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. `5sart5//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 with charges \( q, -q, 2Q, \) and \( Q \) at corners A, B, C, and D respectively is zero. The side of the square is \( 2b \).

### Step-by-Step Solution:

1. **Understanding the Setup:**
- We have a square with corners A, B, C, and D.
- Charges at the corners are:
- A: \( q \)
- B: \( -q \)
- C: \( 2Q \)
- D: \( Q \)
- The side length of the square is \( 2b \).

2. **Identify the Midpoint M:**
- The midpoint M of side CD is located at coordinates \( (b, 0) \) if we place:
- A at \( (0, 2b) \)
- B at \( (0, 0) \)
- C at \( (2b, 2b) \)
- D at \( (2b, 0) \)

3. **Calculate Distances:**
- The distances from M to each charge are:
- Distance from M to A: \( AM = \sqrt{(b - 0)^2 + (0 - 2b)^2} = \sqrt{b^2 + 4b^2} = \sqrt{5b^2} = b\sqrt{5} \)
- Distance from M to B: \( BM = \sqrt{(b - 0)^2 + (0 - 0)^2} = b \)
- Distance from M to C: \( CM = \sqrt{(b - 2b)^2 + (0 - 2b)^2} = \sqrt{(-b)^2 + (-2b)^2} = \sqrt{5b^2} = b\sqrt{5} \)
- Distance from M to D: \( DM = \sqrt{(b - 2b)^2 + (0 - 0)^2} = b \)

4. **Calculate Electric Fields:**
- The electric field \( E \) due to a point charge is given by \( E = k \frac{|Q|}{r^2} \) where \( k \) is Coulomb's constant.
- Electric field contributions at M:
- Due to charge A:
\[ E_A = k \frac{q}{(b\sqrt{5})^2} = \frac{kq}{5b^2} \]
(Directed away from A)
- Due to charge B:
\[ E_B = k \frac{-q}{b^2} = -\frac{kq}{b^2} \]
(Directed towards B)
- Due to charge C:
\[ E_C = k \frac{2Q}{(b\sqrt{5})^2} = \frac{2kQ}{5b^2} \]
(Directed away from C)
- Due to charge D:
\[ E_D = k \frac{Q}{b^2} = \frac{kQ}{b^2} \]
(Directed away from D)

5. **Setting Up the Electric Field Equation:**
- The net electric field at M must be zero:
\[ E_A + E_B + E_C + E_D = 0 \]
- Substituting the values:
\[ \frac{kq}{5b^2} - \frac{kq}{b^2} + \frac{2kQ}{5b^2} + \frac{kQ}{b^2} = 0 \]

6. **Simplifying the Equation:**
- Combine like terms:
\[ \left(\frac{kq}{5b^2} - \frac{5kq}{5b^2}\right) + \left(\frac{2kQ}{5b^2} + \frac{5kQ}{5b^2}\right) = 0 \]
- This simplifies to:
\[ -\frac{4kq}{5b^2} + \frac{7kQ}{5b^2} = 0 \]
- Rearranging gives:
\[ 4q = 7Q \]

7. **Finding the Ratio:**
- Thus, the ratio \( \frac{q}{Q} \) is:
\[ \frac{q}{Q} = \frac{7}{4} \]

### Final Answer:
\[ \frac{q}{Q} = \frac{7}{4} \]