E-Field and E-Potential

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  • #1
A Charge of 3.2x10-6 sits .025 m from a charge of -1.6x-6

a. Calculate the magnitude and direction of the E-filed at a point directly between the 2 charges

b. calculate the electric potential (relative to V = 0 at infinity) at the point directly between the two charges

c. Calculate the work needed to bring a 1x10-6 charge from infinity to the point between the two charges

19. A 2000 V/m electric field is directed along the +x-axis. If the potential at x = 10 m is 800 V, what is the potential at x = 6m?
A. 8800V
B. 2000V
C. 7200V
D. 1600V

Homework Equations

F = kqQ/r2
V= w/q

The Attempt at a Solution

a. F=9x10^9*-1.6x10^-6/.025^2
=-73.728 N

b. I don't know, v=w/q uses only one charge, and I don't know any other formulas. Some sort of clue, a formula, would be great

c. I think if I knew how to do part b I could do part c

19. I know that as you get closer, the potential increase, so the answer can't be B or D.
However, I don't know any formula that could tell me how to find the answer.

Help please?

Answers and Replies

  • #2
For (a), you have the correct formula but the distance should be half of .025 since the point you are calculating is midway between the two charges. Also, you must use the formula twice to calculate the E due to each of the two charges. Add them together if they are in the same direction.
  • #3
Thank you, for part c would it be v=w/q with my anwser from part b?

Also do you know to do 19?

Thank very much
  • #4
Yes on part c.
For 19, you must use E = V/d where the V should really be delta V, the potential difference between the two points separated by distance d.

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