Trista
Mar14-06, 12:52 PM
Here is the problem:
Three point charges are aligned along the x-axis as shown below. Find the electric field at the position x = +2.0m, y=0.
..............................y
..............................|
--------|<--.50m --->|<--------------.80m-------->|
--------0------------0-----------------------------0--------- X
........-4.0nC.............|5.0nC.......................... ....3.0nC
..............................|
So, I figured that I have to add up the E along the x axis and that should give me my answer. But, I'm not sure what to do with the numbers when they are already an Electrical Field... -4.nC isn't the charge, so, don't I need to find the q (or charge) first? then put it in the form kq/r^2??
The only way I can come up with the answer is wrong... total E = 4,
EA = 4 nC X 2m = 8 nC/m. 3 Charges times 8 nC/m = 24 nC... But that was simply a coincidence, I'm sure. 24 nC is the right answer, just need help getting there.:eek:
Thank you in advance for your help and patience.
Three point charges are aligned along the x-axis as shown below. Find the electric field at the position x = +2.0m, y=0.
..............................y
..............................|
--------|<--.50m --->|<--------------.80m-------->|
--------0------------0-----------------------------0--------- X
........-4.0nC.............|5.0nC.......................... ....3.0nC
..............................|
So, I figured that I have to add up the E along the x axis and that should give me my answer. But, I'm not sure what to do with the numbers when they are already an Electrical Field... -4.nC isn't the charge, so, don't I need to find the q (or charge) first? then put it in the form kq/r^2??
The only way I can come up with the answer is wrong... total E = 4,
EA = 4 nC X 2m = 8 nC/m. 3 Charges times 8 nC/m = 24 nC... But that was simply a coincidence, I'm sure. 24 nC is the right answer, just need help getting there.:eek:
Thank you in advance for your help and patience.