Please see the attachment about electric field:
How do you know to substitute 6N/C for charge 2?
Thanks,
CR video Electrostatic
CR video Electrostatic
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Re: CR video Electrostatic
I think you meant to ask how we know about the 2 N/C. Remember that these are field strengths, and fields are not uniform in strength, but fall off as an inverse square of distance. So what we are doing in this problem is finding field strengths at point A.
The easiest way to do this is to calculate each field strength and direction individually:
Field 1 (from +Q) must be pointing to the right (since two positive charges repel each other) and is given as 6 N/C.
Field 2 must be pointing to the left, but we do not yet know it's magnitude. We do however know it is 3 times further away from 3Q than from Q, so we can simply adjust:
So Field 2 must be pointing to the left with a magnitude of 2 N/C.
We add these vectors up and we get a total of 4 N/C pointing to the right. Then we just solve by F=qE and find that this charge must be experiencing a force of 4 N.
The easiest way to do this is to calculate each field strength and direction individually:
Field 1 (from +Q) must be pointing to the right (since two positive charges repel each other) and is given as 6 N/C.
Field 2 must be pointing to the left, but we do not yet know it's magnitude. We do however know it is 3 times further away from 3Q than from Q, so we can simply adjust:
Code: Select all
6N/C * 3 / (3^2) = 2 N/C
We add these vectors up and we get a total of 4 N/C pointing to the right. Then we just solve by F=qE and find that this charge must be experiencing a force of 4 N.