Electric Fields

We started the day understanding more about electricity behavior. We predicted about where electron will hit when it passes through a negative and positive plate. The electron will be attracted to go to the positive plate as it has a negative charge. The distance between negative and positive is 2a because each a starts from half way between the 2 charges. Then, we predicted where forces of 2 different position of charges will go when there are electric field going horizontally through them. Each of these forces will go the opposite  way since the positive and negative pole is on different sides of the electric field. Thus, it creates a torque.

Torque that we learned in rigid bodies behaves the same way as the torque that we will learn in this class. Since there are 2 charges, positive and negative, the torque net then will be added which makes makes the torque net becomes 2 of them. As we previously known, F = qE & p = 2aq, we can substitute into the equation and simplify it as the cross product between dipole moments and electric field.

Here we can see how we use torque to determine the change of electric potential energy by setting up an integral of torque in terms of theta. Assumed that the initial potential is zero at 90 degrees of angle, we can simplify the equation to solve for final potential energy which is just the multiplication of dipole moment and electric field. Since the change of potential energy also equals to the work done, then we could also solve the same that way.

Flux is the amount of charge enclosed. In a closed system, the net flux, which is the change of electric flux that comes in minus the flux going out will be zero. Flux can be determined using the electric field multiplied by the cross sectional area and the cosine of the angle of the charge shooting towards the surface. In a case of zero degree of angle, cosine of angle will be 1, and the flux can be simplified to just the electric field and area of cross section.

As I mentioned before, the flux of a closed system will be zero. Thus, for a unit cube that has a charge with a electric field moving from left to right, we can see that the flux will be zero by observing that the left side will receive negative charge with a magnitude of EL^2, and the right side will release positive charge with an equal but opposite magnitude. As we add up the incoming flux to the leaving flux, the net will be zero.