Electric Field

The electric field due to a point charge is given by Coulomb's law: Where:
- is the electric field,
- is Coulomb's constant,
- is the charge, - is the distance from the charge to the point where the electric field is being calculated.
Since the charges are located on a circular arc, the electric field at the center due to each charge will have both a radial and tangential component.

The components from each charge can be added vectorially to find the resultant electric field. Given: - , - , - , - , - The angle between the charges is . Since the field vectors due to the charges and will cancel out in the tangential direction and add up in the radial direction, and the field due to charge contributes directly along the axis, the net electric field at the center will be the resultant of these components.
After vector addition, the magnitude of the net electric field is approximately .

In this question, the electric field at the center of the pentagon due to the charges at each vertex is asked. The geometry of the system is symmetric, and we can use the principle of superposition to find the electric field contributions at point .

1. Case 1: Without charge at Without the charge at , the electric fields at due to the charges at the other four vertices (i.e., at and ) will contribute to the net electric field at . The magnitude of each of these fields is denoted as , and due to symmetry, the horizontal components cancel out, leaving only a net vertical component.

2. Case 2: With charge at replaced with an equal and opposite charge. When the charge at is replaced with an equal and opposite charge, the electric field contributions at from all five charges need to be considered. The field due to the charge at now has the opposite direction to that of the other charges at and . Due to symmetry, the electric field components due to charges and add up, while the field from adds in the opposite direction.

3. Conclusion: The net electric field in the second case will be half of the field in the first case, so the ratio of the magnitudes of the electric fields is . Thus, the ratio is .

The electric field is the negative gradient of the potential. The gradient of the potential function in three dimensions is: We are given the potential: Now, calculate the partial derivatives:
1. 2. 3.
At the point : 1.
2.
3.
Thus, the electric field is:
Therefore, the electric field at point is .