**Problem statement:**

Two equal negative point charges q_{1} = q_{2} = -q are respectively located at points (0,a) and (0,-a). A positive point charge q_{0} of mass m moves from point B (-b,0) to the origin of the Cartesian coordinate system with an initial velocity **v**_{0}.

- Draw the net electric field
**E**due to the charges q_{1}and q_{2}at point B and the force vector exerted on q_{0}when it is located at this point. - Calculate the electrostatic potential due to the charges q
_{1}and q_{2}both at B and the origin of the Cartesian coordinate system. - Calculate the velocity of q
_{0}when it reaches the origin of the Cartesian coordinate system. - Calculate the distance traveled by q
_{0}from the origin of the Cartesian coordinate system before it comes to a rest.

__Givens__: k = 1/(4πε_{0}); ε_{0} = 8.854 10^{-12} C^{2}/(N·m^{2})

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**Solution:**

The situation described in the problem statement is represented in the following figure:

First we are going to represent the net electric field due to the charges q_{1} and q_{2} at point B to visualise the force exerted on q_{0}. We will name **E**_{1} the field due to the charge q_{1} and **E**_{2} the field due to the charge q_{2}.

To find the direction of the vector **E**_{1} we perform a “thought experiment” that consists in placing a **positive** test charge at point *B* and to find out the direction of the force it would experience in presence of charge q_{1}. Since q_{1} is negative the test charge would be attracted, and therefore **E**_{1} points towards q_{1}. Recall that negative charges are sinks of electric field lines. We find the direction of the vector **E**_{2} by performing the same “thought experiment” for q_{2}.

The net field **E** is the sum of both vectors and we can determine it graphically with the help of the parallelogram law. The three vectors are represented on the left side of the next figure.

If we place the charge q_{0} at point B, the force exerted on it is given by:

And, as q_{0} is positive, the vector **F** is parallel to **E**, as seen on the right side of the previous figure.

The electric potential at B is the sum of the potentials due to q_{1} and q_{2} at this point:

Where r is the distance from each charge to point B. We use Pythagoras’ theorem to find r_{1} and r_{2}:

And after substituting in the potential expression we obtain:

We use the same method to calculate the electric potential at the origin of the Cartesian coordinate system *O*:

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The electrostatic force is conservative, therefore the net energy of charge q_{0} is the same at both points B and O. If we name K the kinetic energy and v the velocity of the charge at point *O*, the sum of the kinetic energy and the electrostatic potential energy is constant:

After substituting the potential and isolating v^{2} we obtain:

The charge q_{0} will come to a rest at point C of coordinates (c,0) when its velocity becomes zero. The potential at point C is:

And after applying the conservation of energy between point B and C, we get:

Finally, after substituting the potential at both points we obtain:

And we can isolate the value of c from the previous expression.

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