An electric dipole is placed at an angle `60^(@)` with an electric field of strength `4xx10^(5)` N/C. It experiences a torque equal to `8sqrt(3)` Nm. Calculate the charge on the dipole, if dipole is of length 4 cm

Options

A. `10^(-1)` C
B. `10^(-2)` C
C. `10^(-3)` C
D. `10^(-4)` C

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
We have the following information:
- Angle \( \theta = 60^\circ \)
- Electric field strength \( E = 4 \times 10^5 \, \text{N/C} \)
- Torque \( \tau = 8 \sqrt{3} \, \text{Nm} \)
- Length of the dipole \( 2a = 4 \, \text{cm} = 4 \times 10^{-2} \, \text{m} \)

### Step 2: Calculate the dipole moment \( p \)
The dipole moment \( p \) is given by:
\[ p = Q \cdot (2a) \]
where \( Q \) is the charge on the dipole.

### Step 3: Use the formula for torque
The torque \( \tau \) experienced by a dipole in an electric field is given by:
\[ \tau = p \cdot E \cdot \sin(\theta) \]
Substituting the expression for \( p \):
\[ \tau = (Q \cdot 2a) \cdot E \cdot \sin(\theta) \]

### Step 4: Substitute the known values
Substituting the known values into the torque equation:
\[ 8 \sqrt{3} = (Q \cdot 4 \times 10^{-2}) \cdot (4 \times 10^5) \cdot \sin(60^\circ) \]
Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \), we can write:
\[ 8 \sqrt{3} = (Q \cdot 4 \times 10^{-2}) \cdot (4 \times 10^5) \cdot \left(\frac{\sqrt{3}}{2}\right) \]

### Step 5: Simplify the equation
Now, let's simplify the equation:
\[ 8 \sqrt{3} = Q \cdot 4 \times 10^{-2} \cdot 4 \times 10^5 \cdot \frac{\sqrt{3}}{2} \]
This simplifies to:
\[ 8 \sqrt{3} = Q \cdot 8 \times 10^3 \cdot \frac{\sqrt{3}}{2} \]
\[ 8 \sqrt{3} = Q \cdot 4 \times 10^3 \sqrt{3} \]

### Step 6: Cancel out \( \sqrt{3} \)
Dividing both sides by \( \sqrt{3} \):
\[ 8 = Q \cdot 4 \times 10^3 \]

### Step 7: Solve for \( Q \)
Now, solving for \( Q \):
\[ Q = \frac{8}{4 \times 10^3} = \frac{2}{10^3} = 2 \times 10^{-3} \, \text{C} \]

### Step 8: Final answer
Thus, the charge on the dipole is:
\[ Q = 10^{-3} \, \text{C} \]