A positive charge kept at one of the vertices of a regular hexagon produces electric intensity E at the centre of the hexagon. If the charge is moved to an adjacent vertex, the magnitude of change in the electric intensity will be

Options

A. zero
B. `sqrt(3)E`
C. `E`
D. `E//sqrt(3)`

Explanation

To solve the problem, we need to analyze the electric field produced by a positive charge at the vertices of a regular hexagon and how it changes when the charge is moved to an adjacent vertex. Here's a step-by-step solution:

### Step 1: Understanding the Setup
We have a regular hexagon with a positive charge \( +q \) placed at one of its vertices. This charge produces an electric field \( E \) at the center of the hexagon.

**Hint:** Visualize the hexagon and label the vertices and the center clearly.

### Step 2: Electric Field at the Center
The electric field \( E \) at the center of the hexagon due to the charge at vertex A (let's say) can be expressed as:
\[ E_1 = \frac{kq}{a^2} \]
where \( k \) is Coulomb's constant and \( a \) is the distance from the charge to the center of the hexagon.

**Hint:** Remember that the electric field due to a point charge is inversely proportional to the square of the distance.

### Step 3: Moving the Charge to an Adjacent Vertex
When the charge is moved to an adjacent vertex (let's say from A to B), it will produce a new electric field \( E_2 \) at the center of the hexagon:
\[ E_2 = \frac{kq}{a^2} \]
The magnitudes of \( E_1 \) and \( E_2 \) are the same since the distance from the charge to the center remains unchanged.

**Hint:** The magnitudes of the electric fields due to the charge at both vertices are equal.

### Step 4: Analyzing the Direction of the Electric Fields
The direction of the electric field \( E_1 \) due to the charge at vertex A will point towards the center, while the electric field \( E_2 \) due to the charge at vertex B will also point towards the center but at an angle of \( 60^\circ \) from \( E_1 \) because of the geometry of the hexagon.

**Hint:** Use symmetry and geometry to determine the angles between the electric field vectors.

### Step 5: Calculating the Change in Electric Field
To find the change in electric field when the charge is moved, we can use the vector addition of the electric fields:
\[ \Delta E = |E_2 - E_1| \]
Using the cosine rule for the angle between the two vectors:
\[ |\Delta E| = \sqrt{E_1^2 + E_2^2 - 2E_1E_2\cos(60^\circ)} \]
Since \( E_1 = E_2 = E \):
\[ |\Delta E| = \sqrt{E^2 + E^2 - 2E^2 \cdot \frac{1}{2}} = \sqrt{2E^2 - E^2} = \sqrt{E^2} = E \]

**Hint:** Remember that \( \cos(60^\circ) = \frac{1}{2} \).

### Step 6: Conclusion
Thus, the magnitude of the change in electric intensity when the charge is moved to an adjacent vertex is:
\[ \Delta E = E \]

**Final Answer:** The change in electric intensity is \( E \).