Two point charges 5uc and 3uc are fixed 4cm apart . calculate the distance in between them at which the resultant is zero
2 answers
show calculation
5/x^2 = 3/y^2
x+y= 4 so y = (4-x)
5/x^2 = 3 / (4-x)^2
5(16 - 8 x + x^2) = 3 x^2
80 - 40 x + 5 x^2 = 3 x^2
2 x^2 - 40 x + 80 = 0
x^2 - 20 x + 40 = 0
realistic root at 2.254
check
x = 2.254
y = 4-2.254 = 1.746
5/x^2 = .984
3/y^2 = .984
yeah, that works
x+y= 4 so y = (4-x)
5/x^2 = 3 / (4-x)^2
5(16 - 8 x + x^2) = 3 x^2
80 - 40 x + 5 x^2 = 3 x^2
2 x^2 - 40 x + 80 = 0
x^2 - 20 x + 40 = 0
realistic root at 2.254
check
x = 2.254
y = 4-2.254 = 1.746
5/x^2 = .984
3/y^2 = .984
yeah, that works