To determine the relative strength of the electric fields created by different pairs of objects, we can use Coulomb's law, which states that the electric force \( F \) between two charges is given by:
\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \]
Where:
- \( F \) is the force between the charges,
- \( k \) is Coulomb’s constant (approximately \( 8.99 \times 10^9 , \text{N m}^2/\text{C}^2 \)),
- \( |q_1| \) and \( |q_2| \) are the magnitudes of the charges,
- \( r \) is the distance between the charges.
Analyzing Each Case
-
Cotton/Hand: Charge (+1)(-1), Distance 10 cm
- Magnitude of Charges: \( |q_1| = 1 \), \( |q_2| = 1 \)
- Distance: \( r = 10 \text{ cm} = 0.1 \text{ m} \)
- Force relationship: \( F = k \frac{1 \cdot 1}{(0.1)^2} = k \frac{1}{0.01} = k \cdot 100 \)
- Therefore, the strength of the field is Strong.
-
Cotton/Iron: Charge (+1)(-5), Distance 3 cm
- Magnitude of Charges: \( |q_1| = 1 \), \( |q_2| = 5 \)
- Distance: \( r = 3 \text{ cm} = 0.03 \text{ m} \)
- Force relationship: \( F = k \frac{1 \cdot 5}{(0.03)^2} = k \frac{5}{0.0009} = k \cdot 5555.56 \)
- Therefore, the strength of the field is Strong.
-
Iron/Iron: Charge (+5)(-2), Distance 7 cm
- Magnitude of Charges: \( |q_1| = 5 \), \( |q_2| = 2 \)
- Distance: \( r = 7 \text{ cm} = 0.07 \text{ m} \)
- Force relationship: \( F = k \frac{5 \cdot 2}{(0.07)^2} = k \frac{10}{0.0049} = k \cdot 2040.82 \)
- Therefore, the strength of the field is Strong.
Summary Table
| Object Material | Charge | Distance | Strength of Field | |-----------------|----------------|----------|--------------------| | Cotton/Hand | (+1)(-1) | 10 cm | Strong | | Cotton/Iron | (+1)(-5) | 3 cm | Strong | | Iron/Iron | (+5)(-2) | 7 cm | Strong |
All three cases result in a Strong electric field based on the charge magnitudes and distances provided.