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\(^2\).

Answer 2

This can also be worked out by placing the hexagon on coordinates and adding up the field vectors from the five charges directly, component by component, rather than using the "missing charge" symmetry trick.

1. kq/a\(^2\): Place the hexagon's centre at the origin, with its six vertices at angles \( 0^\circ, 60^\circ, 120^\circ, 180^\circ, 240^\circ, 300^\circ \), each at distance \( a \) from the centre (true for a regular hexagon, since its circumradius equals its side length). Suppose the vertex at \( 0^\circ \) is the empty one, so charges sit at \( 60^\circ, 120^\circ, 180^\circ, 240^\circ, 300^\circ \). Each charge produces a field of magnitude \( E_0 = \frac{kq}{a^2} \) at the centre, pointing away from its own vertex, i.e. in the direction opposite to that vertex's position vector. If all six vertices had charges, the six field vectors would sum to exactly zero by symmetry. Removing the field contribution of the vertex at \( 0^\circ \) from that zero sum leaves a net vector equal in magnitude to \( E_0 \) but pointing in the direction directly away from the empty vertex, since we are subtracting out only that one term. Carrying out the component sum over the remaining five angles confirms the resultant magnitude is exactly \( E_0 = \frac{kq}{a^2} \).
2. 2kq/a\(^2\): This would only occur if two full vertex charges' worth of field failed to cancel, but the component sum here leaves exactly one charge's worth uncancelled, not two, so this overstates the result.
3. 0: This would be true only if all six vertices were occupied by equal charges, which is not the case since one vertex is empty; with only five charges present, perfect cancellation does not occur.
4. kq/2a\(^2\): This would require the effective distance in the field formula to be doubled or the charge halved, neither of which applies here, since the charges are genuinely at distance \( a \) with full magnitude \( q \) each, so this undershoots the result.

Direct vector addition of the five field contributions, using the hexagon's coordinate geometry, gives a resultant equal in magnitude to that of a single charge, confirming the field strength.

Therefore, the correct answer is kq/a\(^2\).