Answer :
Sure, let's go through the details of the calculation step-by-step.
Given values:
- Charge \( q_1 = -75.8 \times 10^{-6} \) C
- Charge \( q_2 = 75.8 \times 10^{-6} \) C (same magnitude as \( q_1 \))
- Charge \( q_3 = 75.8 \times 10^{-6} \) C (same magnitude as \( q_1 \))
We also assume:
- The distance between \( q_1 \) and \( q_3 \) is \( 0.1 \) m
- The distance between \( q_2 \) and \( q_3 \) is \( 0.1 \) m
Coulomb's constant \( k = 8.99 \times 10^9 \, \text{N m}^2 / \text{C}^2 \).
### Step 1: Calculate Force \( F_1 \)
Force \( F_1 \) is the force exerted on \( q_3 \) by \( q_1 \).
Using Coulomb's law:
[tex]\[ F_1 = k \frac{|q_1 q_3|}{r^2} \][/tex]
Substitute the given values:
[tex]\[ F_1 = 8.99 \times 10^9 \, \frac{|(-75.8 \times 10^{-6}) \times (75.8 \times 10^{-6})|}{(0.1)^2} \][/tex]
Result:
[tex]\[ F_1 = 5165.33 \text{ N (rounding to 2 decimal places for simplicity)} \][/tex]
Since \( q_1 \) is negative and \( q_3 \) is positive, the force \( F_1 \) will be attractive, pulling \( q_3 \) to the left, hence:
[tex]\[ \vec{F}_1 = -5165.33 \text{ N} \][/tex]
### Step 2: Calculate Force \( F_2 \)
Force \( F_2 \) is the force exerted on \( q_3 \) by \( q_2 \).
Using Coulomb's law:
[tex]\[ F_2 = k \frac{|q_2 q_3|}{r^2} \][/tex]
Substitute the given values:
[tex]\[ F_2 = 8.99 \times 10^9 \, \frac{|(75.8 \times 10^{-6}) \times (75.8 \times 10^{-6})|}{(0.1)^2} \][/tex]
Result:
[tex]\[ F_2 = 5165.33 \text{ N (rounding to 2 decimal places for simplicity)} \][/tex]
Since both \( q_2 \) and \( q_3 \) are positive, the force \( F_2 \) will be repulsive, pushing \( q_3 \) to the right, hence:
[tex]\[ \vec{F}_2 = 5165.33 \text{ N} \][/tex]
### Step 3: Calculate Net Force \( F \) on \( q_3 \)
The net force \( \vec{F} \) is the sum of the forces \( \vec{F}_1 \) and \( \vec{F}_2 \):
[tex]\[ \vec{F} = \vec{F}_1 + \vec{F}_2 \][/tex]
Substitute the forces:
[tex]\[ \vec{F} = -5165.33 \text{ N} + 5165.33 \text{ N} \][/tex]
Result:
[tex]\[ \vec{F} = 10330.66 \text{ N} \][/tex]
### Summary
[tex]\[ \begin{aligned} \vec{F}_1 & = -5165.33 \text{ N (left)} \\ \vec{F}_2 & = 5165.33 \text{ N (right)} \\ \vec{F} & = 10330.66 \text{ N (net force to the right)} \\ \end{aligned} \][/tex]
Given values:
- Charge \( q_1 = -75.8 \times 10^{-6} \) C
- Charge \( q_2 = 75.8 \times 10^{-6} \) C (same magnitude as \( q_1 \))
- Charge \( q_3 = 75.8 \times 10^{-6} \) C (same magnitude as \( q_1 \))
We also assume:
- The distance between \( q_1 \) and \( q_3 \) is \( 0.1 \) m
- The distance between \( q_2 \) and \( q_3 \) is \( 0.1 \) m
Coulomb's constant \( k = 8.99 \times 10^9 \, \text{N m}^2 / \text{C}^2 \).
### Step 1: Calculate Force \( F_1 \)
Force \( F_1 \) is the force exerted on \( q_3 \) by \( q_1 \).
Using Coulomb's law:
[tex]\[ F_1 = k \frac{|q_1 q_3|}{r^2} \][/tex]
Substitute the given values:
[tex]\[ F_1 = 8.99 \times 10^9 \, \frac{|(-75.8 \times 10^{-6}) \times (75.8 \times 10^{-6})|}{(0.1)^2} \][/tex]
Result:
[tex]\[ F_1 = 5165.33 \text{ N (rounding to 2 decimal places for simplicity)} \][/tex]
Since \( q_1 \) is negative and \( q_3 \) is positive, the force \( F_1 \) will be attractive, pulling \( q_3 \) to the left, hence:
[tex]\[ \vec{F}_1 = -5165.33 \text{ N} \][/tex]
### Step 2: Calculate Force \( F_2 \)
Force \( F_2 \) is the force exerted on \( q_3 \) by \( q_2 \).
Using Coulomb's law:
[tex]\[ F_2 = k \frac{|q_2 q_3|}{r^2} \][/tex]
Substitute the given values:
[tex]\[ F_2 = 8.99 \times 10^9 \, \frac{|(75.8 \times 10^{-6}) \times (75.8 \times 10^{-6})|}{(0.1)^2} \][/tex]
Result:
[tex]\[ F_2 = 5165.33 \text{ N (rounding to 2 decimal places for simplicity)} \][/tex]
Since both \( q_2 \) and \( q_3 \) are positive, the force \( F_2 \) will be repulsive, pushing \( q_3 \) to the right, hence:
[tex]\[ \vec{F}_2 = 5165.33 \text{ N} \][/tex]
### Step 3: Calculate Net Force \( F \) on \( q_3 \)
The net force \( \vec{F} \) is the sum of the forces \( \vec{F}_1 \) and \( \vec{F}_2 \):
[tex]\[ \vec{F} = \vec{F}_1 + \vec{F}_2 \][/tex]
Substitute the forces:
[tex]\[ \vec{F} = -5165.33 \text{ N} + 5165.33 \text{ N} \][/tex]
Result:
[tex]\[ \vec{F} = 10330.66 \text{ N} \][/tex]
### Summary
[tex]\[ \begin{aligned} \vec{F}_1 & = -5165.33 \text{ N (left)} \\ \vec{F}_2 & = 5165.33 \text{ N (right)} \\ \vec{F} & = 10330.66 \text{ N (net force to the right)} \\ \end{aligned} \][/tex]
