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Mila
3 months ago
14

A proton starts from rest and gains 8.35 x 10^-14 joule of kinetic energy as it accelerates between points A and B in an electri

c field. The potential difference between points A and B in the electric field is approximately

Physics
2 answers:
ValentinkaMS [3.4K]3 months ago
8 0

Answer: The electric potential difference between points A and B is, 5.22\times 10^{5}V

Explanation:

The relationship between kinetic energy and potential difference can be expressed as:

K.E=q\times V

where,

K.E = kinetic energy = 8.35\times 10^{-14}J

q = charge on the proton = 1.602\times 10^{-19}C

V = potential difference =?

Substituting the provided values into the equation, we arrive at:

8.35\times 10^{-14}J=1.602\times 10^{-19}C\times V

V=5.22\times 10^{5}V

Hence, the potential difference between points A and B is, 5.22\times 10^{5}V

Ostrovityanka [3.2K]3 months ago
6 0

Answer:

5.22 x 10^5 V

Explanation:

I made an educated guess on castle learning and ended up being correct

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