# Field and electric potential at the center of a rectangle

Problem Statement:

Four point charges are located at the vertices (corners) of a rectangle of width a = 4 m and height b = 2 m. The center of the Cartesian coordinate system is located at the center of the rectangle. Calculate:

1. the electric field at the center of the rectangle (A).
2. the electric potential at the center of the rectangle (A) and at point (B), the middle point of the rectangle base.
3. the work done by the electric force to move a charge q0 from point B to infinity. Givens:|q| = 1 nC; q0 = -2 μC; k = 9 109 Nm2/C2

### Ad blocker detected

Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!

Solution:

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

First we will draw the electric field due to each one of the charges at the center of the rectangle. To find the direction of the field vector due to a charge at any point, we perform a “thought experiment” that consists in placing a positive test charge at this point. The direction of the electric field is the same as the direction of the electric force exerted on the positive test charge.

The positive charges are sources of field lines (the field lines are directed away from the positive charges) and the negative charges are sinks of field lines (the field lines are directed toward the negative charges).

The electric field due to each charge at point A is represented in the next figure. The unit vectors we will use to calculate the fields are represented in red and we have also included the distances r between each charge and point A. The electric fields E2 and E3 have the same magnitude, direction and orientation. They are therefore represented next to each other, in green and blue respectively. The same happens for fields E1 and E4.

We are now going to see how to calculate the four electric fields.

The field due to each charge is given by: Where r is the distance between each charge and point A. We will use Pythagoras’ theorem to find r1, r2, r3 and r4. The unit vectors ur are directed away from the charge creating the field and toward the point where the field is calculated.

The coordinates of the charges creating the field are represented in the next figure. The unit vector ur1 is calculated by dividing the vector A that goes from the location of q1 to the point A by its magnitude: We repeat the process to calculate the unit vectors associated with the three other charges: We can now substitute these data in the expression of the electric field due to each one of the charges to get: The net field at point A is the sum of these four vectors: The net electric field at point A is a vector directed toward the negative direction of the y axis. We can verify it graphically by adding the vectors together with the help of the parallelogram law. The resulting fields from charges 1 and 2 (in green) and from charges 3 and 4 (in blue) are represented separately to improve the visibility. The net field is the sum of the light green and light blue vectors. ### Ad blocker detected

Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!

The electric potential due to the four charges at point A is given by: This potential is zero, because r has the same value for the all the charges and two of them are positive while the other two are negative: As you can see, just because the electric potential is zero at a particular point, it does not necessarily mean that the electric field is zero at that point.

The distances between the charges and point B are represented in the next figure. Its value is calculated using Pythagoras’ theorem: And the potential at point B is: Finally, after substituting the numerical values we obtain: The work done by the electric force to move q0 from point B to infinity is equal to the value of the charge multiplied by the potential difference between the two points. The zero of the potential is chosen at infinity r = ∞, therefore we have: And after substituting we obtain: You can 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.

The post Field and electric potential at the center of a rectangle appeared first on YouPhysics