ActivPhysics: Electric Field, Point Charge
To observe and gain a better understanding of electric fields, we used ActivPhysics to perform a few simulations involving point charges and electric fields. This board shows the answers to the six questions that were asked in the simulation. I will go over these questions one by one.
1) In this part we observed the nature of electric charges and the electric fields that they cause. In this case, we are using two positive point charges.
The first observation to note is that the electric field points away from the charges. This means that two positive charges repel each other.
Next we observed that the closer the point charges were to each other, the magnitude of the electric field increased.
2) In this part, we set the first charge (q1) to +10.0E-8, the second charge (q2) to 4.0E-8, and the seperation distance was 100 cm. We calculated the electric field caused by q1 as shown on the whiteboard picture. The electric field was 900 N/C.
3) In this simulation, we examined how an electric field changes as a point charge changes.
We saw that with a positive charge, the field lines point outward while with a negative charge they point inward. The lines seemed to start and end in the same place relative to the different charges we observed. However, we did notice that the larger the charge, the larger the line density of the field.
4,5,6) In this simulation we observed a uniform field with two plates. We saw that the electric field was spaced uniformly and that the magnitude was the same between the planes and it was independent of the position.
The Electric Field from an Extruded Charge Distribution
In this problem, there is a rod with uniform charge throughout. The charges of the rod are divided into ten 1.00 cm "point charge" segments. We are to calculate the electric field from the rod to two points in space: P which is located 5.00 cm to the left and on axis to the rod, and P' which is located 5.00 cm perpendicular the the center of the rod. We used the formula E=Q/(r^2) to find the electric field of each rod segment. We were given that the entire rod has a charge of 5.00E-8 C. The charge is divided into ten 1.00 cm pieces therefore we used a charge of 5.00E-9 C to calculate the electric field of each segment. We used excel as shown below to calculate the charge on each segment. The initial distance (r) was 0.055 m (point P to the center of the segment) and the distance increased by 0.01 m. We calculated the charges with excel for all of the segments and added them together to get the electric field of the rod on point P, which came out to be 5.96E4 N/C.
Electric Field from a Uniformly Charged Rod
Here we calculated the electric field on the point P' from the same rod as the previous part. In this part, the direct distance from the point to the center of a segment piece (r) is needed to calculate the electric field on the point. We know that the vertical distance from P' to the rod is 0.05 m. We then complete the triangle using the variable x as the horizontal distance from the point on the rod just mentioned to "r". Solving for the distance we get that r=/sqrt(0.05^2+x^2). Now we use our equation for electric field to get E=(2k(C/10)
Electric Field vs. Distance
Electric Field Hockey
In this game, we place charges on the screen to move a charged puck to a goal.
