Day 9: Electric Field

4 Statements

At the beginning of the class, we are told to change four statements from gravity situations to electric situation.

We combined the equation F = k*q1*q2/r^2 and E = F/q to get E = k*q/^2

This shows the electric field of a proton.

We used 4 step to find the electric field equation.

This is our prediction for the vpython code. 

This is the actual pictures when we solved for the electric field equations.

Electric Field from Two Point Charges

This is a problem of electric field from two point charges. We need to find Etot.

Here is another example problem where the point charge changed places.

E-Field Vectors from a Uniformly Charged Rod

We also did a problem of electric field vectors from a uniformly charged rod.

The first point where there is no y-axis, the Etot is about 60000 N/C

In another example problem, we have another position where we need to find Ex and Ey. We found that the Ex is 0 and Ey is 110210 N/C

Electric Field Due to a Differential Charge

We used the method of electric field due to a differential charge to calculate the Etot, which turned out to be 60000 N/C. The result is close to our first result which was also about 60000 N/C.

Conclusion

In today's class, we first predicted the electric field vectors of a proton, then we used vpython to see the actual picture. Then we practiced with few electric field problems to calculated Etot between two charges.