Question
A +5 C lies on a line in between two other charges. In which case will the +5 C charge experience
no net electrostatic force? (1 point)
when a -9 C charge is placed 3 cm to its right and a -1 C is placed 1 cm to its left
when a -3 C charge is placed 3 cm to its right and a -1 C is placed 1 cm to its left
when a -2 C charge is placed 1 cm to its right and a -2 C is placed 2 cm to its left
when a -4 C charge is placed 1 cm to its right and a -2 C is placed 2 cm to its left
no net electrostatic force? (1 point)
when a -9 C charge is placed 3 cm to its right and a -1 C is placed 1 cm to its left
when a -3 C charge is placed 3 cm to its right and a -1 C is placed 1 cm to its left
when a -2 C charge is placed 1 cm to its right and a -2 C is placed 2 cm to its left
when a -4 C charge is placed 1 cm to its right and a -2 C is placed 2 cm to its left
Answers
GPT-4o mini
To determine in which scenario the +5 C charge experiences no net electrostatic force, we need to apply Coulomb's law. The electrostatic force \( F \) between two charges \( Q_1 \) and \( Q_2 \) separated by a distance \( r \) is given by the formula:
\[
F = k \frac{|Q_1 Q_2|}{r^2}
\]
where \( k \) is Coulomb's constant and the force is attractive if the charges have opposite signs and repulsive if they have the same sign.
We will analyze each case to see if the forces on the +5 C charge balance each other out, resulting in no net force.
### Case 1
- **Charge on the right**: -9 C (3 cm away)
- **Charge on the left**: -1 C (1 cm away)
1. **Force due to -1 C** (left):
\[
F_1 = k \frac{5 \times 1}{1^2} = 5k \, \text{(Attractive to the left)}
\]
2. **Force due to -9 C** (right):
\[
F_2 = k \frac{5 \times 9}{3^2} = k \frac{45}{9} = 5k \, \text{(Attractive to the right)}
\]
The forces are equal and opposite. **Result**: Net force = 0.
### Case 2
- **Charge on the right**: -3 C (3 cm away)
- **Charge on the left**: -1 C (1 cm away)
1. **Force due to -1 C** (left):
\[
F_1 = k \frac{5 \times 1}{1^2} = 5k \, \text{(Attractive to the left)}
\]
2. **Force due to -3 C** (right):
\[
F_2 = k \frac{5 \times 3}{3^2} = k \frac{15}{9} = \frac{5}{3}k \, \text{(Attractive to the right)}
\]
The forces do not balance. **Result**: Net force ≠ 0.
### Case 3
- **Charge on the right**: -2 C (1 cm away)
- **Charge on the left**: -2 C (2 cm away)
1. **Force due to -2 C** (right, 1 cm away):
\[
F_1 = k \frac{5 \times 2}{1^2} = 10k \, \text{(Attractive to the right)}
\]
2. **Force due to -2 C** (left, 2 cm away):
\[
F_2 = k \frac{5 \times 2}{2^2} = k \frac{10}{4} = 2.5k \, \text{(Attractive to the left)}
\]
The forces do not balance. **Result**: Net force ≠ 0.
### Case 4
- **Charge on the right**: -4 C (1 cm away)
- **Charge on the left**: -2 C (2 cm away)
1. **Force due to -2 C** (left, 2 cm away):
\[
F_1 = k \frac{5 \times 2}{2^2} = k \frac{10}{4} = 2.5k \, \text{(Attractive to the left)}
\]
2. **Force due to -4 C** (right, 1 cm away):
\[
F_2 = k \frac{5 \times 4}{1^2} = 20k \, \text{(Attractive to the right)}
\]
The forces do not balance. **Result**: Net force ≠ 0.
### Conclusion
The charges and distances in **Case 1** yield a balance of forces, leading to no net electrostatic force on the +5 C charge. Thus, the correct answer is:
**when a -9 C charge is placed 3 cm to its right and a -1 C is placed 1 cm to its left.**
\[
F = k \frac{|Q_1 Q_2|}{r^2}
\]
where \( k \) is Coulomb's constant and the force is attractive if the charges have opposite signs and repulsive if they have the same sign.
We will analyze each case to see if the forces on the +5 C charge balance each other out, resulting in no net force.
### Case 1
- **Charge on the right**: -9 C (3 cm away)
- **Charge on the left**: -1 C (1 cm away)
1. **Force due to -1 C** (left):
\[
F_1 = k \frac{5 \times 1}{1^2} = 5k \, \text{(Attractive to the left)}
\]
2. **Force due to -9 C** (right):
\[
F_2 = k \frac{5 \times 9}{3^2} = k \frac{45}{9} = 5k \, \text{(Attractive to the right)}
\]
The forces are equal and opposite. **Result**: Net force = 0.
### Case 2
- **Charge on the right**: -3 C (3 cm away)
- **Charge on the left**: -1 C (1 cm away)
1. **Force due to -1 C** (left):
\[
F_1 = k \frac{5 \times 1}{1^2} = 5k \, \text{(Attractive to the left)}
\]
2. **Force due to -3 C** (right):
\[
F_2 = k \frac{5 \times 3}{3^2} = k \frac{15}{9} = \frac{5}{3}k \, \text{(Attractive to the right)}
\]
The forces do not balance. **Result**: Net force ≠ 0.
### Case 3
- **Charge on the right**: -2 C (1 cm away)
- **Charge on the left**: -2 C (2 cm away)
1. **Force due to -2 C** (right, 1 cm away):
\[
F_1 = k \frac{5 \times 2}{1^2} = 10k \, \text{(Attractive to the right)}
\]
2. **Force due to -2 C** (left, 2 cm away):
\[
F_2 = k \frac{5 \times 2}{2^2} = k \frac{10}{4} = 2.5k \, \text{(Attractive to the left)}
\]
The forces do not balance. **Result**: Net force ≠ 0.
### Case 4
- **Charge on the right**: -4 C (1 cm away)
- **Charge on the left**: -2 C (2 cm away)
1. **Force due to -2 C** (left, 2 cm away):
\[
F_1 = k \frac{5 \times 2}{2^2} = k \frac{10}{4} = 2.5k \, \text{(Attractive to the left)}
\]
2. **Force due to -4 C** (right, 1 cm away):
\[
F_2 = k \frac{5 \times 4}{1^2} = 20k \, \text{(Attractive to the right)}
\]
The forces do not balance. **Result**: Net force ≠ 0.
### Conclusion
The charges and distances in **Case 1** yield a balance of forces, leading to no net electrostatic force on the +5 C charge. Thus, the correct answer is:
**when a -9 C charge is placed 3 cm to its right and a -1 C is placed 1 cm to its left.**