gravitational field, electric field model, electric field charges, ActivPhysics

Gravitational field -- analogous to electric field

The electric field is almost identical to the gravitational field.  We showed this by taking four statements, morphing them by changing a few words, and making them from statements about gravity to statements about the electric field!











Electric Field Model

Positive electric field buldges outward and negative concaves inward.  When a positive charges moves inward towards negatively charged electric field, the positive charge falls to the middle of the negative electric field.  Now with positive and negative electric field next to eachother: when you put the positive charge on top of positively charged electric field it will be repelled from the positively charged field and towards negative.

Electric Field Vectors from Two Point Charges

Two charges of opposite sign were placed on a graph and four points were also placed on the graph and we were to find the electric field at the four points.  The equation we used to find the electric field was kq/r^2.  We then made a spreadsheet on excel and calculated the electric field at distences of .5 cm to 10 cm.  We then found the electric field from the positive charge and the negative charge in the x and y directions by looking at the spreadsheet and filling in the value of the electric field at that certain distance.  After we found all of the electric fields, we added them together to find the final sum of all the electric fields.  It was found that all of the y parts of the electric field canceled and the resultant electric field was 52600 N/C in the x direction.  

Electric Field Charges from a Uniformly Charged Rod


A uniformly charged rod is placed with a point is placed underneath the rod and to the side of the rod.  The rod is split into ten pieces so it is easier to calculate the net electric field.  The electric field is drawn from the two points to the ten sections on the rod.  Then the electric field is calculated using the equation kQ/r^2.  The values of r and the equation were put into a spreadsheet and then the sum was calculated to find the resultant electric field at the two points.  For the point next to the rod, the resultant electric field was accurate because all of the electric field lines lie in the x axis.  As for the point beneath the rod, it is more difficult to calculate the net resultant electric field because the radius had to be calculated by the sqrt of  (X^2 + Y^2).  A better way of finding the electric field for the point underneath the bar is by using an integral.  When we found the electric field by calculating all 10 points and sum it up we got it to be  5.95 x10^4 and when we calculated it using an integral, we found it to be 6.0 x10^4 giving us a .47% error which is very good.  The little bit of error we got was due to the fact we had to calculate the radius and that we only cut the rod into 10 pieces instead of more.

ActivPhysics

ActivPhysics answers for electric fields