An electric dipole is along a uniform electric field. If it is deflected by `60^(@)`, work done by agent is `2 xx 10^(-19)J`. Then the work done by an agent if it is deflected by `30^(@)` further is

Options

A. `2.5 xx 10^(-19)J`
B. `2 xx 10^(-19)J`
C. `4 xx 10^(-19)J`
D. `2 xx 10^(-16)J`

Answer

To solve the problem, we need to calculate the work done by the agent when the electric dipole is deflected further by 30 degrees after already being deflected by 60 degrees.

### Step-by-Step Solution:

1. **Understanding the Work Done by the Electric Dipole**:
The work done \( W \) in moving an electric dipole in a uniform electric field is given by the formula:
\[ W = -pE \cos(\theta) \]
where:
- \( p \) is the dipole moment,
- \( E \) is the electric field strength,
- \( \theta \) is the angle between the dipole moment and the electric field.

2. **Calculating Work Done for 60 Degrees**:
From the problem, the work done when the dipole is deflected by \( 60^\circ \) is given as:
\[ W_1 = -pE \cos(60^\circ) = 2 \times 10^{-19} \text{ J} \]
Since \( \cos(60^\circ) = \frac{1}{2} \), we can express this as:
\[ W_1 = -pE \cdot \frac{1}{2} \]
Rearranging gives:
\[ pE = -2 \times 10^{-19} \times 2 = -4 \times 10^{-19} \text{ J} \]

3. **Calculating Work Done for Further Deflection of 30 Degrees**:
Now, if the dipole is deflected further by \( 30^\circ \), the total angle becomes \( 60^\circ + 30^\circ = 90^\circ \).
Using the work done formula again:
\[ W_2 = -pE \cos(90^\circ) \]
Since \( \cos(90^\circ) = 0 \):
\[ W_2 = -pE \cdot 0 = 0 \]

4. **Finding the Work Done by the Agent**:
The work done by the external agent in moving the dipole from \( 60^\circ \) to \( 90^\circ \) is given by:
\[ W_{\text{agent}} = W_2 - W_1 \]
Substituting the values we have:
\[ W_{\text{agent}} = 0 - (-2 \times 10^{-19}) = 2 \times 10^{-19} \text{ J} \]

5. **Final Answer**:
Therefore, the work done by the agent when the dipole is deflected by an additional \( 30^\circ \) is:
\[ \boxed{2 \times 10^{-19} \text{ J}} \]