Two charges of `-4 muC and +4muC` are placed at the points `A (1,0,4) and B (2,-1,5)` location in an electric field `vec(E) = 0.20 hat(i) V//cm`. Calculate the torque acting on the dipole.

Explanation

To calculate the torque acting on the dipole formed by two charges of `-4 µC` and `+4 µC`, we will follow these steps:

### Step 1: Identify the Position Vectors
The position vectors of the charges are given as:
- Charge `-4 µC` at point A (1, 0, 4):
\[ \vec{r_1} = 1\hat{i} + 0\hat{j} + 4\hat{k} \]
- Charge `+4 µC` at point B (2, -1, 5):
\[ \vec{r_2} = 2\hat{i} - 1\hat{j} + 5\hat{k} \]

### Step 2: Calculate the Dipole Moment
The dipole moment \(\vec{p}\) is given by the formula:
\[ \vec{p} = q \cdot \vec{d} \]
where \(\vec{d}\) is the displacement vector from the negative charge to the positive charge.

First, calculate \(\vec{d}\):
\[ \vec{d} = \vec{r_2} - \vec{r_1} = (2\hat{i} - 1\hat{j} + 5\hat{k}) - (1\hat{i} + 0\hat{j} + 4\hat{k}) = (2-1)\hat{i} + (-1-0)\hat{j} + (5-4)\hat{k} = 1\hat{i} - 1\hat{j} + 1\hat{k} \]

Now, substituting \(q = 4 \times 10^{-6} C\):
\[ \vec{p} = 4 \times 10^{-6} \cdot (1\hat{i} - 1\hat{j} + 1\hat{k}) = 4 \times 10^{-6}\hat{i} - 4 \times 10^{-6}\hat{j} + 4 \times 10^{-6}\hat{k} \]

### Step 3: Electric Field Vector
The electric field vector is given as:
\[ \vec{E} = 0.20\hat{i} \text{ V/cm} = 20\hat{i} \text{ V/m} \]

### Step 4: Calculate the Torque
The torque \(\vec{\tau}\) acting on the dipole in an electric field is given by:
\[ \vec{\tau} = \vec{p} \times \vec{E} \]

Substituting the values of \(\vec{p}\) and \(\vec{E}\):
\[ \vec{\tau} = (4 \times 10^{-6}\hat{i} - 4 \times 10^{-6}\hat{j} + 4 \times 10^{-6}\hat{k}) \times (20\hat{i}) \]

Using the properties of the cross product:
\[ \vec{\tau} = 4 \times 10^{-6} \times 20 (\hat{i} \times \hat{i}) - 4 \times 10^{-6} \times 20 (\hat{j} \times \hat{i}) + 4 \times 10^{-6} \times 20 (\hat{k} \times \hat{i}) \]

Since \(\hat{i} \times \hat{i} = 0\), we have:
\[ \vec{\tau} = 0 - 4 \times 10^{-6} \times 20 (-\hat{k}) + 4 \times 10^{-6} \times 20 \hat{j} \]
\[ = 80 \times 10^{-6} \hat{k} + 80 \times 10^{-6} \hat{j} = 8 \times 10^{-5} \hat{j} + 8 \times 10^{-5} \hat{k} \]

### Step 5: Magnitude of the Torque
The magnitude of the torque vector is given by:
\[ |\vec{\tau}| = \sqrt{(8 \times 10^{-5})^2 + (8 \times 10^{-5})^2} = 8 \times 10^{-5} \sqrt{2} \]
\[ = 8\sqrt{2} \times 10^{-5} \text{ Nm} \]

Calculating the numerical value:
\[ |\vec{\tau}| \approx 1.13 \times 10^{-4} \text{ Nm} \]

### Final Answer
The torque acting on the dipole is approximately:
\[ \boxed{1.13 \times 10^{-4} \text{ Nm}} \]