Question
Charges of −q and +2q are fixed in place, with a distance of 4.65 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.
Answers
Damon
potential due to -q = -kq/L
potential due to 2Q = +2k q/(L^2+4.65^2)^.5
so if
1/L = 2/(L^2+4.65^2)^.5
2 L = sqrt(L^2 + 4.65^2)
4 L^2 = L^2 + 4.65^2
3 L^2 = 21.6
L = 2.68
potential due to 2Q = +2k q/(L^2+4.65^2)^.5
so if
1/L = 2/(L^2+4.65^2)^.5
2 L = sqrt(L^2 + 4.65^2)
4 L^2 = L^2 + 4.65^2
3 L^2 = 21.6
L = 2.68