Two charges each equal to `eta q(eta^(-1) lt sqrt(3))` are placed at the corners of an equilateral triangle of side `a`. The electric field at the third corner is `E_(3)` where`(E_(0)= q//4piepsilon_(0)a^(2))`

Options

A. `E_(3)=E_(0)`
B. `E_(3)lt E_(0)`
C. `E_(3)gt E_(0)`
D. `E_(3)ge E_(0)`

Answer

To solve the problem step by step, we will analyze the electric fields produced by the two charges at the corners of an equilateral triangle and find the resultant electric field at the third corner.

### Step 1: Understand the Setup
We have two charges, each equal to \( \eta q \), placed at two corners of an equilateral triangle with side length \( a \). We need to find the electric field at the third corner due to these charges.

### Step 2: Identify the Electric Fields
The electric field \( E_B \) at corner A due to charge at B is directed away from charge B, and the electric field \( E_C \) at corner A due to charge at C is directed away from charge C. Both electric fields have the same magnitude because the charges are equal.

### Step 3: Calculate the Magnitude of the Electric Fields
The magnitude of the electric field due to a point charge is given by:
\[ E = \frac{k \cdot |q|}{r^2} \]
where \( k = \frac{1}{4 \pi \epsilon_0} \) and \( r \) is the distance from the charge to the point where the field is being calculated.

For both charges:
\[ E_B = E_C = \frac{1}{4 \pi \epsilon_0} \cdot \frac{\eta q}{a^2} = \eta E_0 \]
where \( E_0 = \frac{q}{4 \pi \epsilon_0 a^2} \).

### Step 4: Determine the Angle Between the Electric Fields
Since the triangle is equilateral, the angle between \( E_B \) and \( E_C \) is \( 60^\circ \).

### Step 5: Calculate the Resultant Electric Field
The resultant electric field \( E_3 \) at corner A can be found using the formula for the resultant of two vectors:
\[ E_3 = \sqrt{E_B^2 + E_C^2 + 2 E_B E_C \cos(60^\circ)} \]
Substituting \( E_B = E_C = \eta E_0 \):
\[ E_3 = \sqrt{(\eta E_0)^2 + (\eta E_0)^2 + 2 (\eta E_0)(\eta E_0) \cdot \frac{1}{2}} \]
\[ = \sqrt{2 (\eta E_0)^2 + (\eta E_0)^2} = \sqrt{3 (\eta E_0)^2} = \sqrt{3} \cdot \eta E_0 \]

### Step 6: Relate \( E_3 \) to \( E_0 \)
From the expression derived:
\[ E_3 = \sqrt{3} \cdot \eta E_0 \]

### Step 7: Analyze the Given Condition
We are given that \( \eta^{-1} < \sqrt{3} \). Taking the reciprocal gives:
\[ \eta > \frac{1}{\sqrt{3}} \]

### Step 8: Substitute \( \eta \) in the Resultant Electric Field
Substituting \( \eta \) into the equation for \( E_3 \):
\[ E_3 > \sqrt{3} \cdot \frac{1}{\sqrt{3}} E_0 = E_0 \]

### Conclusion
Thus, we conclude that:
\[ E_3 > E_0 \]

### Final Answer
The correct option is \( E_3 \) is greater than \( E_0 \).

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