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Two charges +5 nC and -3nC are located at x=0 cm and x=15cm. Why do we find the zero potential at two different points ?

8 years ago

The key is that electric fields from indivudal point charges can be calculated seperately, and added. That''s called the superposition principle.

magnitude of E= q/r^2, but you also have to keep in mind the direction. E points away from the charge if q>0, and towards it if q<0.

So, E1 = q1 / (x^2) if x > 0
E1 = -q1/ (x^2) if x<0

E2 = q2/(x-15)^2 if x>15
E2 = -q2/(x-15)^2 if x<15

where q1 and q2 are signed.

You want to find E = E1+E2 = 0
or E1 = -E2
you''ll have to consider each of 3 regions seperately.
x<0 or x>1
-q2/(x-15)^2 = q1/ (x^2)
or q1*(x-15)^2 = -q2(x^2)

x>0 and x<15
q2/(x-15)^2 = q1/ (x^2)

given the signs of charges, it should be obvious that only the first case is possible. So, expand out (x-15)^2, and then solve using the quadratic equation. There will be two solutions, however only one will be >15 or < 0, which was a condition for solving the first set of equations.

you could also just take the square root of both sides, as long as you include a +/- sign, and rearrange terms. You end up with:
x = sqrt(q1)/(sqrt(q1) +/- sqrt(q2) ), and choose the one that gives you a value that is either >15 or <0.