Question

Five equal point charges, each +q, are placed at five of the six vertices of a regular hexagon of side length a. What is the magnitude of the electric field at the center of the hexagon?

• A. kq/a\(^2\)
• B. 2kq/a\(^2\)
• C. 0
• D. kq/2a\(^2\)

Hint:

For symmetric arrangements of charges, if one charge is missing, the net field at the center is equal in magnitude and opposite in direction to the field that would have been produced by that missing charge alone. This shortcut significantly simplifies calculations.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to calculate the magnitude of the electric field at the center of a regular hexagon where five out of six vertices have identical point charges.

Step 2: Key Formula or Approach:
1. Electric field due to a point charge: The magnitude of the electric field ($E$) due to a point charge $q$ at a distance $r$ is $E = k \frac{|q|}{r^2}$, where $k = \frac{1}{4\pi\epsilon_0}$.
2. Superposition Principle: The net electric field at a point due to multiple charges is the vector sum of the electric fields due to individual charges.
3. Symmetry for a Hexagon: For a regular hexagon, the distance from the center to any vertex is equal to the side length $a$. If identical charges were placed at all six vertices, the electric field at the center would be zero due to perfect symmetry.

Step 3: Detailed Explanation:
Let the vertices of the regular hexagon be $V_1, V_2, V_3, V_4, V_5, V_6$.
Let the charge at each vertex be $+q$.
The distance from the center $O$ to any vertex $V_i$ is $a$.
The magnitude of the electric field at the center due to a single charge $+q$ at any vertex is $E_0 = k \frac{q}{a^2}$.
The direction of this field is radially outward from the charge.
Hypothetical Scenario (all 6 charges present):
If all six vertices ($V_1$ to $V_6$) had a charge of $+q$, then by symmetry, the net electric field at the center of the hexagon would be zero:
\[ \vec{E}_{total, 6q} = \vec{E}_{V1} + \vec{E}_{V2} + \vec{E}_{V3} + \vec{E}_{V4} + \vec{E}_{V5} + \vec{E}_{V6} = \vec{0} \]
Actual Scenario (5 charges present):
In the given problem, five charges are present, and one vertex (let's assume $V_6$) is empty. The electric field due to the five charges ($\vec{E}_{5q}$) is:
\[ \vec{E}_{5q} = \vec{E}_{V1} + \vec{E}_{V2} + \vec{E}_{V3} + \vec{E}_{V4} + \vec{E}_{V5} \]
From the hypothetical scenario, we can write:
\[ \vec{E}_{5q} = \vec{0} - \vec{E}_{V6} \]
This means that the electric field due to the five charges is equal in magnitude and opposite in direction to the electric field that would have been produced by a single charge at the empty vertex ($V_6$).
The electric field vector $\vec{E}_{V6}$ would point radially outwards from $V_6$ (i.e., from the center $O$ towards $V_6$). Its magnitude is $E_0 = k \frac{q}{a^2}$.
Therefore, the net electric field due to the five charges, $\vec{E}_{5q}$, will have a magnitude of $k \frac{q}{a^2}$ and will be directed from $V_6$ towards the center $O$.

Step 4: Final Answer:
The magnitude of the electric field at the center of the hexagon is kq/a\(^2\).