When is Electric Potential equal to 0?

[dwk17]

Homework Statement

A 3.03μC and a -1.93μC charge are placed 4.14 cm apart. At what point along the line joining them is the electric field zero? Assume that the first charge is at the origin and the second charge is at +4.14 cm. what points along the line joining them is the potential zero? Let V = 0 at r = infinity.

Homework Equations

E=kq/r^2 and V=kq/r

The Attempt at a Solution

I already solved for the part about the electric field, the answer was r=2.05cm. For the second part of the question I tried...
Vtotal=V1+V2
0=kq1/r1+kq2/r2
-q2/r2= q1/r1
but I am not sure what the two distances are supposed to be. Also how would this give me two different answers?

 

[BruceW]

Hi man, welcome to physicsforums.

The problem does not make sense to me... There is supposed to be a positive charge and a negative charge, and we are told to find where between them the electric field is zero? Try drawing some field lines, does it make sense that the electric field could be zero between them? Are you sure there was meant to be one negative and one positive? Or are they meant to be both the same charge?

Edit: or hi woman, I can't tell from the name

 

[dwk17]

Hi man, welcome to physicsforums.

The problem does not make sense to me... There is supposed to be a positive charge and a negative charge, and we are told to find where between them the electric field is zero? Try drawing some field lines, does it make sense that the electric field could be zero between them? Are you sure there was meant to be one negative and one positive? Or are they meant to be both the same charge?

Edit: or hi woman, I can't tell from the name

Yes that is my mistake, I meant to write down that I solved for the electric field to be 20.5cm not 2.05cm **** sorry, that makes a lot more sense

 

[BruceW]

Haha, actually I was also confused. I was thinking that the problem wanted us to find a point between the two charges where the electric field is zero. But of course, the question says along the line joining the two charges, not along the line segment (which is what I interpreted it to be, for some reason).

Anyway, I agree with your answer of 20.5cm. Now, for the part of the problem calculating potential, the distances mean the same as they did for the electric field. They are just distance from the point you are calculating the potential at, to the point charge which is contributing to the potential.

 

[Nickie]

The electric potential is equal to 0 when the distance between the two charges is equal to the sum of their individual distances from the point in question. In this case, the two charges are placed 4.14 cm apart, so the point where the potential is 0 would be at a distance of 2.07 cm from each charge (since 2.07 cm + 2.07 cm = 4.14 cm). This means that there are two points along the line joining the two charges where the potential is 0, one at 2.07 cm from the first charge and one at 2.07 cm from the second charge.

To find the potential at these points, you can use the equation V = kq/r. Plugging in the values, we get V1 = (9 x 10^9 Nm^2/C^2)(3.03 x 10^-6 C)/(2.07 x 10^-2 m) = 4.14 V and V2 = (9 x 10^9 Nm^2/C^2)(-1.93 x 10^-6 C)/(2.07 x 10^-2 m) = -3.03 V. Therefore, the potential at the point 2.07 cm from the first charge is 4.14 V and the potential at the point 2.07 cm from the second charge is -3.03 V. This means that the total potential at these points is 4.14 V - 3.03 V = 1.11 V.

In summary, the potential is 0 at two points along the line joining the two charges, one at a distance of 2.07 cm from the first charge and one at a distance of 2.07 cm from the second charge. The total potential at these points is 1.11 V.