**Problem statement:**

Two point charges q_{1} = q_{2} = 10^{-6} C are respectively located at the points of coordinates (-1, 0) y (1, 0) (the coordinates are expressed in meters). Calculate:

- The electric field due to the charges at a point P of coordinates (0, 1).
- The force that a charge q
_{0}= – 2 10^{-9}C situated at the point P would experience. - The value of a point charge q
_{3}situated at the origin of the cartesian coordinate system in order for the electric field to be zero at point P.

__Givens__: k = 9 10^{9} N m^{2}/C^{2}

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**Solution:**

In this problem we are going to see step by step how to calculate the electric field due to multiple point charges at any point P.

You can see how to calculate step by step the electrostatic potential due to the point charges q_{1} and q_{2} in this page.

First, we will represent the charges and point *P* in a cartesian coordinate system.

The unit vectors we are going to use to calculate the electric fields **E**_{1} and **E**_{2} are represented in red in the figure.

To find the direction of the vector **E**_{1} we perform a “thought experiment” that consists in placing a **positive** test charge at point *P* and to identify the direction of the force it would experience in presence of charge q_{1}. The test charge would be repelled as q_{1} is positive, therefore **E**_{1} originates from q_{1}. Recall that positive charges are sources of electric field lines. We find the direction of the vector **E**_{2} by performing the same “thought experiment” for q_{2}.

The fields **E**_{1} and **E**_{2} are respectively given by:

Where r is the distance between each charge and point P. We apply Pythagoras’ theorem to calculate r_{1} and r_{2}.

To calculate the unit vector **u**_{r1} we will divide the vector **A** that goes from the location of q_{1} to the point P by its magnitude. This **unit vector always goes from the point charge that creates the field to the point where we want to calculate the field**:

And the vector **A** is calculated by subtracting the coordinates of the location of q_{1} from the coordinates of point P. Using the unit vector notation, we get:

We repeat the same procedure to calculate **u**_{r2}:

We calculate the vector **B** that goes from the location of q_{2} to the point P and we divide it by its magnitude:

We can now substitute the unit vectors, the distance between the charges and point P in the expression of the electric field to get:

And the net field in point P is the sum of the two previous vectors:

As you can see in the expression, the net field only has a vertical component. We can verify it graphically by adding the vectors **E**_{1} and **E**_{2} together with the help of the parallelogram law as seen in the next figure.

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If a charge q_{0} is placed at point P, it will experience an electrostatic force given by:

This force is represented in the next figure:

We will now calculate the value that a point charge located in the origin of the cartesian coordinate system should have to zero the field at point P.

We have represented the charge q_{3} located at the origin of the cartesian coordinate system and the electric field **E**_{3} it has to create in point P to zero the field at this point.

We can deduce from the previous figure that the charge q_{3} has to be negative, because the field **E**_{3} has to be oriented toward the point charge (recall that negative charges are sinks of field lines).

Furthermore, for the net field to be zero in point P, the vectors **E** and **E**_{3} must have the same magnitude, therefore the following must be fulfilled:

And after isolating the absolute value of q_{3} we have:

Therefore q_{3} should be:

Check the units of measurement page to know more about the prefixes used in Physics to express the multiples and submultiples of the SI units.

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