Question

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

Answer 1

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

Explanation :

The relation between kinetic energy and potential difference is:

$K.E=q\times V$

where,

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

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

V = potential difference = ?

Now put all the given values in the above formula, we get:

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

$V=5.22\times 10^{5}V$

Therefore, the potential difference between points A and B in the electric field is, $5.22\times 10^{5}V$

Answer 2

Answer:

5.22 x 10^5 V

Explanation:

guessed on castle learning and got it right