Question

the magnitude of the electrical force acting between a +2.4 × 10⁻⁸ c charge and a +1.8 × 10⁻⁶ c charge that are separated by 0.008 m is n, rounded to the tenths place.

Explanation:

To find the magnitude of the electrical force between two charges, we use Coulomb's Law, which is given by the formula:

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

where:

• \( k = 8.988 \times 10^9 \, \text{N·m}^2/\text{C}^2 \) (Coulomb's constant),
• \( q_1 = 2.4 \times 10^{-8} \, \text{C} \),
• \( q_2 = 1.8 \times 10^{-6} \, \text{C} \),
• \( r = 0.008 \, \text{m} \) (distance between the charges).

Step 1: Identify the values

We have:

• \( q_1 = 2.4 \times 10^{-8} \, \text{C} \)
• \( q_2 = 1.8 \times 10^{-6} \, \text{C} \)
• \( r = 0.008 \, \text{m} \)
• \( k = 8.988 \times 10^9 \, \text{N·m}^2/\text{C}^2 \)

Step 2: Plug the values into Coulomb's Law

First, calculate the product of the charges:
\[
q_1 q_2 = (2.4 \times 10^{-8}) \times (1.8 \times 10^{-6}) = 4.32 \times 10^{-14} \, \text{C}^2
\]

Next, calculate the square of the distance:
\[
r^2 = (0.008)^2 = 6.4 \times 10^{-5} \, \text{m}^2
\]

Now, substitute these values into Coulomb's Law:
\[
F = (8.988 \times 10^9) \times \frac{4.32 \times 10^{-14}}{6.4 \times 10^{-5}}
\]

Step 3: Simplify the expression

First, simplify the fraction:
\[
\frac{4.32 \times 10^{-14}}{6.4 \times 10^{-5}} = \frac{4.32}{6.4} \times 10^{-14 + 5} = 0.675 \times 10^{-9} = 6.75 \times 10^{-10}
\]

Then, multiply by \( k \):
\[
F = 8.988 \times 10^9 \times 6.75 \times 10^{-10}
\]

\[
F = 8.988 \times 6.75 \times 10^{9 - 10}
\]

\[
F = 8.988 \times 6.75 \times 10^{-1}
\]

Calculate \( 8.988 \times 6.75 \):
\[
8.988 \times 6.75 \approx 60.6792
\]

Then multiply by \( 10^{-1} \):
\[
F \approx 60.6792 \times 0.1 = 6.06792 \, \text{N}
\]

Step 4: Round to the tenths place

Rounding \( 6.06792 \) to the tenths place gives \( 6.1 \).

Answer:

\( 6.1 \)