Asked by SHUBHANGI
Two point charges, 6.0 µC and -2.0 µC, are placed 6.0 cm apart on the x axis, such that the -2.0 µC charge is at x = 0 and the 6.0 µC charge is at x = 6.0 cm.
(a) At what point(s) along the x axis is the electric field zero? (If there is no point where E = 0 in a region, enter "0" in that box.)
x < 0 ___________
0 < x < 6.0 cm____________
6.0 cm < x____________
(b) At what point(s) along the x axis is the potential zero? Let V = 0 at r = . (If there is no point where V = 0 in a region, enter "0" in that box.)
x < 0 ____________
0 < x < 6.0 cm______________
6.0 cm < x________________
(a) At what point(s) along the x axis is the electric field zero? (If there is no point where E = 0 in a region, enter "0" in that box.)
x < 0 ___________
0 < x < 6.0 cm____________
6.0 cm < x____________
(b) At what point(s) along the x axis is the potential zero? Let V = 0 at r = . (If there is no point where V = 0 in a region, enter "0" in that box.)
x < 0 ____________
0 < x < 6.0 cm______________
6.0 cm < x________________
Answers
Answered by
bobpursley
so figure the E from each charge, and add them, set to zero, and solve.
It is obvious that between the charges, E will point to the left, and there is no solution there.
I will set up the problem for the x<0
E=0=-2k/x^2 +6k/(absx+6)^2
2k/x^2=6k/(x+6)^2
take the sqrt of each side.
x=(abs x+6)/sqrt 3
absx *sqrt3-absx=6
solve for x, remembering x is the distance from zero to the left.
It is obvious that between the charges, E will point to the left, and there is no solution there.
I will set up the problem for the x<0
E=0=-2k/x^2 +6k/(absx+6)^2
2k/x^2=6k/(x+6)^2
take the sqrt of each side.
x=(abs x+6)/sqrt 3
absx *sqrt3-absx=6
solve for x, remembering x is the distance from zero to the left.
Answered by
SHUBHANGI
I don't get it..
could u please explain it
could u please explain it
Answered by
bobpursley
so figure the E from each charge, and add them, set to zero, and solve.
I am not going to do all that algebra for you. I will be happy to critique or check your work.
I am not going to do all that algebra for you. I will be happy to critique or check your work.
Answered by
You couldn't be less helpful
You couldn't be less helpful
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