Answer :
To solve the problem, we follow these steps:
1. Identify the given values:
- Charge [tex]\(q_1 = 8.4 \times 10^{-6} \text{ C}\)[/tex]
- Charge [tex]\(q_2 = -3.5 \times 10^{-6} \text{ C}\)[/tex]
- Charge [tex]\(q_3 = -1.21 \times 10^{-6} \text{ C}\)[/tex]
- Distance between [tex]\(q_1\)[/tex] and [tex]\(q_3\)[/tex], [tex]\(r_{12} = 0.14 \text{ m}\)[/tex]
- Distance between [tex]\(q_2\)[/tex] and [tex]\(q_3\)[/tex], [tex]\(r_{23} = 0.18 \text{ m}\)[/tex]
- Coulomb's constant, [tex]\(k = 8.99 \times 10^9 \, \text{N} \text{m}^2 \text{C}^{-2}\)[/tex]
2. Calculate the force exerted on [tex]\(q_3\)[/tex] by [tex]\(q_1\)[/tex] ([tex]\(F_1\)[/tex]):
Using Coulomb's law:
[tex]\[ F_1 = k \frac{|q_1 q_3|}{r_{12}^2} \][/tex]
3. Where [tex]\(q_1\)[/tex] and [tex]\(q_3\)[/tex] attract (opposite charges), [tex]\(F_1\)[/tex] direction is right (positive):
The calculated value of [tex]\(F_1\)[/tex] is:
[tex]\[ F_1 \approx 4.6619571428571405 \, \text{N} \][/tex]
4. Calculate the force exerted on [tex]\(q_3\)[/tex] by [tex]\(q_2\)[/tex] ([tex]\(F_2\)[/tex]):
Using Coulomb's law again:
[tex]\[ F_2 = k \frac{|q_2 q_3|}{r_{23}^2} \][/tex]
5. Where [tex]\(q_2\)[/tex] and [tex]\(q_3\)[/tex] repel (same charges), [tex]\(F_2\)[/tex] direction is right (positive):
The calculated value of [tex]\(F_2\)[/tex] is:
[tex]\[ F_2 \approx 4.6619571428571405 \, \text{N} \][/tex]
6. Calculate the net force on [tex]\(q_3\)[/tex]:
The net force is the vector sum of [tex]\(F_1\)[/tex] and [tex]\(F_2\)[/tex]. Since both forces act to the right and are positive, we have:
[tex]\[ \text{Net force} = F_1 + F_2 \][/tex]
7. The result for the net force is:
[tex]\[ \text{Net force} = 3.4868753527336835 \, \text{N} \][/tex]
In conclusion:
- [tex]\(F_1\)[/tex] and [tex]\(F_2\)[/tex] are both approximately [tex]\(4.6619571428571405 \, \text{N}\)[/tex], directed to the right.
- The net force on [tex]\(q_3\)[/tex] is approximately [tex]\(3.4868753527336835 \, \text{N}\)[/tex] directed to the right.
1. Identify the given values:
- Charge [tex]\(q_1 = 8.4 \times 10^{-6} \text{ C}\)[/tex]
- Charge [tex]\(q_2 = -3.5 \times 10^{-6} \text{ C}\)[/tex]
- Charge [tex]\(q_3 = -1.21 \times 10^{-6} \text{ C}\)[/tex]
- Distance between [tex]\(q_1\)[/tex] and [tex]\(q_3\)[/tex], [tex]\(r_{12} = 0.14 \text{ m}\)[/tex]
- Distance between [tex]\(q_2\)[/tex] and [tex]\(q_3\)[/tex], [tex]\(r_{23} = 0.18 \text{ m}\)[/tex]
- Coulomb's constant, [tex]\(k = 8.99 \times 10^9 \, \text{N} \text{m}^2 \text{C}^{-2}\)[/tex]
2. Calculate the force exerted on [tex]\(q_3\)[/tex] by [tex]\(q_1\)[/tex] ([tex]\(F_1\)[/tex]):
Using Coulomb's law:
[tex]\[ F_1 = k \frac{|q_1 q_3|}{r_{12}^2} \][/tex]
3. Where [tex]\(q_1\)[/tex] and [tex]\(q_3\)[/tex] attract (opposite charges), [tex]\(F_1\)[/tex] direction is right (positive):
The calculated value of [tex]\(F_1\)[/tex] is:
[tex]\[ F_1 \approx 4.6619571428571405 \, \text{N} \][/tex]
4. Calculate the force exerted on [tex]\(q_3\)[/tex] by [tex]\(q_2\)[/tex] ([tex]\(F_2\)[/tex]):
Using Coulomb's law again:
[tex]\[ F_2 = k \frac{|q_2 q_3|}{r_{23}^2} \][/tex]
5. Where [tex]\(q_2\)[/tex] and [tex]\(q_3\)[/tex] repel (same charges), [tex]\(F_2\)[/tex] direction is right (positive):
The calculated value of [tex]\(F_2\)[/tex] is:
[tex]\[ F_2 \approx 4.6619571428571405 \, \text{N} \][/tex]
6. Calculate the net force on [tex]\(q_3\)[/tex]:
The net force is the vector sum of [tex]\(F_1\)[/tex] and [tex]\(F_2\)[/tex]. Since both forces act to the right and are positive, we have:
[tex]\[ \text{Net force} = F_1 + F_2 \][/tex]
7. The result for the net force is:
[tex]\[ \text{Net force} = 3.4868753527336835 \, \text{N} \][/tex]
In conclusion:
- [tex]\(F_1\)[/tex] and [tex]\(F_2\)[/tex] are both approximately [tex]\(4.6619571428571405 \, \text{N}\)[/tex], directed to the right.
- The net force on [tex]\(q_3\)[/tex] is approximately [tex]\(3.4868753527336835 \, \text{N}\)[/tex] directed to the right.