'Two short dipoles `p hat(k)` and `(p)/(2) hat(k)` are located at `(0,0,0)` and `(1 (~m), 0,2 (~m))`, respectively. The resultant electric field due to the two dipoles at the point `(1, m, 0,0)` is `(4 n p)/(32 pi varepsilon_(0)) k`. Find `n`.

Answer

To solve the problem, we need to calculate the electric field at the point (1, 0, 0) due to two dipoles located at (0, 0, 0) and (1, 0, 2). The dipole moments are given as **p** and **p/2** respectively.

### Step-by-Step Solution:

1. **Identify the Positions of the Dipoles**:
- Dipole 1 (p) is located at (0, 0, 0).
- Dipole 2 (p/2) is located at (1, 0, 2).

2. **Calculate the Distance from Each Dipole to the Point (1, 0, 0)**:
- For Dipole 1: The distance \( r_1 \) from (0, 0, 0) to (1, 0, 0) is:
\[ r_1 = \sqrt{(1-0)^2 + (0-0)^2 + (0-0)^2} = 1 \, \text{m} \]
- For Dipole 2: The distance \( r_2 \) from (1, 0, 2) to (1, 0, 0) is:
\[ r_2 = \sqrt{(1-1)^2 + (0-0)^2 + (0-2)^2} = 2 \, \text{m} \]

3. **Calculate the Electric Field due to Each Dipole**:
- The electric field \( E \) due to a dipole at a point along the axial line is given by:
\[ E = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2p}{r^3} \]
- For Dipole 1 (p):
\[ E_1 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2p}{1^3} = \frac{2p}{4 \pi \epsilon_0} \hat{k} \]
- For Dipole 2 (p/2):
\[ E_2 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2(p/2)}{2^3} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{p}{8} = \frac{p}{32 \pi \epsilon_0} \hat{k} \]

4. **Determine the Resultant Electric Field**:
- Since both electric fields are in the same direction (along +k), we can add them:
\[ E_{\text{total}} = E_1 + E_2 = \frac{2p}{4 \pi \epsilon_0} + \frac{p}{32 \pi \epsilon_0} \]
- To combine these, we need a common denominator:
\[ E_{\text{total}} = \frac{16p}{32 \pi \epsilon_0} + \frac{p}{32 \pi \epsilon_0} = \frac{17p}{32 \pi \epsilon_0} \hat{k} \]

5. **Compare with Given Result**:
- We know from the problem statement that the resultant electric field is:
\[ E_{\text{total}} = \frac{4n p}{32 \pi \epsilon_0} \hat{k} \]
- Setting the two expressions for the electric field equal gives:
\[ \frac{17p}{32 \pi \epsilon_0} = \frac{4n p}{32 \pi \epsilon_0} \]
- Canceling \( \frac{p}{32 \pi \epsilon_0} \) from both sides (assuming \( p \neq 0 \)):
\[ 17 = 4n \]
- Solving for \( n \):
\[ n = \frac{17}{4} = 4.25 \]

### Final Answer:
\[ n = 4.25 \]