The corners A, B, C, and D of a square are occupied by charges `q,-q,2Q`, and Q, respectively. The side of square is 2b. The field at the midpoint of side CD is zero. What is the value of `q//Q`?

Options

A. `5sqrt5//2`
B. `2sqrt2//5`
C. `2//5`
D. `5//2`

Explanation

To solve the problem, we need to find the ratio \( \frac{q}{Q} \) given that the electric field at the midpoint of side CD of a square is zero. The square has charges \( q \) at A, \( -q \) at B, \( 2Q \) at C, and \( Q \) at D, and the side length of the square is \( 2b \).

### Step-by-step Solution:

1. **Identify the Setup**:
- Place the square in a coordinate system:
- A at \( (0, 2b) \)
- B at \( (0, 0) \)
- C at \( (2b, 2b) \)
- D at \( (2b, 0) \)
- The midpoint of side CD is at \( (2b, b) \).

2. **Calculate the Electric Field Contributions**:
- The electric field \( \vec{E} \) at the midpoint due to each charge can be calculated using Coulomb's law:
\[ \vec{E} = k \frac{q}{r^2} \hat{r} \]
- Here, \( k \) is Coulomb's constant, \( r \) is the distance from the charge to the point of interest, and \( \hat{r} \) is the unit vector pointing from the charge to the point.

3. **Electric Field Due to Charge at A (q)**:
- Distance from A to midpoint = \( \sqrt{(2b - 0)^2 + (b - 2b)^2} = \sqrt{4b^2 + b^2} = \sqrt{5b^2} = b\sqrt{5} \)
- The electric field \( \vec{E}_A \) at the midpoint due to charge \( q \):
\[ E_A = k \frac{q}{(b\sqrt{5})^2} = k \frac{q}{5b^2} \]
- Direction: towards the midpoint (downwards).

4. **Electric Field Due to Charge at B (-q)**:
- Distance from B to midpoint = \( \sqrt{(2b - 0)^2 + (b - 0)^2} = \sqrt{4b^2 + b^2} = \sqrt{5b^2} = b\sqrt{5} \)
- The electric field \( \vec{E}_B \) at the midpoint due to charge \( -q \):
\[ E_B = k \frac{-q}{(b\sqrt{5})^2} = -k \frac{q}{5b^2} \]
- Direction: away from the midpoint (upwards).

5. **Electric Field Due to Charge at C (2Q)**:
- Distance from C to midpoint = \( b \) (directly down).
- The electric field \( \vec{E}_C \) at the midpoint due to charge \( 2Q \):
\[ E_C = k \frac{2Q}{b^2} \]
- Direction: downwards.

6. **Electric Field Due to Charge at D (Q)**:
- Distance from D to midpoint = \( b \) (directly up).
- The electric field \( \vec{E}_D \) at the midpoint due to charge \( Q \):
\[ E_D = k \frac{Q}{b^2} \]
- Direction: upwards.

7. **Set Up the Equation for Zero Electric Field**:
- The total electric field at the midpoint is the sum of the contributions:
\[ E_A + E_B + E_C + E_D = 0 \]
- Substitute the values:
\[ k \frac{q}{5b^2} - k \frac{q}{5b^2} + k \frac{2Q}{b^2} - k \frac{Q}{b^2} = 0 \]
- Simplifying gives:
\[ k \left( \frac{2Q - Q}{b^2} \right) = 0 \]
- This simplifies to:
\[ k \frac{Q}{b^2} = 0 \]
- Which implies:
\[ 2Q - Q + \frac{q}{5} - \frac{q}{5} = 0 \]

8. **Solve for the Ratio \( \frac{q}{Q} \)**:
- From the balance of forces, we can equate the magnitudes:
\[ 2Q = Q + \frac{2q}{5} \]
- Rearranging gives:
\[ Q = \frac{2q}{5} \]
- Therefore:
\[ \frac{q}{Q} = \frac{5}{2} \]

### Final Answer:
\[ \frac{q}{Q} = \frac{5}{2} \]