(a) Two point charges totaling 7.00 µC exert a repulsive force of 0.250 N on one another when separated by 0.586 m. What is the charge on each?

To solve for the charges, we can use Coulomb's law, which states:

\[ F = k \frac{|q_1 q_2|}{r^2} \]

where:

• \( F \) is the force between the charges (in Newtons),
• \( k \) is Coulomb's constant (\( k \approx 8.99 \times 10^9 , \text{N m}^2/\text{C}^2 \)),
• \( q_1 \) and \( q_2 \) are the magnitudes of the charges (in coulombs),
• \( r \) is the distance between the charges (in meters).

Part (a): Repulsive Force

Given:

• \( F = 0.250 , \text{N} \)
• \( r = 0.586 , \text{m} \)
• Total charge \( q_1 + q_2 = 7.00 , \mu\text{C} = 7.00 \times 10^{-6} , \text{C} \)

Let \( q_1 = q \) and \( q_2 = 7.00 \times 10^{-6} - q \).

Using Coulomb’s law:

\[ 0.250 = k \frac{q (7.00 \times 10^{-6} - q)}{(0.586)^2} \]

Substituting the value of \( k \):

\[ 0.250 = (8.99 \times 10^9) \frac{q (7.00 \times 10^{-6} - q)}{0.343396} \]

Now, we multiply both sides by \( 0.343396 \):

\[ 0.250 \cdot 0.343396 = (8.99 \times 10^9) q (7.00 \times 10^{-6} - q) \]

Calculating \( 0.250 \cdot 0.343396 \):

\[ 0.085849 = (8.99 \times 10^9) q (7.00 \times 10^{-6} - q) \]

Now we divide by \( 8.99 \times 10^9 \):

\[ \frac{0.085849}{8.99 \times 10^9} = q (7.00 \times 10^{-6} - q) \]

Calculating \( \frac{0.085849}{8.99 \times 10^9} \):

\[ 9.554 \times 10^{-12} = q (7.00 \times 10^{-6} - q) \]

Rearranging gives us the quadratic equation:

\[ q^2 - (7.00 \times 10^{-6})q + 9.554 \times 10^{-12} = 0 \]

Using the quadratic formula \( q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 1, b = -7.00 \times 10^{-6}, c = 9.554 \times 10^{-12} \).

Calculating the discriminant:

\[ D = (-7.00 \times 10^{-6})^2 - 4 \cdot 1 \cdot 9.554 \times 10^{-12} \] \[ D = 4.9 \times 10^{-11} - 3.8216 \times 10^{-11} \] \[ D = 1.0784 \times 10^{-11} \]

Now applying the quadratic formula:

\[ q = \frac{7.00 \times 10^{-6} \pm \sqrt{1.0784 \times 10^{-11}}}{2} \] \[ q \approx \frac{7.00 \times 10^{-6} \pm 1.039 \times 10^{-6}}{2} \]

Calculating the two possible values:

1. \( q_1 \approx \frac{8.039 \times 10^{-6}}{2} \approx 4.0195 \times 10^{-6} \)
2. \( q_2 \approx \frac{5.9605 \times 10^{-6}}{2} \approx 2.98025 \times 10^{-6} \)

Thus, the charges are approximately:

\[ q_1 \approx 4.02 , \mu\text{C}, \quad q_2 \approx 2.98 , \mu\text{C}. \]

Part (b): Attractive Force

If the force is attractive, then one charge is positive and the other is negative, meaning their magnitudes still sum to 7.00 µC but their product is negative. We can set \( q_1 = q \) and \( q_2 = -(-7.00 \times 10^{-6}) + q \) which is equivalent to:

Let’s denote the negative charge as \( -q_2 \):

1. \( q_1 = q \)
2. \( q_2 = - (7.00 \times 10^{-6} - q) \)

\(-q_1 (q_2)\):

\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \Rightarrow 0.250 = k \frac{|-q(q + 7.00 \times 10^{-6})|}{(0.586)^2} \]

The equations will follow a similar pattern to the repulsive case, leading us to a similar quadratic equation:

This leads us back to the same values, giving us \( q \) values falling along the same paths.

Using this set up, we can end up with \( q_1 \approx 4.02, \quad q_2 \approx -2.98 µC \) or \( q_1 \approx -4.02, \quad q_2 \approx 2.98 µC \).

So we can write:

• The charges are approximately \( (4.02 , \mu C, -2.98 , \mu C) \) if the force is attractive.

In summary:

• For (a), \( 4.02 , \mu C\) and \( 2.98 , \mu C\) are the charges causing repulsion.
• For (b), \( 4.02 , \mu C\) and \(-2.98 , \mu C\) are the charges causing attraction.