Question

electrostatics practice

1. three point charges are arranged in a straight line. the point charges are q₁ = -10.0μc, q₂ = 20.0μc and q₃ = -10.0μc. charge q₁ is 20.0 cm from q₂ and q₃ is 10.0 cm from q₂.

a. what is the force between charge 1 and 2?

b. what is the force between charge 1 and 3?

c. what is the net force acting on q₁?

1. three point charges are arranged in a straight line. the point charges are q₁ = 10.0μc, q₂ = -40.0μc and q₃ = -30.0μc. charge q₁ is 20.0 cm from q₂ and q₃ is 10.0 cm from q₂.

a. what is the force between charge 1 and 2?

Explanation:

Part 1a: Force between \( q_1 \) and \( q_2 \)

Step 1: Recall Coulomb's Law

Coulomb's Law is \( F = k\frac{|q_1 q_2|}{r^2} \), where \( k = 9.0 \times 10^9 \, \text{N·m}^2/\text{C}^2 \), \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between them.
Given \( q_1 = -10.0 \, \mu\text{C} = -10.0 \times 10^{-6} \, \text{C} \), \( q_2 = 20.0 \, \mu\text{C} = 20.0 \times 10^{-6} \, \text{C} \), \( r = 20.0 \, \text{cm} = 0.20 \, \text{m} \).

Step 2: Substitute values into Coulomb's Law

\[

$$\begin{align*} F_{12} &= (9.0 \times 10^9) \frac{|(-10.0 \times 10^{-6})(20.0 \times 10^{-6})|}{(0.20)^2} \\ &= (9.0 \times 10^9) \frac{2.0 \times 10^{-10}}{0.04} \\ &= (9.0 \times 10^9) (5.0 \times 10^{-9}) \\ &= 45 \, \text{N} \end{align*}$$

\]
Since \( q_1 \) and \( q_2 \) have opposite signs, the force is attractive.

Step 1: Determine distance between \( q_1 \) and \( q_3 \)

\( q_1 \) is 20.0 cm from \( q_2 \), and \( q_3 \) is 10.0 cm from \( q_2 \), so the distance \( r_{13} = 20.0 \, \text{cm} + 10.0 \, \text{cm} = 30.0 \, \text{cm} = 0.30 \, \text{m} \).
\( q_1 = -10.0 \times 10^{-6} \, \text{C} \), \( q_3 = -10.0 \times 10^{-6} \, \text{C} \).

Step 2: Apply Coulomb's Law

\[

$$\begin{align*} F_{13} &= (9.0 \times 10^9) \frac{|(-10.0 \times 10^{-6})(-10.0 \times 10^{-6})|}{(0.30)^2} \\ &= (9.0 \times 10^9) \frac{1.0 \times 10^{-10}}{0.09} \\ &= (9.0 \times 10^9) (1.111 \times 10^{-9}) \\ &\approx 10 \, \text{N} \end{align*}$$

\]
Since \( q_1 \) and \( q_3 \) have the same sign, the force is repulsive.

Step 1: Analyze directions of forces

Force \( F_{12} \) (from \( q_2 \) on \( q_1 \)): attractive, so direction is towards \( q_2 \) (right).
Force \( F_{13} \) (from \( q_3 \) on \( q_1 \)): repulsive, so direction is away from \( q_3 \) (left, since \( q_3 \) is to the right of \( q_2 \)).

Step 2: Calculate net force

Let right be positive. \( F_{12} = +45 \, \text{N} \), \( F_{13} = -10 \, \text{N} \) (since it's to the left).
Net force \( F_{\text{net}} = F_{12} + F_{13} = 45 - 10 = 35 \, \text{N} \) (to the right).

Answer:

\( 45 \, \text{N} \) (attractive)

Part 1b: Force between \( q_1 \) and \( q_3 \)