Five identical point charges +Q are placed at the five corners of a regular hexagon of edge length a. One corner is empty. Electric field intensity at the centre of the hexagon is

Options

A. Zero
B. `Q/(4piepsilon_0d^2)`, towards the empty corner
C. `Q/(4piepsilon_0d^2)`, away from the empty corner
D. `(5Q)/(4piepsilon_0a^2)` towards the empty corner

Answer

To find the electric field intensity at the center of a regular hexagon with five identical point charges +Q placed at its corners (with one corner empty), we can follow these steps:

### Step 1: Understand the Configuration
We have a regular hexagon with five charges +Q at five of its corners. The sixth corner is empty. The electric field at the center of the hexagon will be influenced by the charges at the corners.

### Step 2: Identify the Electric Field Contribution from Each Charge
The electric field \( E \) due to a point charge \( Q \) at a distance \( r \) is given by the formula:
\[ E = \frac{kQ}{r^2} \]
where \( k \) is Coulomb's constant (\( k = \frac{1}{4\pi\epsilon_0} \)).

### Step 3: Determine the Distance from Charges to the Center
In a regular hexagon, the distance from each corner to the center can be calculated. For a hexagon with edge length \( a \), the distance \( d \) from the center to any vertex is given by:
\[ d = \frac{a}{\sqrt{3}} \]

### Step 4: Calculate the Magnitude of Electric Field from Each Charge
Since all charges are identical and equidistant from the center, the magnitude of the electric field due to each charge at the center is:
\[ E_i = \frac{kQ}{d^2} = \frac{kQ}{\left(\frac{a}{\sqrt{3}}\right)^2} = \frac{3kQ}{a^2} \]

### Step 5: Analyze the Direction of Electric Fields
The electric field vectors due to the charges will point away from the charges since they are positive. We have:
- \( E_1 \) and \( E_4 \) will cancel each other out.
- \( E_2 \) and \( E_5 \) will also cancel each other out.

### Step 6: Consider the Remaining Charge
The only charge that does not have a counterpart to cancel it is \( E_3 \). Therefore, the net electric field at the center will be equal to the electric field due to charge \( E_3 \).

### Step 7: Determine the Direction of the Net Electric Field
The direction of the net electric field will be directed towards the empty corner of the hexagon, as the other charges are pushing away from the center.

### Final Result
The magnitude of the electric field at the center of the hexagon is:
\[ E = \frac{3kQ}{a^2} \]
and the direction is towards the empty corner.