An electric dipole of moment `vecP=(hati+3hatj-5hatk)xx10^(-19)Cm` is kept in a electric field `vecE=(2hati-3hatj+2hatk)xx10^(5)V/m`. Find torque experienced by dipole.

Options

A. `(9hati+12hatj-9hatk)xx10^(-14)N-m`
B. `(9hati+12hatj+9hatk)xx10^(-14)N-m`
C. `(-9hati-12hatj-9hatk)xx10^(-14)N-m`
D. `(-9hati+12hatj-9hatk)xx10^(-14)N-m`

Explanation

To find the torque experienced by an electric dipole in an electric field, we can use the formula for torque (\(\vec{\tau}\)) given by the cross product of the dipole moment (\(\vec{P}\)) and the electric field (\(\vec{E}\)):

\[ \vec{\tau} = \vec{P} \times \vec{E} \]

### Step-by-Step Solution:

1. **Identify the dipole moment and electric field vectors:**
- Given dipole moment:
\[ \vec{P} = (1\hat{i} + 3\hat{j} - 5\hat{k}) \times 10^{-19} \, \text{Cm} \]
- Given electric field:
\[ \vec{E} = (2\hat{i} - 3\hat{j} + 2\hat{k}) \times 10^{5} \, \text{V/m} \]

2. **Set up the cross product:**
- We will calculate \(\vec{P} \times \vec{E}\) using the determinant of a matrix:
\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 \times 10^{-19} & 3 \times 10^{-19} & -5 \times 10^{-19} \\ 2 \times 10^{5} & -3 \times 10^{5} & 2 \times 10^{5} \end{vmatrix} \]

3. **Calculate the determinant:**
- Expanding the determinant:
\[ \vec{\tau} = \hat{i} \begin{vmatrix} 3 \times 10^{-19} & -5 \times 10^{-19} \\ -3 \times 10^{5} & 2 \times 10^{5} \end{vmatrix} - \hat{j} \begin{vmatrix} 1 \times 10^{-19} & -5 \times 10^{-19} \\ 2 \times 10^{5} & 2 \times 10^{5} \end{vmatrix} + \hat{k} \begin{vmatrix} 1 \times 10^{-19} & 3 \times 10^{-19} \\ 2 \times 10^{5} & -3 \times 10^{5} \end{vmatrix} \]

4. **Calculate each of the 2x2 determinants:**
- For \(\hat{i}\):
\[ = (3 \times 10^{-19})(2 \times 10^{5}) - (-5 \times 10^{-19})(-3 \times 10^{5}) = 6 \times 10^{-14} - 15 \times 10^{-14} = -9 \times 10^{-14} \]
- For \(\hat{j}\):
\[ = (1 \times 10^{-19})(2 \times 10^{5}) - (-5 \times 10^{-19})(2 \times 10^{5}) = 2 \times 10^{-14} + 10 \times 10^{-14} = 12 \times 10^{-14} \]
- For \(\hat{k}\):
\[ = (1 \times 10^{-19})(-3 \times 10^{5}) - (3 \times 10^{-19})(2 \times 10^{5}) = -3 \times 10^{-14} - 6 \times 10^{-14} = -9 \times 10^{-14} \]

5. **Combine the results:**
- Thus, the torque vector is:
\[ \vec{\tau} = (-9 \times 10^{-14})\hat{i} - (12 \times 10^{-14})\hat{j} + (-9 \times 10^{-14})\hat{k} \]
- Therefore, we can write:
\[ \vec{\tau} = -9\hat{i} - 12\hat{j} - 9\hat{k} \times 10^{-14} \, \text{N m} \]

### Final Answer:
\[ \vec{\tau} = (-9\hat{i} - 12\hat{j} - 9\hat{k}) \times 10^{-14} \, \text{N m} \]