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Sagot :
To find the kinetic energy of the charge at point B, we can use the principle of conservation of energy. According to this principle, the total energy at point A must be equal to the total energy at point B. The total energy at any point is the sum of the potential energy and the kinetic energy at that point.
Given data:
- Potential energy at point A ([tex]\(PE_A\)[/tex]) = [tex]\(5.6 \times 10^{-10}\)[/tex] joules
- Potential energy at point B ([tex]\(PE_B\)[/tex]) = [tex]\(2.3 \times 10^{-10}\)[/tex] joules
- Kinetic energy at point A ([tex]\(KE_A\)[/tex]) = 0 joules
Applying the conservation of energy:
[tex]\[ PE_A + KE_A = PE_B + KE_B \][/tex]
Substitute the known values:
[tex]\[ (5.6 \times 10^{-10} \, \text{joules}) + (0 \, \text{joules}) = (2.3 \times 10^{-10} \, \text{joules}) + KE_B \][/tex]
To solve for [tex]\(KE_B\)[/tex], rearrange the equation:
[tex]\[ KE_B = (5.6 \times 10^{-10} \, \text{joules}) - (2.3 \times 10^{-10} \, \text{joules}) \][/tex]
[tex]\[ KE_B = 3.3 \times 10^{-10} \, \text{joules} \][/tex]
So, the kinetic energy at point B is [tex]\(3.3 \times 10^{-10}\)[/tex] joules.
The correct answer is:
D. [tex]\(3.3 \times 10^{-10}\)[/tex] joules
Given data:
- Potential energy at point A ([tex]\(PE_A\)[/tex]) = [tex]\(5.6 \times 10^{-10}\)[/tex] joules
- Potential energy at point B ([tex]\(PE_B\)[/tex]) = [tex]\(2.3 \times 10^{-10}\)[/tex] joules
- Kinetic energy at point A ([tex]\(KE_A\)[/tex]) = 0 joules
Applying the conservation of energy:
[tex]\[ PE_A + KE_A = PE_B + KE_B \][/tex]
Substitute the known values:
[tex]\[ (5.6 \times 10^{-10} \, \text{joules}) + (0 \, \text{joules}) = (2.3 \times 10^{-10} \, \text{joules}) + KE_B \][/tex]
To solve for [tex]\(KE_B\)[/tex], rearrange the equation:
[tex]\[ KE_B = (5.6 \times 10^{-10} \, \text{joules}) - (2.3 \times 10^{-10} \, \text{joules}) \][/tex]
[tex]\[ KE_B = 3.3 \times 10^{-10} \, \text{joules} \][/tex]
So, the kinetic energy at point B is [tex]\(3.3 \times 10^{-10}\)[/tex] joules.
The correct answer is:
D. [tex]\(3.3 \times 10^{-10}\)[/tex] joules
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