lets put the first one at 0 and the second one at 1
the answer HAS TO BE left of zero ! (why ? :)
( right one pushes left, left one pulls right :)
only ratio of charges matter, call it one and three coulombs
right pull by left one = left push by right one
magnitude of right pull = 2/(0-x)^2
magnitude of left push = 6/(1+x)^2
so
2/x^2 = 6/(x^2 + 2 x + 1)
2 x^2 + 4 x + 2 = 6 x^2
4 x^2 - 4 x - 2 = 0
2 x^2 - 2 x - 1 = 0
x = -.366 or+ 1.366
I think it is on the left of zero so -.366
NOW move it all 4 units right because we did it for x = 0 and one instead of 4 and 5
4 - .366 = 3.634
of course they could have intended x = -4 and +5, it is hard to say
Two charges of magnitudes Q1 = -2μc and Q2 = 6μc lie along the x-axis at a distance 4m and 5m from the origin respectively. At what point from the origin does the net electrostatic field be zero.
1 answer