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A charge q moves from point A to point B in an electric field. The potential energy of the charge at point A is 9.4 × 10-13 joules, and its kinetic energy is zero. The kinetic energy at point B is 6.5 × 10-13 joules. What is the potential energy at B?

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Respuesta :

By energy conservation we will say

[tex]PE_i + KE_i = PE_f + KE_f[/tex]

as we know that

[tex]PE_i = 9.4 \times 10^{-13} J[/tex]

[tex]KE_i = 0 J[/tex]

also we have

[tex]KE_f = 6.5 \times 10^{-13}J[/tex]

now we will have

[tex]9.4 \times 10^{-13} + 0 = 6.5 \times 10^{-13} + PE[/tex]

[tex](9.4 - 6.5)\times 10^{-13} J = PE[/tex]

[tex]PE = 2.9 \times 10^{-13} J[/tex]

Answer:

A.  

5.6 × 10-12 joules

B.  

1.3 × 10-12 joules

C.  

2.3 × 10-12 joules

D.  

3.3 × 10-10 joules

E.  

5.6 × 10-10 joules

Explanation:

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