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

Question

7 days ago

14

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

2 answers:

ValentinkaMS [1.1K] · Answer 1 · 2026-02-26T01:48:33+03:00

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 [942] · Answer 2 · 2026-02-26T01:45:51+03:00

6 0

Answer:

5.22 x 10^5 V

Explanation:

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