At the corners A, B, C of a square ABCD, charges `10mC, -20 m C and 10mC` are placed. The electric intensity at the centre of the square to become zero, the charge to be placed at the corner D is

Options

A. `-20mC`
B. `+20mC`
C. `30 mC`
D. `-30 mC`

Answer

To solve the problem, we need to find the charge \( q \) that should be placed at corner D of the square ABCD such that the electric intensity at the center of the square is zero.

### Step-by-step Solution:

1. **Understand the Configuration**:
- We have a square ABCD.
- Charges are placed at corners A, B, and C:
- Charge at A (\( Q_A \)) = \( +10 \, \text{mC} \)
- Charge at B (\( Q_B \)) = \( -20 \, \text{mC} \)
- Charge at C (\( Q_C \)) = \( +10 \, \text{mC} \)
- We need to find the charge at D (\( Q_D \)) such that the electric field at the center of the square is zero.

2. **Electric Field Calculation**:
- The electric field \( E \) due to a point charge is given by:
\[ E = k \frac{|Q|}{r^2} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge to the point where the field is being calculated.

3. **Distance to the Center**:
- The distance from each corner of the square to the center is \( r = \frac{x}{\sqrt{2}} \), where \( x \) is the length of the side of the square.

4. **Electric Field Contributions**:
- The electric field at the center due to charge at A:
\[ E_A = k \frac{10 \times 10^{-3}}{\left(\frac{x}{\sqrt{2}}\right)^2} = k \frac{10 \times 10^{-3} \cdot 2}{x^2} = \frac{20k \times 10^{-3}}{x^2} \]
- The electric field at the center due to charge at B:
\[ E_B = k \frac{20 \times 10^{-3}}{\left(\frac{x}{\sqrt{2}}\right)^2} = k \frac{20 \times 10^{-3} \cdot 2}{x^2} = \frac{40k \times 10^{-3}}{x^2} \]
(Note: This field is directed towards B since B is negatively charged)
- The electric field at the center due to charge at C:
\[ E_C = k \frac{10 \times 10^{-3}}{\left(\frac{x}{\sqrt{2}}\right)^2} = \frac{20k \times 10^{-3}}{x^2} \]

5. **Net Electric Field at the Center**:
- The net electric field at the center due to charges at A, B, and C is:
\[ E_{\text{net}} = E_A + E_C - E_B \]
\[ E_{\text{net}} = \frac{20k \times 10^{-3}}{x^2} + \frac{20k \times 10^{-3}}{x^2} - \frac{40k \times 10^{-3}}{x^2} \]
\[ E_{\text{net}} = \frac{40k \times 10^{-3}}{x^2} - \frac{40k \times 10^{-3}}{x^2} = 0 \]

6. **Condition for Zero Electric Field**:
- To achieve zero electric field at the center, the electric field due to charge \( Q_D \) at D must balance the net field due to A, B, and C.
- The electric field due to charge \( Q_D \) at the center is:
\[ E_D = k \frac{|Q_D|}{\left(\frac{x}{\sqrt{2}}\right)^2} = k \frac{|Q_D| \cdot 2}{x^2} \]
- Setting \( E_D \) equal to the net electric field from A, B, and C:
\[ k \frac{|Q_D| \cdot 2}{x^2} = \frac{40k \times 10^{-3}}{x^2} \]
- Simplifying gives:
\[ |Q_D| \cdot 2 = 40 \times 10^{-3} \]
\[ |Q_D| = 20 \times 10^{-3} \, \text{mC} \]

7. **Determining the Sign of Charge \( Q_D \)**:
- Since the net electric field due to A and C is positive (both are positive charges), \( Q_D \) must be negative to balance it out.
- Therefore, \( Q_D = -20 \, \text{mC} \).

### Final Answer:
The charge to be placed at corner D is \( Q_D = -20 \, \text{mC} \).