- #1
Linus Pauling
- 190
- 0
1. Estimate the electric field (strength and direction) at the point A.
5cmx5cm gridL
2. E = -dV/ds
3. I had four options, each a combination of 500 or 1000V/m for the E field strength and then left or right for the direction. I knew it must be to the left because the field points in the direction of decreasing potential, but how was the strength obtained (the answer is 500V/m)?
I did E = dV/ds = 12.5V/.00625m = 2000V/m, because point A is 1/8(5cm) from the left side of the grid...
5cmx5cm gridL
2. E = -dV/ds
3. I had four options, each a combination of 500 or 1000V/m for the E field strength and then left or right for the direction. I knew it must be to the left because the field points in the direction of decreasing potential, but how was the strength obtained (the answer is 500V/m)?
I did E = dV/ds = 12.5V/.00625m = 2000V/m, because point A is 1/8(5cm) from the left side of the grid...