Three charges each +q, are placed at the corners of an isosceles triangle ABC of sides BC and AC, 2a D and E are the mid point of BC and CA. The work done in taking a charge Q and D to E is :

Options

A. `(3qQ)/(4pi in_(0)a)`
B. `(3qQ)/(8pi in_(0)a)`
C. `(qQ)/(4pi in_(0)a)`
D. Zero

Explanation

To solve the problem of finding the work done in moving a charge \( Q \) from point \( D \) to point \( E \) in the presence of three charges \( +q \) located at the corners of an isosceles triangle, we can follow these steps:

### Step 1: Understand the Configuration
We have an isosceles triangle \( ABC \) with \( AB = AC = 2a \) and \( BC \) as the base. The points \( D \) and \( E \) are the midpoints of sides \( BC \) and \( AC \) respectively.

### Step 2: Calculate the Potential at Points \( D \) and \( E \)
The electric potential \( V \) at a point due to a point charge is given by the formula:
\[ V = k \frac{q}{r} \]
where \( k \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the distance from the charge to the point where the potential is being calculated.

#### Potential at Point \( D \):
1. The distance from charge \( A \) to point \( D \) is \( AD = \sqrt{a^2 + a^2} = a\sqrt{2} \).
2. The distance from charge \( B \) to point \( D \) is \( BD = a \).
3. The distance from charge \( C \) to point \( D \) is \( CD = a \).

Thus, the potential at \( D \) is:
\[ V_D = k \frac{q}{AD} + k \frac{q}{BD} + k \frac{q}{CD} = k \frac{q}{a\sqrt{2}} + k \frac{q}{a} + k \frac{q}{a} \]
\[ V_D = k \left( \frac{q}{a\sqrt{2}} + \frac{2q}{a} \right) \]

#### Potential at Point \( E \):
1. The distance from charge \( A \) to point \( E \) is \( AE = a \).
2. The distance from charge \( B \) to point \( E \) is \( BE = \sqrt{a^2 + a^2} = a\sqrt{2} \).
3. The distance from charge \( C \) to point \( E \) is \( CE = a \).

Thus, the potential at \( E \) is:
\[ V_E = k \frac{q}{AE} + k \frac{q}{BE} + k \frac{q}{CE} = k \frac{q}{a} + k \frac{q}{a\sqrt{2}} + k \frac{q}{a} \]
\[ V_E = k \left( \frac{2q}{a} + \frac{q}{a\sqrt{2}} \right) \]

### Step 3: Compare the Potentials
From the calculations:
- \( V_D = k \left( \frac{q}{a\sqrt{2}} + \frac{2q}{a} \right) \)
- \( V_E = k \left( \frac{2q}{a} + \frac{q}{a\sqrt{2}} \right) \)

Notice that both expressions for \( V_D \) and \( V_E \) are equal:
\[ V_D = V_E \]

### Step 4: Calculate the Work Done
The work done \( W \) in moving a charge \( Q \) from point \( D \) to point \( E \) is given by:
\[ W = Q (V_E - V_D) \]
Since \( V_E = V_D \):
\[ W = Q (0) = 0 \]

### Final Answer
The work done in taking charge \( Q \) from point \( D \) to point \( E \) is:
\[ \boxed{0} \]