Charges and coordinate system?

Question

A charge of -2.93 x 10^-9 C is at the origin of a coordinate system and a charge of 8.03 x 10^-9 C is on the x-axis at 2.47 m. There are two locations on the x-axis, where the electric potential is zero.

A) Find the location of the point between the charges. answer in m.

B) Find the location of the point to the left of the y-axis. answer in m.

I would appreciate any help =)

Answer 1

The electric potential from a point charge is q/4πe0r, where r is the distance from the charge. A point between the charges is a distance x from the - charge and a distance (r-x) from the + charge (r is the charge separation). The potentials are opposite in sign, so the zero point occurs when the magnitudes of the potential from each charge are the same.

Let q1 and q2 be the absolute magnitude of the charges.

q1/(4πe0x) = q2/[4πe0(r-x)]

q1/x = q2/(r-x)

q1*(r-x) = q2*x

x(q1 + q2) = q1*r

x = q1*r/(q1+q2)

On the left of the origin, the distances are x and r+x and the solution

q1/x = q2/(r+x)

q1*(r+x) = q2*x

x(q2 - q1) = q1*r

x = q1*r/(q2 - q1)

Note this will only happen if |q2| > |q1|, which is the case.

Answer 2

Think of the x-axis as kind of a long slinky. When you put the negative charge down at the origin,it's as if you're pulling down on the slinky at that point; when you put the positive charge down at 2.47 m, it's like you're pulling up on it at that point (so now you've got this slinky that's being pushed down at one point and pulled up at another; essentially, the question is asking about where those two effects cancel each other out).

How much is it pulled down or up? q/(4*pi*epsilon*distance from the charge).

so the pulled down is:
-2.93 x 10^-9 /(4*pi*epsilon*absolute value(x))

the pull up is
8.03 x 10^-9/ (4 * pi* epsilon*absolute value (2.47-x))

add the two together and solve for x.