Help figuring out Electric Potential

[Dudemanword]

Homework Statement

Three charges are laid out in a Cartesian Plane:
+q: (0,a)
-q: (0,0)
+q: (0,-a)

There is some point p located on the x-axis in the positive x direction

Show that the points where the Electric potential is 0 is x= +/- a/sqrt(3)

Homework Equations

I believe I am to use
$$\begin{flalign*} V = \frac{1}{4 \pi \epsilon_0} \ \frac{q}{r}\\ \end{flalign*}$$
because they are point charges

The Attempt at a Solution

I created right triangles using "P" as a point to start pythagorizing( is that a word?). Eventually I simplified it to q/(x(sqrt(a^2+x^2))
No idea how to get it to +/- a/sqrt(3)

 

[anathema]

I appreciate your attempt at solving this problem. However, I believe there may be some errors in your approach.

Firstly, the electric potential at a point due to a point charge is given by:

\begin{flalign*}
V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r}
\end{flalign*}

where q is the magnitude of the charge, r is the distance from the point charge, and ε0 is the permittivity of free space.

In this case, we have three point charges, so the total electric potential at point P (located on the x-axis) can be found by adding the individual potentials at that point:

\begin{flalign*}
V_{total} = V_1 + V_2 + V_3 = \frac{1}{4 \pi \epsilon_0} \left(\frac{q}{r_1} + \frac{-q}{r_2} + \frac{q}{r_3}\right)
\end{flalign*}

where r1, r2, and r3 are the distances from each charge to point P.

To find the points where the electric potential is 0, we need to set Vtotal equal to 0 and solve for x. This gives us the following equation:

\begin{flalign*}
0 = \frac{1}{4 \pi \epsilon_0} \left(\frac{q}{\sqrt{x^2 + a^2}} - \frac{q}{x} + \frac{q}{\sqrt{x^2 + a^2}}\right)
\end{flalign*}

Simplifying this equation, we get:

\begin{flalign*}
0 = \frac{1}{4 \pi \epsilon_0} \left(\frac{q}{x} + \frac{2q}{\sqrt{x^2 + a^2}}\right)
\end{flalign*}

Multiplying both sides by 4πε0 and then by x, we get:

\begin{flalign*}
0 = q + 2q\sqrt{\frac{a^2}{x^2}+1}
\end{flalign*}

Solving for x, we get:

\begin{flalign*}
x = \pm \frac{a}{\sqrt

 

[kerimek]

First, let's define some variables:
- q: charge of each point charge
- a: distance between the charges (in the y-direction)
- x: distance between the point p and the origin (in the x-direction)

To find the electric potential at point p, we can use the equation you provided:
V = (1/4πε0) * (q/x)

To find the potential at point p, we need to add the potential contributions from each of the three point charges. Since the potential is a scalar quantity, we can simply add the individual potentials together.

The potential contribution from the first +q charge is:
V1 = (1/4πε0) * (q/(x-a))

The potential contribution from the second -q charge is:
V2 = (1/4πε0) * (-q/x)

The potential contribution from the third +q charge is:
V3 = (1/4πε0) * (q/(x+a))

Now, we can add these three potentials together to get the total potential at point p:
V = V1 + V2 + V3
V = (1/4πε0) * (q/(x-a) - q/x + q/(x+a))

We want to find the value of x where the potential is 0, so we can set V=0 and solve for x:
0 = (1/4πε0) * (q/(x-a) - q/x + q/(x+a))

Multiplying both sides by 4πε0 and simplifying, we get:
0 = q/(x-a) - q/x + q/(x+a)
0 = q(x+a)(x-a) - q(x-a)(x+a) + q(x-a)x
0 = q(x^2+a^2-x^2+a^2) - q(x^2-a^2+x^2-a^2) + q(x^2-ax)
0 = q(2a^2) - q(2a^2) + q(x^2-ax)
0 = q(x^2-ax)
0 = x(x-a)

This gives us two solutions for x:
x = 0 and x = a

However, we know that x cannot be 0 because that would place point p on top of the -q charge. Therefore, the only valid solution for x is x =