Answer :
1.) To determine if two charges are attracting or repelling, we need to look at their signs.
- The first charge \( q_1 = -5.0 \mu C \)
- The second charge \( q_2 = -3.0 \mu C \)
Both charges are negative. Charges with the same sign repel each other. Therefore, these two charges are repelling each other.
Next, we need to calculate the electric force between the two charges. We will use Coulomb's Law for this calculation, which is given by:
[tex]\[ F = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{|q_1 \cdot q_2|}{r^2} \][/tex]
where:
- \( F \) is the magnitude of the force between the charges,
- \( \varepsilon_0 \) is the permittivity of free space (\( \varepsilon_0 = 8.854 \times 10^{-12} \) F/m),
- \( q_1 \) and \( q_2 \) are the magnitudes of the two charges,
- \( r \) is the distance between the centers of the two charges.
Given:
- \( q_1 = -5.0 \times 10^{-6} \) C (converting \(\mu C\) to C)
- \( q_2 = -3.0 \times 10^{-6} \) C (converting \(\mu C\) to C)
- Distance \( r = 35 \) cm = 0.35 m (converting cm to m)
Now substituting the given values into Coulomb’s Law:
[tex]\[ F = \frac{1}{4 \pi \cdot 8.854 \times 10^{-12}} \cdot \frac{|(-5.0 \times 10^{-6}) \cdot (-3.0 \times 10^{-6})|}{(0.35)^2} \][/tex]
After performing the calculations, we find that the electric force \( F \) is approximately \( 1.1005 \) N.
Therefore, the two charges are repelling each other with an electric force of approximately [tex]\( 1.1005 \)[/tex] Newtons.
- The first charge \( q_1 = -5.0 \mu C \)
- The second charge \( q_2 = -3.0 \mu C \)
Both charges are negative. Charges with the same sign repel each other. Therefore, these two charges are repelling each other.
Next, we need to calculate the electric force between the two charges. We will use Coulomb's Law for this calculation, which is given by:
[tex]\[ F = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{|q_1 \cdot q_2|}{r^2} \][/tex]
where:
- \( F \) is the magnitude of the force between the charges,
- \( \varepsilon_0 \) is the permittivity of free space (\( \varepsilon_0 = 8.854 \times 10^{-12} \) F/m),
- \( q_1 \) and \( q_2 \) are the magnitudes of the two charges,
- \( r \) is the distance between the centers of the two charges.
Given:
- \( q_1 = -5.0 \times 10^{-6} \) C (converting \(\mu C\) to C)
- \( q_2 = -3.0 \times 10^{-6} \) C (converting \(\mu C\) to C)
- Distance \( r = 35 \) cm = 0.35 m (converting cm to m)
Now substituting the given values into Coulomb’s Law:
[tex]\[ F = \frac{1}{4 \pi \cdot 8.854 \times 10^{-12}} \cdot \frac{|(-5.0 \times 10^{-6}) \cdot (-3.0 \times 10^{-6})|}{(0.35)^2} \][/tex]
After performing the calculations, we find that the electric force \( F \) is approximately \( 1.1005 \) N.
Therefore, the two charges are repelling each other with an electric force of approximately [tex]\( 1.1005 \)[/tex] Newtons.
