Analyzing the Electrostatic Force Between Point Charges

This problem requires us to apply Coulomb's Law to find the new electrostatic force between two point charges after they have been modified. We are given the initial charges, the initial force between them, and the change applied to each charge. We need to find the magnitude and nature (attractive or repulsive) of the final force.

Understanding Coulomb's Law for Point Charges

Coulomb's Law is the fundamental principle governing the force between two stationary electric charges. It states that the magnitude of the electrostatic force ($F$) between two point charges ($q_1$ and $q_2$) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance ($r$) between them. The mathematical expression is:

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

Here, $k$ is Coulomb's constant. The nature of the force depends on the signs of the charges:

• If the charges have opposite signs (one positive, one negative), the force is attractive.
• If the charges have the same sign (both positive or both negative), the force is repulsive.

Calculating the New Force and Its Nature

Let's calculate the new force step-by-step using the information provided.

Step 1: Analyze the Initial State

We are given the initial charges and the force between them:

• Initial charge $Q_{1,initial} = +3 \mu C = +3 \times 10^{-6} C$.
• Initial charge $Q_{2,initial} = -8 \mu C = -8 \times 10^{-6} C$.
• The initial force $F_1 = 48 N$. Since the charges have opposite signs, this force is attractive.
• The product of the absolute values of the initial charges is: $|Q_{1,initial} Q_{2,initial}| = |(+3 \mu C) \times (-8 \mu C)| = |-24 (\mu C)^2| = 24 \times 10^{-12} C^2$.

Using Coulomb's Law for the initial state:

$F_1 = k \frac{|Q_{1,initial} Q_{2,initial}|}{r^2} = 48 N$

Step 2: Calculate the New Charges

Each charge is given an additional charge of $+6 \mu C$. Let's find the new values of the charges:

• New charge $Q_{1,new} = Q_{1,initial} + 6 \mu C = +3 \mu C + 6 \mu C = +9 \mu C$.
• New charge $Q_{2,new} = Q_{2,initial} + 6 \mu C = -8 \mu C + 6 \mu C = -2 \mu C$.

Step 3: Calculate the Product of the New Charges

Now, we find the product of the absolute values of the new charges:

• $|Q_{1,new} Q_{2,new}| = |(+9 \mu C) \times (-2 \mu C)| = |-18 (\mu C)^2| = 18 \times 10^{-12} C^2$.

Step 4: Calculate the New Force Magnitude

The distance $r$ between the charges remains the same. We can find the new force $F_2$ using the ratio of the forces:

$\frac{F_2}{F_1} = \frac{k \frac{|Q_{1,new} Q_{2,new}|}{r^2}}{k \frac{|Q_{1,initial} Q_{2,initial}|}{r^2}} = \frac{|Q_{1,new} Q_{2,new}|}{|Q_{1,initial} Q_{2,initial}|}$

Substitute the known values:

$\frac{F_2}{48 N} = \frac{18 \times 10^{-12} C^2}{24 \times 10^{-12} C^2} = \frac{18}{24}$

Simplify the fraction:

$\frac{18}{24} = \frac{3 \times 6}{4 \times 6} = \frac{3}{4}$

Now, solve for $F_2$:

$F_2 = 48 N \times \frac{3}{4}$

$F_2 = \frac{48}{4} \times 3 N = 12 \times 3 N = 36 N$

So, the magnitude of the new force is $36 N$.

Step 5: Determine the Nature of the New Force

We look at the signs of the new charges:

• $Q_{1,new} = +9 \mu C$ (positive)
• $Q_{2,new} = -2 \mu C$ (negative)

Since the charges have opposite signs, the force between them is attractive.

Final Result Summary

The new force between the charges has a magnitude of $36 N$ and is attractive in nature.