Charges of −q and +2q are fixed in place, with a distance of 5.49 m between them. A dashed line is drawn throu?
Charges of −q and +2q are fixed in place, with a distance of 5.49 m between them. A dashed line is drawn through the negative charge, perpendicular to the line between the charges. On the dashed line, at a distance L from the negative charge, there is at least one spot where the total potential is zero. Find L.
they give the hint that there are 2 points where ther is zero potential. one above the -q charge and one that is below it. but no matter how much i try i cant figure out what L is. please help
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The potentials of both charges at that point must be:
V1 + V2 = 0
Draw a triangle with vertices: -q (point A), 2q (point B), and a point on the dashed line (I will call it P).
AP = r........................................what we need to find
BP = sqrt(r^2 + 5.49^2) .............from Pythagorean theorem
Now use the formula: V = q / (4 pi eps R).....(I wrote eps for the epsilon sub 0).
So my first equation turns into:
- q / (4 pi eps r) + 2q / (4 pi eps sqrt(r^2 + 5.49^2)) = 0
Take the first term to the right and multiply both sides by (4 pi eps):
2q / sqrt(r^2 + 5.49^2) = q / r
Divide by q and then cross multiply (as a proportion):
2r = sqrt(r^2 + 5.49^2)) ........square both sides
4r^2 = r^2 + 5.49^2
4r^2 - r^2 = 5.49^2
3r^2 = 5.49^2...................... divide by 3
r^2 = 5.49^2 / 3................... sqrt
r = +/ - 3.1697 ................... that is rounded, and it's the answer!
Now it makes sence: the two points are on the number line (the dashed line described) pointing up (if you choose so), so the positive answer is the point on one side of the charge (above), and the negative on the other.