Answer :
To solve this problem, we will use Coulomb's Law to find the forces between the charges and determine the net force on [tex]\( q_1 \)[/tex]. Here are the step-by-step details:
### Given Data:
Charges:
- [tex]\( q_1 = 4.44 \times 10^{-6} \, \text{C} \)[/tex]
- [tex]\( q_2 = -4.44 \times 10^{-6} \, \text{C} \)[/tex] (opposite charge)
- [tex]\( q_3 = 4.44 \times 10^{-6} \, \text{C} \)[/tex] (same charge as [tex]\( q_1 \)[/tex])
Distances:
- Distance between [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex], [tex]\( r_{12} = 0.02 \, \text{m} \)[/tex]
- Distance between [tex]\( q_1 \)[/tex] and [tex]\( q_3 \)[/tex], [tex]\( r_{13} = 0.04 \, \text{m} \)[/tex]
Coulomb's Constant:
- [tex]\( k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)[/tex]
### Step 1: Calculate the Force [tex]\( F_2 \)[/tex] Exerted by [tex]\( q_2 \)[/tex] on [tex]\( q_1 \)[/tex]
Using Coulomb's Law:
[tex]\[ F = k \cdot \frac{|q_1 \cdot q_2|}{r_{12}^2} \][/tex]
Given:
[tex]\[ q_1 = 4.44 \times 10^{-6} \, \text{C}, \quad q_2 = -4.44 \times 10^{-6} \, \text{C}, \quad r_{12} = 0.02 \, \text{m} \][/tex]
We get:
[tex]\[ F_2 = 8.99 \times 10^9 \cdot \frac{|4.44 \times 10^{-6} \cdot (-4.44 \times 10^{-6})|}{(0.02)^2} \][/tex]
Calculating the magnitude of [tex]\( F_2 \)[/tex], we find [tex]\( F_2 = 443.06316 \, \text{N} \)[/tex].
Since [tex]\( q_2 \)[/tex] is negative and [tex]\( q_1 \)[/tex] is positive, the force [tex]\( F_2 \)[/tex] is attractive, which means it is directed to the left. Therefore:
[tex]\[ F_2 = -443.06316 \, \text{N} \][/tex]
### Step 2: Calculate the Force [tex]\( F_3 \)[/tex] Exerted by [tex]\( q_3 \)[/tex] on [tex]\( q_1 \)[/tex]
Again, using Coulomb's Law:
[tex]\[ F = k \cdot \frac{|q_1 \cdot q_3|}{r_{13}^2} \][/tex]
Given:
[tex]\[ q_1 = 4.44 \times 10^{-6} \, \text{C}, \quad q_3 = 4.44 \times 10^{-6} \, \text{C}, \quad r_{13} = 0.04 \, \text{m} \][/tex]
We get:
[tex]\[ F_3 = 8.99 \times 10^9 \cdot \frac{|4.44 \times 10^{-6} \cdot 4.44 \times 10^{-6}|}{(0.04)^2} \][/tex]
Calculating the magnitude of [tex]\( F_3 \)[/tex], we find [tex]\( F_3 = 110.76579 \, \text{N} \)[/tex].
Since both [tex]\( q_1 \)[/tex] and [tex]\( q_3 \)[/tex] are positive, the force [tex]\( F_3 \)[/tex] is repulsive, which means it is directed to the right. Therefore:
[tex]\[ F_3 = 110.76579 \, \text{N} \][/tex]
### Step 3: Calculate the Net Force [tex]\( \vec{F} \)[/tex] on [tex]\( q_1 \)[/tex]
The net force is the sum of the individual forces, taking direction into account:
[tex]\[ \vec{F} = F_2 + F_3 \][/tex]
Substituting the values:
[tex]\[ \vec{F} = -443.06316 \, \text{N} + 110.76579 \, \text{N} \][/tex]
Calculating the net force, we get:
[tex]\[ \vec{F} = -332.29737 \, \text{N} \][/tex]
Hence, the solution is:
- [tex]\( \vec{F}_2 = -443.06316 \, \text{N} \)[/tex]
- [tex]\( \vec{F}_3 = 110.76579 \, \text{N} \)[/tex]
- Net force on [tex]\( q_1 \)[/tex], [tex]\( \vec{F} = -332.29737 \, \text{N} \)[/tex]
This means the net force on [tex]\( q_1 \)[/tex] is directed to the left with a magnitude of [tex]\( 332.29737 \, \text{N} \)[/tex].
