Question

find the net force on q₁.
+4.44 μc +7.28 μc +2.25 μc
q₁ q₂ q₃
←0.100 m→ ←0.100 m→
\\(\vec{f}_2\\) = force exerted on q₁ by q₂
\\(\vec{f}_3\\) = force exerted on q₁ by q₃
\\(\vec{f}_2 = ?\\) n \\(\vec{f}_3 = \square\\) n \\(\sigma \vec{f} = \square\\) n
remember like charges repel;
opposite charges attract.

Explanation:

To find the force exerted on \( q_1 \) by \( q_2 \) (\( \vec{F}_2 \)), we use Coulomb's Law, which is given by:

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

where:

• \( k = 8.988 \times 10^9 \, \text{N·m}^2/\text{C}^2 \) (Coulomb's constant),
• \( q_1 = +4.44 \, \mu\text{C} = 4.44 \times 10^{-6} \, \text{C} \),
• \( q_2 = +7.28 \, \mu\text{C} = 7.28 \times 10^{-6} \, \text{C} \),
• \( r = 0.100 \, \text{m} \) (distance between \( q_1 \) and \( q_2 \)).

Step 1: Identify the values

We have:

• \( k = 8.988 \times 10^9 \, \text{N·m}^2/\text{C}^2 \)
• \( q_1 = 4.44 \times 10^{-6} \, \text{C} \)
• \( q_2 = 7.28 \times 10^{-6} \, \text{C} \)
• \( r = 0.100 \, \text{m} \)

Step 2: Plug into Coulomb's Law

\[
F_2 = k \frac{|q_1 q_2|}{r^2}
\]
Substitute the values:
\[
F_2 = (8.988 \times 10^9) \frac{(4.44 \times 10^{-6})(7.28 \times 10^{-6})}{(0.100)^2}
\]

Step 3: Calculate the numerator

First, calculate the product of \( q_1 \) and \( q_2 \):
\[
(4.44 \times 10^{-6})(7.28 \times 10^{-6}) = 4.44 \times 7.28 \times 10^{-12} \approx 32.3232 \times 10^{-12} \, \text{C}^2
\]

Step 4: Calculate the denominator

\[
(0.100)^2 = 0.0100 \, \text{m}^2
\]

Step 5: Divide numerator by denominator

\[
\frac{32.3232 \times 10^{-12}}{0.0100} = 3232.32 \times 10^{-12} = 3.23232 \times 10^{-9}
\]

Step 6: Multiply by \( k \)

\[
F_2 = (8.988 \times 10^9) \times (3.23232 \times 10^{-9})
\]
\[
F_2 \approx 8.988 \times 3.23232 \approx 29.05 \, \text{N}
\]

Since both charges are positive, they repel each other. So the force \( \vec{F}_2 \) on \( q_1 \) by \( q_2 \) is directed to the left (away from \( q_2 \)).

Answer:

The magnitude of \( \vec{F}_2 \) is approximately \( \boxed{29.1} \, \text{N} \) (directed to the left).