Answer:
Explanation:
The magnitude of electric field = 8 x 10⁴ V /m
there is a potential difference of 8 x 10⁴ V on a separation of 1 m
so on a separation of .5 m , potential drop or change in potential will be equal to
.5 x 8 x 10⁴
= 4 x 10⁴ V .
The increase in kinetic energy of proton = V X Q
V is potential drop x Q is charge on proton
= 4 x 10⁴ x 1.6 x 10⁻¹⁹
= 6.4 x 10⁻¹⁵ J
potential energy of the proton-field system will be correspondingly decreased by the same amount or by an amount of
- 6.4 x 10⁻¹⁵ J .
