kq₁/x²=kq₂/(6-x)²
2/x²=6/(6-x)²
36-12x+x²=3x²
2x²+12x-36=0
x²+6x-18=0
x = - 3±sqrt(9+18) =
= - 3 ± 5.2
If q₁ >0 and q₂>0 x= 2.2 cm (to the right from the left charge (between the charges))
If q₁ <0 and q₂>0 x=-8.2 cm (to the left from the left (negative) charge)
Two particles are fixed along a straight line, separated by a distance 6.0 cm. The left one has a charge +2.0 nC and the right one has a charge of +6.0 nC. You must try to locate a point where the electric field is zero.
(a) What is the location of the point?
(b) How far from the left charge is the point?
(c) Suppose that the left charge is -2 nC. Now where would you find a point where the electric field is zero??
(d) How far is the point measured from the left charge?
1 answer
1 answer