Lab

Electric Potential Energy

Here we drew a few diagrams of circuits with light bulbs in series and parallel; From experimentation we found that the bulbs were brightest when the bulbs were parallel and the batteries were in series. we found them to be least brilliant when the batteries were parallel and the bulbs were in series. We also discussed switches and described what pole and throw are.

Here we define voltage as potential energy over q and we described potential energy lines.

This formula relates electric potential difference with potential energy work and the formal definition of work.

Circuits

Today, we were introduced to circuits, we were thought how circuits are similar to rivers. First we had current, the flow of the electric charge measured in amperes. Then we had voltage, the electrical potential difference measure in volts. Then we found power, the ability to do work in watts. Next we went on to define Resistance as opposition to the passage of the electric current in ohms.

Here we define the current density (J), or current per cross sectional area (I/A) measure in amps/meter^2. Vd= delx/delt, I=delQ/delt. pn or charge density, charge/density. I=pn*Volume*q/delt. Therefore I=pnA (Vd * t) q /delt = pn*A*Vd*q
Note used on hw:(Finally J= I/A=pn*A*Vd*q/A = n*Vd*q.)

At the end of today we looked at wires and how their thickness effects the flow of electric charge (I). 

Electric Field Lines and Flux

E=kq/r^2 is proven to be true for point charges and spheres.

The surface are of a cylinder as well as its charge ratio.

Here we go ahead and find the electric field of a cylinder that is infinitely long.

Comparison of gravity to electric field

Flux and Gauss Law

Introduction, Today we learned about Gauss Law. We learned that flux = E.A or flux is equal to the closed surface integral of the electric field dotted with dA (the areas that makes up the entire surface)

or the charge enclosed over some constant epsilon not(q/epsilon). 

Note net electric force is always the same anywhere in the field defined so we can relate F=qE=ma

Here we have two charges to which we find the electric dipole moment, which is similar to moment of inertia. So what we did was let F= qEsin(theta) and then say r= d/2. so torque is F(d/2) + F(d/2), or qEdsin(theta) or pxE. potential energy is qrE. (think m=q, h=r, E=g)

In conclusion the Flux is dependent on the surface area or area perpindicular to field and the strength of the field or E.A. The electric dipole moment can be said to equal pxF.

Electric Field

Introduction: In today's lecture we learned about electric fields. We learned about their properties, and how to derive this field E. Where E is Force/Q2 and can be rewritten as E=kQ/r^2.

Answers to the questions from online. Answers to basic questions, to test understanding of the Electric Field and how they react differently to positive and negative charges.

We are asked to find the E-field at these 4 points.

Here we show the total E at each point. E=kq/r^2. We wrote an equation so that wherever you place your two point charges (x,y) and (x1,y1) for any charge q1 and q2 it can find the E-field at any point (Px,Py).

In Conclusion, Every particle with a charge emits an electric field, no matter how far. However it is inversely proportional to distance, so many times the electric field is negligible if the distance is really large and the charge is small. There are two types of charges that play a role in electric fields and that would be a positive charge and a negative charge the negative charge looks like a bunch of vectors pointing towards it, if you were to look at how the electric field around a charge would behave. A positive charge would have vectors pointing radially outward.

Electric Force

Chris Minaya, Michael Pryor, Andrew. Pete

Introduction: In this lab, we will observe the force between two charged objects. We will examine a video where a ball hanging from a string is slowly subjected to another ball that has the same charge.then the ball hanging from the string is repelled.

Here we derive an equation for the Force acting on the ball hanging from the string.

Conclusion:
1.) Yes, we were able to tell immediately that it was an inverse relationship. Then by looking at the slope we figured it must be and inverse square due to the steepness of the graph.
2a.)  They are the same value, so the percent difference is 0. However our equation for the function does not accurately describe the function too 100%.
2b.) A=kq^2 => q = sqrt(A/k)= 5.25 * 10^-8
2c.)A=k(q/2)(q)=k(q^2/2) => q = sqrt(2A/k)= 7.43 * 10^-8 and q/2 = 3.72*10^-8
3.) No, we derived the force in the x direction from using a fbd and equations for static equillibrium. Since this is what we used for our force equation our force will only be positive since distance is positive (measure from origin). Hence the best we can do is find the absolute value of q.
4.)Uncertainty for force can be found, systematic error can be found for not accounting for the mass of the string string and small things like friction and minute spinning of the string. We also assume that acceleration is equal to zero.