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 the electric dipole in the given electric field, we can use the formula for torque (\(\vec{\tau}\)) on a dipole, which is 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 1: Identify the vectors
Given:
\[ \vec{P} = (1\hat{i} + 3\hat{j} - 5\hat{k}) \times 10^{-19} \, \text{Cm} \]
\[ \vec{E} = (2\hat{i} - 3\hat{j} + 2\hat{k}) \times 10^{5} \, \text{V/m} \]

### Step 2: Write the cross product
We need to calculate the cross product \(\vec{P} \times \vec{E}\). First, we can factor out the constants:
\[ \vec{\tau} = \left((1\hat{i} + 3\hat{j} - 5\hat{k}) \times (2\hat{i} - 3\hat{j} + 2\hat{k})\right) \times 10^{-14} \]

### Step 3: Set up the determinant for the cross product
We can use the determinant method to calculate the cross product:
\[ \vec{P} \times \vec{E} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -5 \\ 2 & -3 & 2 \end{vmatrix} \]

### Step 4: Calculate the determinant
Calculating the determinant, we have:
\[ \vec{P} \times \vec{E} = \hat{i} \begin{vmatrix} 3 & -5 \\ -3 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -5 \\ 2 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 3 \\ 2 & -3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \(\hat{i}\):
\[ \begin{vmatrix} 3 & -5 \\ -3 & 2 \end{vmatrix} = (3 \cdot 2) - (-5 \cdot -3) = 6 - 15 = -9 \]

2. For \(\hat{j}\):
\[ \begin{vmatrix} 1 & -5 \\ 2 & 2 \end{vmatrix} = (1 \cdot 2) - (-5 \cdot 2) = 2 + 10 = 12 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 1 & 3 \\ 2 & -3 \end{vmatrix} = (1 \cdot -3) - (3 \cdot 2) = -3 - 6 = -9 \]

### Step 5: Combine the results
Putting it all together:
\[ \vec{P} \times \vec{E} = -9\hat{i} - 12\hat{j} - 9\hat{k} \]

### Step 6: Multiply by the constant factor
Now, we multiply the result by \(10^{-14}\):
\[ \vec{\tau} = (-9\hat{i} - 12\hat{j} - 9\hat{k}) \times 10^{-14} \]

### Final Answer
Thus, the torque experienced by the dipole is:
\[ \vec{\tau} = -9 \times 10^{-14} \hat{i} - 12 \times 10^{-14} \hat{j} - 9 \times 10^{-14} \hat{k} \, \text{Nm} \]