Six equal positive charges +Q sit at the vertices of a regular hexagon with sides of length R (note that R is also the distance from each charge to the center of the hexagon). The electric field at the center of the hexagon (point P) is given by (assume that e,points upward): R R O +Q | +Q 1 2Q а. Е Ey 4πεο R b. E = 1 2Q Ey - Απεο R 1 Q С. Е Ey 4πεο R2 1 d. E : Ey - Απεο R e. E = 0
Six equal positive charges +Q sit at the vertices of a regular hexagon with sides of length R (note that R is also the distance from each charge to the center of the hexagon). The electric field at the center of the hexagon (point P) is given by (assume that e,points upward): R R O +Q | +Q 1 2Q а. Е Ey 4πεο R b. E = 1 2Q Ey - Απεο R 1 Q С. Е Ey 4πεο R2 1 d. E : Ey - Απεο R e. E = 0
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Transcribed Image Text:**Title: Understanding Electric Fields in a Regular Hexagon Configuration**
**Introduction:**
Six equal positive charges (+Q) are placed at the vertices of a regular hexagon with sides of length R. The point of interest is the center of the hexagon (point P), which is also at a distance R from each charge. Our goal is to determine the electric field at point P, assuming the vertical unit vector \(e_y\) points upward.
**Diagram Explanation:**
The diagram visually represents the hexagon with six charges marked as +Q. Each charge is evenly spaced at the vertices. Dashed lines indicate the symmetry in distance (R) from point P to each charge.
**Electric Field Calculation:**
The problem asks for the electric field at the center of the hexagon due to the arrangement of the charges. The electric field is given in terms of \(\frac{1}{{4\pi\epsilon_0}}\), where \(\epsilon_0\) is the permittivity of free space. The possible expressions for the electric field \(E\) at point P are:
a. \(E = \frac{1}{{4\pi\epsilon_0}} \frac{2Q}{{R^2}} e_y\)
b. \(E = -\frac{1}{{4\pi\epsilon_0}} \frac{2Q}{{R^2}} e_y\)
c. \(E = \frac{1}{{4\pi\epsilon_0}} \frac{Q}{{R^2}} e_y\)
d. \(E = -\frac{1}{{4\pi\epsilon_0}} \frac{Q}{{R^2}} e_y\)
e. \(E = 0\)
The answer to the problem is choice **c**, \(E = \frac{1}{{4\pi\epsilon_0}} \frac{Q}{{R^2}} e_y\), indicating that the electric field at the center is directed upward with a certain magnitude.
**Conclusion:**
In this symmetrical configuration of equal charges in a hexagonal arrangement, the resultant electric field at the center is non-zero and directed along \(e_y\), demonstrating the principles of superposition and symmetry in electrostatics.
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