Discover the best answers at Westonci.ca, where experts share their insights and knowledge with you. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
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.
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.