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Step-by-Step Solution
Step 1: Identify the Charges and Their Locations
We have two charges:
• A negative charge −4 μC located at A(1, 0, 4) m.
• A positive charge +4 μC located at B(2, −1, 5) m.
These two charges together form an electric dipole whose dipole moment vector will point from the negative charge to the positive charge.
Step 2: Determine the Displacement Vector (r)
The displacement vector from A to B is:
\mathbf{r} = (2 - 1,\, -1 - 0,\, 5 - 4) = (1,\, -1,\, 1)\text{ m}.
Step 3: Calculate the Dipole Moment (p)
Each charge has magnitude 4 \times 10^{-6}\text{ C} .
Hence, the dipole moment vector is
\mathbf{p} = q \,\mathbf{r} = 4 \times 10^{-6} \times (1,\, -1,\, 1).
Therefore,
\mathbf{p} = \bigl(4 \times 10^{-6},\, -4 \times 10^{-6},\, 4 \times 10^{-6}\bigr)\,\text{C}\cdot\text{m}.
Step 4: Convert the Electric Field to SI Units
The electric field given is 0.20\,\hat{i}\,\text{V/cm} .
In SI units (V/m), we have:
0.20\,\text{V/cm} = 0.20 \div 10^{-2} = 20\,\text{V/m}.
Thus, the electric field vector is
\mathbf{E} = (20,\, 0,\, 0)\,\text{V/m}.
Step 5: Compute the Torque (τ) on the Dipole
The torque on an electric dipole in a uniform electric field is given by
\boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}.
Let
\mathbf{p} = \bigl(4\times 10^{-6},\, -4\times 10^{-6},\, 4\times 10^{-6}\bigr),
\quad
\mathbf{E} = (20,\, 0,\, 0).
The cross product components are:
\tau_x = p_y E_z - p_z E_y, \qquad
\tau_y = p_z E_x - p_x E_z, \qquad
\tau_z = p_x E_y - p_y E_x.
Substituting:
\tau_x = \bigl(-4 \times 10^{-6}\bigr) \cdot 0 - \bigl(4 \times 10^{-6}\bigr) \cdot 0 = 0,
\tau_y = \bigl(4 \times 10^{-6}\bigr) \cdot 20 - \bigl(4 \times 10^{-6}\bigr) \cdot 0 = 8 \times 10^{-5},
\tau_z = \bigl(4 \times 10^{-6}\bigr) \cdot 0 - \bigl(-4 \times 10^{-6}\bigr) \cdot 20
= 8 \times 10^{-5}.
Hence,
\boldsymbol{\tau} = \bigl(0,\, 8 \times 10^{-5},\, 8 \times 10^{-5}\bigr)\,\text{N}\cdot\text{m}.
Step 6: Find the Magnitude of the Torque
The magnitude of the torque vector is
\tau = \sqrt{(0)^2 + \bigl(8\times 10^{-5}\bigr)^2 + \bigl(8\times 10^{-5}\bigr)^2}
= 8 \times 10^{-5} \,\sqrt{0 + 1 + 1}
= 8 \times 10^{-5} \,\sqrt{2}.
Step 7: Compare with the Given Form
The problem states that the torque is in the form
8 \,\sqrt{\alpha} \,\times 10^{-5}\,\text{N}\cdot\text{m}.
We have found it to be
8 \times 10^{-5} \,\sqrt{2}\,\text{N}\cdot\text{m}.
Therefore, we identify
\sqrt{\alpha} = \sqrt{2}
\quad \Rightarrow \quad
\alpha = 2.
Final Answer
The value of \alpha is 2.
