Three charge `-q,+q` and `-q` are kept at the veticles of an equllateral triangle of `10cm` side.The potential at the mid point in between `-q,-q,` if `q=5muC` is

Options

A. `-6.4xx10^(5)V`
B. `-12.8xx10^(4)V`
C. `-6.4xx10^(4)V`
D. `-12.8xx10^(5)V`

Explanation

To find the electric potential at the midpoint between the two charges \(-q\) in an equilateral triangle configuration, we can follow these steps:

### Step 1: Understand the Configuration
We have three charges located at the vertices of an equilateral triangle:
- Charge \(A = -q\)
- Charge \(B = +q\)
- Charge \(C = -q\)

The side length of the triangle is \(L = 10 \, \text{cm} = 0.1 \, \text{m}\). We need to find the potential at the midpoint \(D\) of the line segment joining charges \(B\) and \(C\).

### Step 2: Calculate Distances
The distance from point \(D\) to charge \(B\) and charge \(C\) is:
\[ BD = CD = \frac{L}{2} = \frac{0.1}{2} = 0.05 \, \text{m} \]
The distance from point \(D\) to charge \(A\) can be calculated using the properties of the equilateral triangle:
\[ AD = L \cdot \sin(60^\circ) = 0.1 \cdot \frac{\sqrt{3}}{2} = 0.05\sqrt{3} \, \text{m} \]

### Step 3: Calculate the Electric Potential
The electric potential \(V\) at point \(D\) due to a point charge is given by:
\[ V = k \frac{q}{r} \]
where \(k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\) is Coulomb's constant, \(q\) is the charge, and \(r\) is the distance from the charge to the point where we are calculating the potential.

#### Potential at \(D\) due to each charge:
1. **Due to charge \(A\) (which is \(-q\)):**
\[ V_A = k \frac{-q}{AD} = 9 \times 10^9 \cdot \frac{-5 \times 10^{-6}}{0.05\sqrt{3}} = \frac{-45 \times 10^3}{\sqrt{3}} \, \text{V} \]

2. **Due to charge \(B\) (which is \(+q\)):**
\[ V_B = k \frac{q}{BD} = 9 \times 10^9 \cdot \frac{5 \times 10^{-6}}{0.05} = 9 \times 10^9 \cdot 10^{-4} = 90 \, \text{V} \]

3. **Due to charge \(C\) (which is \(-q\)):**
\[ V_C = k \frac{-q}{CD} = 9 \times 10^9 \cdot \frac{-5 \times 10^{-6}}{0.05} = -90 \, \text{V} \]

### Step 4: Total Potential at Point \(D\)
Now, we sum the potentials due to all three charges:
\[ V_D = V_A + V_B + V_C \]
Substituting the values:
\[ V_D = \left(\frac{-45 \times 10^3}{\sqrt{3}} + 90 - 90\right) \, \text{V} \]
Since \(V_B\) and \(V_C\) cancel each other out:
\[ V_D = \frac{-45 \times 10^3}{\sqrt{3}} \, \text{V} \]

### Step 5: Calculate the Numerical Value
Calculating \(V_D\):
\[ V_D \approx -45 \times 10^3 \cdot 0.577 \approx -26.9 \times 10^3 \, \text{V} \approx -26.9 \, \text{kV} \]

### Final Answer
Thus, the potential at the midpoint \(D\) is approximately:
\[ \boxed{-26.9 \, \text{kV}} \]