Dipole Moments and Flux

We started the day discussing about electric field. Just like gravitational field that we have learned in 4A. There are 4 understandings of gravitational field, and all of those are related to masses. In electrical field, these understandings are exactly the same, except that it is related to charges as shown below.

Recall from last lecture about the force of 2 charges in the equation below. Today we learn about electrical force as force divided by charge. This charge is the source charge; thus, we can get a new equation of electrical force as affecting charge multiplied by constant divided by difference in charge distance squared.

To calculate electric field, we can basically use the basic equation using force divided by the charge of a particle. Another way to do it by using superposition, which basically just means the sum of the electrical fields created by each particle present. The best way to calculate this is by using the electric field equation which involves the charge, radius between particles, and electric constant.

To project the picture above into a phython, there are 4 steps as shown below. We want to find r first by subtracting r2-r1. Then, find the magnitude. Entering the scalar, then lastly, multiply it by r.

Given a program, we first predict how the projection would be and we predicted to be as below.

When its projected in the computer, it was close to what we predicted. The cylinder go on the x, y, and z-axis infinitely. There are 6 arrows as predicted going on x direction.

Given 2 charges with the same distance from the origin. Using the equation we learned from earlier, we can find 2 different electrical force. By adding these electrical forces, we can find the total electrical force.

We can now find the relation between electrical field and length. By deriving the change of electrical force and, with limit from d to d+L, we can find the linear density charge, which is the electric charge divided by the length of the vector magnitude.