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Step-by-Step Solution
Step 1: Identify the Physical Situation
We have a line charge (on the z-axis) with linear charge density
$ \lambda = 3.0 \times 10^{-6} \,\text{C/m} $,
and an electric dipole placed along the x-axis.
The positive and negative charges of the dipole are at distances of 10 mm ($0.01\,\text{m}$)
and 12 mm ($0.012\,\text{m}$) from the origin, respectively.
The net force on the dipole is given as $4\,\text{N}$.
Step 2: Write the Expression for the Electric Field Due to a Line Charge
The magnitude of the electric field at a distance $r$ from an infinite line charge with linear charge density
$ \lambda $ is
$$
E = \frac{2k \lambda}{r},
$$
where $k = 9 \times 10^{9}\,\text{N}\,\text{m}^2\,\text{C}^{-2}$.
Step 3: Note the Fields at the Two Charges of the Dipole
Let
$ E_1 = \frac{2k\lambda}{r_1} = \frac{2 \times 9 \times 10^9 \times 3 \times 10^{-6}}{0.01}
\quad\text{and}\quad
E_2 = \frac{2k\lambda}{r_2} = \frac{2 \times 9 \times 10^9 \times 3 \times 10^{-6}}{0.012}.
$
Since the two charges lie at different distances, the net force on the dipole is related to the difference in
the fields $ E_1 $ and $ E_2 $, multiplied by the magnitude of the charge $ q $ on each end of the dipole.
Step 4: Express the Net Force on the Dipole
The net force on the dipole can be written as
$$
F = q \,(E_1 - E_2).
$$
We know $F = 4\,\text{N}$.
Step 5: Substitute the Numerical Values and Solve for $ q $
Using the expression given in the provided reference solution:
$$
4 = q \times 2 \times 9 \times 10^9 \times (3 \times 10^{-6})
\left[ \frac{2}{(10 \times 12 \times 10^{-3})} \right].
$$
Simplify everything carefully:
1. Combine constants:
$$
2 \times 9 \times 10^9 \times 3 \times 10^{-6} =
2 \times 9 \times 3 \times 10^{9 - 6} = 54 \times 10^3 = 5.4 \times 10^4.
$$
2. The bracketed term:
$$
\left[ \frac{2}{120 \times 10^{-3}} \right]
= \frac{2}{120 \times 10^{-3}}
= \frac{2}{0.12}
= \frac{2}{0.12}
= 16.666\ldots
$$
Putting it together,
$$
4 = q \times (5.4 \times 10^4) \times 16.666\ldots
$$
Solving for $ q $ gives approximately
$$
q = 4.44 \times 10^{-6}\,\text{C} = 4.44\,\mu\text{C}.
$$
Step 6: State the Final Answer
Therefore, the magnitude of the positive or negative charge of the dipole is
$ \boxed{4.44\,\mu\text{C}} $.
