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 use the formula for the torque (\(\tau\)) experienced by an electric dipole in an electric field, which is given by:

\[ \tau = p \cdot E \cdot \sin(\theta) \]

Where:
- \(\tau\) is the torque,
- \(p\) is the dipole moment,
- \(E\) is the electric field strength,
- \(\theta\) is the angle between the dipole moment and the electric field.

### Step 1: Identify the given values
- Torque (\(\tau\)) = \(8\sqrt{3}\) Nm
- Electric field strength (\(E\)) = \(4 \times 10^5\) N/C
- Angle (\(\theta\)) = \(60^\circ\)
- Length of the dipole (\(l\)) = \(4\) cm = \(0.04\) m

### Step 2: Convert the angle to radians if necessary
In this case, we can use the sine of \(60^\circ\) directly:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

### Step 3: Substitute the known values into the torque formula
Now we can rearrange the torque formula to solve for the dipole moment \(p\):
\[ p = \frac{\tau}{E \cdot \sin(\theta)} \]

Substituting the known values:
\[ p = \frac{8\sqrt{3}}{(4 \times 10^5) \cdot \left(\frac{\sqrt{3}}{2}\right)} \]

### Step 4: Simplify the expression
Calculating the denominator:
\[ E \cdot \sin(60^\circ) = (4 \times 10^5) \cdot \left(\frac{\sqrt{3}}{2}\right) = 2 \times 10^5 \sqrt{3} \]

Now substituting back into the equation for \(p\):
\[ p = \frac{8\sqrt{3}}{2 \times 10^5 \sqrt{3}} = \frac{8}{2 \times 10^5} = \frac{4}{10^5} = 4 \times 10^{-5} \text{ C m} \]

### Step 5: Calculate the charge \(q\) on the dipole
The dipole moment \(p\) is also given by the formula:
\[ p = q \cdot l \]
Where \(q\) is the charge and \(l\) is the length of the dipole.

Rearranging this formula to find \(q\):
\[ q = \frac{p}{l} \]

Substituting the values:
\[ q = \frac{4 \times 10^{-5}}{0.04} = \frac{4 \times 10^{-5}}{4 \times 10^{-2}} = 10^{-3} \text{ C} \]

### Final Answer
The charge on the dipole is:
\[ q = 1 \text{ mC} \]