Electrostatics home work question

[logearav]

Homework Statement

Two Charges +q and -3q are separated by a distance of 1 m. At what point in between the charges on its axis is the potential is zero.

Homework Equations

I formulated the diagram as given in the attachment

The Attempt at a Solution

At equilibrium V₁ = V₂
q/4*pi*epsilon*x = -3q/4*pi*epsilon*(r-x)
(r-x) = -3x
x = -r/2 -1/2 = -0.5m. I took r = 1m because distance of separation between the charges is 1 m. Am i correct? How to write the answer because i get negative distance. Could you please help where is the location of point where the potential is zero.

 

[MisterX]

You should be finding the value of x where the absolute potentials from the +q and the -3q add to zero, not where these potentials are equal.

 

[logearav]

mr. misterX, so i should use the equation v1+ v2 = 0
am i right, sir?
which gives me 0.5 m as the answer

 

[MisterX]

That's the right equation to use, but 0.5 m is the wrong answer.

 

[logearav]

oh sorry! v1= -v2 so i get 0.25 m as the answer. I hope i am right. Then to how to write the answer. Can i write, the potential is zero at a distance .25m from the right of +q.
Some more doubts.
1) can i put -3q on left and +q on the right and solve the above problem. But i don't get 0.25 m
2) Why we use the equation V1 + V2 = 0 at the point O which is a distance x from A?
Thanks in advance sir

 

[gneill]

oh sorry! v1= -v2 so i get 0.25 m as the answer. I hope i am right. Then to how to write the answer. Can i write, the potential is zero at a distance .25m from the right of +q.
Some more doubts.
1) can i put -3q on left and +q on the right and solve the above problem. But i don't get 0.25 m
2) Why we use the equation V1 + V2 = 0 at the point O which is a distance x from A?
Thanks in advance sir

1) You have to take into account the relative position on of your "test point" with respect to any given charge in your coordinate system.

Imagine that a point charge is surrounded by field arrows all pointing either radially outward (positive charge) or radially inward (negative charge). The field points IN OPPOSITE DIRECTIONS on opposite sides of the charge. This geometry dependence will necessitate a change of sign in your equation for the potential depending upon your test point location.

It's always best to make a sketch of any given setup of charges, noting the directions of the field contributions of each at places where you want to determine the net field. This will guide you in writing the correct signs for the terms your equations.

2) Electric fields obey the superposition principle. That is, the fields from individual charges can be calculated separately and then summed at a given point to find the net potential at that point.