric field (

With Gauss' Law we tend to integrate over a closed surface. Let me explain a little more about this.

Take a point charge anywhere in space. We want to know the electric flux around that point (ie. the intensity of the electric field caused by that charge). Gauss' Law says that we can surround that charge with an imaginary surface (A Gaussian Surface as it is often called). This surface MUST have symmetry if you want easy calculations (...like a sphere, cylinder or flat plane), of course you could take any surface. Now take the point charge and surround it with an imaginary sphere shell with zero thickness. Because of the symmetry we know that the electric flux coming out of the point charge must be equal to the electric flux passing through our shell. If you don't believe this draw a dot surrounded by a circle and draw lines directed radially outward from the dot. See how every line passes at some point through the circle! This is exactly what Gauss' Law states!

If we center our sphere on the point,

Notice here that I have written the area as a vector. This is because in algebra, every flat plane can be defined by its normal vector (ie. a vector that is perpendicular to its surface). With our Gaussian sphere, we can divide it up into tiny little areas which are all flat. Take any one of these and its normal vector points radially outward just like the electric field lines emmanating from the point charge. This is why they are parallel.

So...

integral(

We took the E outside of the integral because it becomes constant because it is the same at all points on the surface.

The integral of dA is simply the sum of all the tiny areas which is equal to the surface area of a sphere.

= E(4*pi*r^2)

Gauss' Law states that the if you perform a closed integral on

So E(4*pi*r^2) = Q/epsilon-naught. Solve for E...

E = Q/(4*pi*epsilon0*r^2).

Notice the 1/r^2 factor?? This makes sense because the electric field obeys the inverse proportionality rule.

Plus this is where Coulomb's Law comes from!!!!!

Remember F = k(q1q2/r^2) (Coulombs Law)

Well E = F/q right!?

So E = [k(q1q2/r^2)]/q = k(Q/r^2) (cancel one of the q's)

We know k is a proportionality constant and is equal to 1/(4*pi*epsilon0)

So E = Q/(4*pi*epsilon0*r^2). Just like above.

This is equal because Coulomb's law ONLY deals with point charges.

However we could easily perform Gauss' Law on any surface like a metal conducting ball! How could we have done this using Coulomb's Law!?!?! Impossible. But with Gauss it is simple! Watch...

Take an conducting sphere which has charge Q distributed evenly throughout. It has a radius of r. We can determine the electric field by performing a closed integral around the ball!

integral(

This result is the same as a point charge because Gauss' Law thinks the sphere is just an enlarged version of the point charge.

Lets do a long conducting wire!

We can calculate the electric field somewhere near the long wire (make sure it is far from the ends as possible). This time we are going to construct a cylindrical Gaussian surface so that it symmetrically encloses the wire!

Again...

integral(

Well that is all I can say about Gauss and Coulomb. But what about Potential?

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