Four point charge `q, - q, 2Q` and Q are placed in order at the corners A, B, C and D of a square. If the field at the midpoint of CD is zero then the value of `q//Q` is `(5 sqrt5)/(x)`. Find the value of x.

Answer

To solve the problem, we need to find the ratio \( \frac{q}{Q} \) such that the electric field at the midpoint of line segment CD is zero. We will analyze the contributions of the electric fields from each charge at that point.

### Step 1: Understand the Configuration
We have four charges placed at the corners of a square:
- Charge \( Q \) at corner A
- Charge \( -Q \) at corner B
- Charge \( 2Q \) at corner C
- Charge \( q \) at corner D

Let the side length of the square be \( a \). The midpoint of CD is the point M, which is located at \( (a, a/2) \).

### Step 2: Calculate Electric Fields at Point M
The electric field \( \vec{E} \) due to a point charge \( Q \) at a distance \( r \) is given by:
\[ \vec{E} = k \frac{Q}{r^2} \hat{r} \]
where \( k \) is Coulomb's constant and \( \hat{r} \) is the unit vector pointing from the charge to the point where the field is being calculated.

**Electric Field due to Charge C (2Q):**
- Distance from C to M: \( \frac{a}{2} \)
- Electric field \( \vec{E}_C \) at M:
\[ \vec{E}_C = k \frac{2Q}{(a/2)^2} \hat{j} = \frac{8kQ}{a^2} \hat{j} \]
(directed upwards)

**Electric Field due to Charge D (q):**
- Distance from D to M: \( \frac{a}{2} \)
- Electric field \( \vec{E}_D \) at M:
\[ \vec{E}_D = k \frac{q}{(a/2)^2} (-\hat{j}) = -\frac{4kq}{a^2} \hat{j} \]
(directed downwards)

**Electric Field due to Charge A (Q):**
- Distance from A to M: \( \frac{a}{\sqrt{2}} \)
- Electric field \( \vec{E}_A \) at M:
\[ \vec{E}_A = k \frac{Q}{(a/\sqrt{2})^2} \hat{r} = \frac{2kQ}{a^2} \left( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \right) \]
(directed towards M)

**Electric Field due to Charge B (-Q):**
- Distance from B to M: \( \frac{a}{\sqrt{2}} \)
- Electric field \( \vec{E}_B \) at M:
\[ \vec{E}_B = k \frac{-Q}{(a/\sqrt{2})^2} \hat{r} = -\frac{2kQ}{a^2} \left( \frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} \right) \]
(directed away from M)

### Step 3: Set the Net Electric Field to Zero
For the electric field at point M to be zero, we need to sum the contributions from all four charges and set them equal to zero.

**In the x-direction:**
\[ E_{Ax} + E_{Bx} = 0 \]
\[ \frac{2kQ}{a^2\sqrt{2}} - \frac{2kQ}{a^2\sqrt{2}} = 0 \]

**In the y-direction:**
\[ E_{Cy} + E_{Dy} + E_{Ay} + E_{By} = 0 \]
\[ \frac{8kQ}{a^2} - \frac{4kq}{a^2} + \frac{2kQ}{a^2\sqrt{2}} - \frac{2kQ}{a^2\sqrt{2}} = 0 \]
This simplifies to:
\[ \frac{8Q - 4q}{a^2} = 0 \]
Thus, we find:
\[ 8Q = 4q \implies q = 2Q \]

### Step 4: Find the Ratio \( \frac{q}{Q} \)
Now substituting \( q = 2Q \):
\[ \frac{q}{Q} = \frac{2Q}{Q} = 2 \]

### Step 5: Find the Value of \( x \)
According to the problem statement, we have:
\[ \frac{q}{Q} = \frac{5\sqrt{5}}{x} \]
Setting this equal to our found ratio:
\[ 2 = \frac{5\sqrt{5}}{x} \implies x = \frac{5\sqrt{5}}{2} \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{5\sqrt{5}/2} \]