The charges each of +Q and -Q coulomb are placed at corners A and B of an equilateral triangle ABC of side 'a'cm. 'D' is the mid point of AB. The work done if a charge of 'q' is moved from D to C is:

Options

A. zero
B. `(1)/(4 pi epis_0 ) xx qQ`
C. `(Q )/( 4 pi epis_0) xx ( sqrt(3))/(2) q `
D. `(Q) /( 4 pi epsi_0 ) xx sqrt(3) q`

Answer

To solve the problem, we need to calculate the work done in moving a charge \( q \) from point \( D \) to point \( C \) in the presence of two charges \( +Q \) at point \( A \) and \( -Q \) at point \( B \) of an equilateral triangle \( ABC \).

### Step-by-step Solution:

1. **Understanding the Setup**:
- We have an equilateral triangle \( ABC \) with each side of length \( a \) cm.
- Charge \( +Q \) is at point \( A \) and charge \( -Q \) is at point \( B \).
- Point \( D \) is the midpoint of \( AB \).

2. **Finding the Potential at Point C**:
- The electric potential \( V_C \) at point \( C \) due to charge \( +Q \) at point \( A \) is given by:
\[ V_C(A) = \frac{1}{4\pi\epsilon_0} \frac{Q}{a} \]
- The electric potential \( V_C \) at point \( C \) due to charge \( -Q \) at point \( B \) is given by:
\[ V_C(B) = \frac{1}{4\pi\epsilon_0} \frac{-Q}{a} \]
- Therefore, the total potential at point \( C \) is:
\[ V_C = V_C(A) + V_C(B) = \frac{1}{4\pi\epsilon_0} \frac{Q}{a} + \frac{1}{4\pi\epsilon_0} \frac{-Q}{a} = 0 \]

3. **Finding the Potential at Point D**:
- The distance from \( D \) to \( A \) and from \( D \) to \( B \) is \( \frac{a}{2} \).
- The electric potential \( V_D \) at point \( D \) due to charge \( +Q \) at point \( A \) is:
\[ V_D(A) = \frac{1}{4\pi\epsilon_0} \frac{Q}{\frac{a}{2}} = \frac{2Q}{4\pi\epsilon_0 a} \]
- The electric potential \( V_D \) at point \( D \) due to charge \( -Q \) at point \( B \) is:
\[ V_D(B) = \frac{1}{4\pi\epsilon_0} \frac{-Q}{\frac{a}{2}} = \frac{-2Q}{4\pi\epsilon_0 a} \]
- Therefore, the total potential at point \( D \) is:
\[ V_D = V_D(A) + V_D(B) = \frac{2Q}{4\pi\epsilon_0 a} + \frac{-2Q}{4\pi\epsilon_0 a} = 0 \]

4. **Calculating the Work Done**:
- The work done \( W \) in moving a charge \( q \) from point \( D \) to point \( C \) is given by:
\[ W = q(V_C - V_D) \]
- Since we found that both \( V_C \) and \( V_D \) are zero:
\[ W = q(0 - 0) = 0 \]

### Final Answer:
The work done in moving the charge \( q \) from point \( D \) to point \( C \) is **0 Joules**.