### Given Data:
Charges:
- [tex]\( q_1 = 4.44 \times 10^{-6} \, \text{C} \)[/tex]
- [tex]\( q_2 = -4.44 \times 10^{-6} \, \text{C} \)[/tex] (opposite charge)
- [tex]\( q_3 = 4.44 \times 10^{-6} \, \text{C} \)[/tex] (same charge as [tex]\( q_1 \)[/tex])
Distances:
- Distance between [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex], [tex]\( r_{12} = 0.02 \, \text{m} \)[/tex]
- Distance between [tex]\( q_1 \)[/tex] and [tex]\( q_3 \)[/tex], [tex]\( r_{13} = 0.04 \, \text{m} \)[/tex]
Coulomb's Constant:
- [tex]\( k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)[/tex]
### Step 1: Calculate the Force [tex]\( F_2 \)[/tex] Exerted by [tex]\( q_2 \)[/tex] on [tex]\( q_1 \)[/tex]
Using Coulomb's Law:
[tex]\[ F = k \cdot \frac{|q_1 \cdot q_2|}{r_{12}^2} \][/tex]
Given:
[tex]\[ q_1 = 4.44 \times 10^{-6} \, \text{C}, \quad q_2 = -4.44 \times 10^{-6} \, \text{C}, \quad r_{12} = 0.02 \, \text{m} \][/tex]
We get:
[tex]\[ F_2 = 8.99 \times 10^9 \cdot \frac{|4.44 \times 10^{-6} \cdot (-4.44 \times 10^{-6})|}{(0.02)^2} \][/tex]
Calculating the magnitude of [tex]\( F_2 \)[/tex], we find [tex]\( F_2 = 443.06316 \, \text{N} \)[/tex].
Since [tex]\( q_2 \)[/tex] is negative and [tex]\( q_1 \)[/tex] is positive, the force [tex]\( F_2 \)[/tex] is attractive, which means it is directed to the left. Therefore:
[tex]\[ F_2 = -443.06316 \, \text{N} \][/tex]
### Step 2: Calculate the Force [tex]\( F_3 \)[/tex] Exerted by [tex]\( q_3 \)[/tex] on [tex]\( q_1 \)[/tex]
Again, using Coulomb's Law:
[tex]\[ F = k \cdot \frac{|q_1 \cdot q_3|}{r_{13}^2} \][/tex]
Given:
[tex]\[ q_1 = 4.44 \times 10^{-6} \, \text{C}, \quad q_3 = 4.44 \times 10^{-6} \, \text{C}, \quad r_{13} = 0.04 \, \text{m} \][/tex]
We get:
[tex]\[ F_3 = 8.99 \times 10^9 \cdot \frac{|4.44 \times 10^{-6} \cdot 4.44 \times 10^{-6}|}{(0.04)^2} \][/tex]
Calculating the magnitude of [tex]\( F_3 \)[/tex], we find [tex]\( F_3 = 110.76579 \, \text{N} \)[/tex].
Since both [tex]\( q_1 \)[/tex] and [tex]\( q_3 \)[/tex] are positive, the force [tex]\( F_3 \)[/tex] is repulsive, which means it is directed to the right. Therefore:
[tex]\[ F_3 = 110.76579 \, \text{N} \][/tex]
### Step 3: Calculate the Net Force [tex]\( \vec{F} \)[/tex] on [tex]\( q_1 \)[/tex]
The net force is the sum of the individual forces, taking direction into account:
[tex]\[ \vec{F} = F_2 + F_3 \][/tex]
Substituting the values:
[tex]\[ \vec{F} = -443.06316 \, \text{N} + 110.76579 \, \text{N} \][/tex]
Calculating the net force, we get:
[tex]\[ \vec{F} = -332.29737 \, \text{N} \][/tex]
Hence, the solution is:
- [tex]\( \vec{F}_2 = -443.06316 \, \text{N} \)[/tex]
- [tex]\( \vec{F}_3 = 110.76579 \, \text{N} \)[/tex]
- Net force on [tex]\( q_1 \)[/tex], [tex]\( \vec{F} = -332.29737 \, \text{N} \)[/tex]
This means the net force on [tex]\( q_1 \)[/tex] is directed to the left with a magnitude of [tex]\( 332.29737 \, \text{N} \)[/tex].
