Physics Lab » Electric Fields

Aug 03 2008

Electric Fields

An electric field surrounds charged particles and represents the force per unit charged felt by other charged particles in that field. If the electric field does not change in time, then the force felt by charged particles in the electric field is given by:

$\vec{F} = \vec{E}(\vec{r})q$

(1)

Where $\vec{E}(\vec{r})$ is the electric field vector at a position $\vec{r}$ from the source producing the electric field and q is the charge being affected by the electric field. The quantity q is positive for positive charges, and negative for negative charges.

Coloumbs Law

Columbs Law descrbes the force felt by two point charges separated by a distance, in scalar form this law is:

$F = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}$

(2)

Where:

$\epsilon_0=8.85418782\times 10^{-12} \frac{A^2 s^4}{m^3 kg}$ is the permittivity of free space.
q₁ and q₂ are the charges of the two particles.
r is the scalar distance between the two particles.

Equation 2 can be slightly simplified by combining the product of coefficients into one constant called Coloumb’s Constant.

$F = k\frac{q_1 q_2}{r^2}$

(3)

Where $k=\frac{1}{4\pi\epsilon_0} = 8.987551787\times 10^9 \frac{Nm}{C^2}$

Equation 2 can be written in vector form in the following way:

$\vec{F} =\frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}\hat{r}$

(4)

Where $\hat{r}$ is a unit vector of the displacement vector between the two charges.

Using equation 4 andn equation 1 we can determine the electric field produced by a point charge. Assume that we are concerned with the electric field produced by charge q_{1 . Then from equation 1 the force felt by charge 2 is:}

$\vec{F}&=\vec{E}(\vec{r})q_2$

Substituting equation 4 into the above equation we get:

$\frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}\hat{r}&=\vec{E}(\vec{r})q_2$

Rearranging the equation to solve for $\vec{E}$ gives us the expression for the electric field produced by a point charge.

$\vec{E}(\vec{r})&=\frac{1}{4\pi\epsilon_0}\frac{q_1 }{r^2}\hat{r}$

(7)

Multiple Charges

When there is more than one charge the electric field produced in the surrounding space is the vector sum of all the individual electric fields produced by each point charge.

$\vec{E}(\vec{r})=\sum_{i=1}^{N}\frac{1}{4\pi\epsilon_0}\frac{q_i }{r_i^2}\hat{r_i}$

(8)

Where $\vec{r_i}$ is the displacement between the charge producing the electric field and the point at which the electric field is to be calculated. $\hat{r_i}$ is the unit vector of the displacement vector.

Electric Potential

By definition, the potential is the negative of the work done in moving an object. The electric potential is defined as the negative of the work done per unit charge by the electric field in moving a point charge from infinity to a distance r away from the source of the field. Starting from the definition of work:

$V=-W=-\int_{\infty}^{r} \frac{\vec{F}}{q}\cdot\vec{dr}$

(9)

The force is defined by equation 4, but we will use a slightly different notation, we will define Q as the charge producing the electric field, and q as the charge that will be moved.

$\vec{F} =\frac{1}{4\pi\epsilon_0}\frac{Qq}{r^2}\hat{r}$

(10)

Substituting equation 10 into equation 9, we get:

$V=-\int_{\infty}^{r}\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\hat{r}\cdot(-\hat{r}dr)$

(11)

The integral operator $\vec{dr}$ was replaced with $-\hat{r}dr$ because the charge moves from infinity towards the charge producing the field.

Rearranging equation 11, we get:

$V=\int_{\infty}^{r}\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\hat{r}\cdot\hat{r}dr$

(12)

Note that $\hat{r}\cdot\hat{r}=1$ because $\hat{r}$ is a unit vector and the scalar product of a unit vector with itself is equal to 1.

$V=\int_{\infty}^{r}\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}dr$

(13)

Integrating equation 13, we get:

$V&=\int_{\infty}^{r}\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}dr \\ V&=\left[ -\frac{1}{4\pi\epsilon_0}\frac{Q}{r} \right]^{r}_{\infty}\\ V&= -\frac{1}{4\pi\epsilon_0}\frac{Q}{r} - \left(-\frac{1}{4\pi\epsilon_0}\frac{Q}{\infty} \right)\\V&= -\frac{1}{4\pi\epsilon_0}\frac{Q}{r}$

(14)

Equation 14 gives the expression for the electric potential at a distance r from the source. Note that if the charge producing the electric field is positive then the electric potential is negative. If the charge is negative, then the electric potential is positive. In the above applet, when calculating the potential, the green gradient represents a postive potential and a red gradient represents a negative potential.

5 responses so far

5 Responses to “Electric Fields”

# chinamanon 18 Sep 2008 at 2:40 am

Hi
I like your posts, It makes me thinking.
# Carteron 01 Oct 2008 at 4:47 pm

Lovely stuff.

Nice Work!
# classifiedon 05 Nov 2008 at 9:46 pm

I tell you a secret.

This electric field is the polarization of the quantum foam, as seen in the Casimir effect, due to the pattern found in the molecular structure of the metal which vibrates from one end to another, and allows for only specific lengths of virtual particles to occurs within the molecular lattices and between the atoms, these creating a domino effect as you look further and further away from within the lattice to outside it, slowly losing their focus and hence strength as the surrounding ‘chaos’ takes hold.

It is same deal for gravity, magnetism and nuclear forces, only different wavelengths and different emergent patterns. Crank up a an em field to 25 megawatts and you will see it bend gravity as well.

Its a secret of course.
# studenton 13 Jan 2009 at 7:39 pm

Great work! I’m really impressed with your applications.
# naveed alamon 09 Dec 2009 at 8:12 am

good job. good idea, i have create some simulation of Modern Physics in C# sharp and open GL.visit the web site http://www.profmsaeed.org.

Electric Fields

Electric Fields

Coloumbs Law

Multiple Charges

Electric Potential

5 Responses to “Electric Fields”

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