Answer:
ΔPE= -4.8×10⁻²⁰J
Explanation:
Given data
Electric field of strength E=3.0 N/C
Charge of proton q=1.60×10⁻¹⁹C
Proton moves distance d=10 cm=0.10 m
To find
Change in electrical potential energy ΔPE
Solution
As we know that:
ΔPE= -qEd
[tex]=-(1.60*10^{-19}C )(3.0N/C)(0.10m)\\=-4.8*10^{-20}J[/tex]
ΔPE= -4.8×10⁻²⁰J
Answer:
Explanation:
Given:
D = 10 cm
= 0.1 m
E = 3.0 N/C
Qp = 1.602 × 10^-19 C
U = Q × V
But,
V = E × D
= 3 × 0.1
= 0.3 V
U = 1.602 × 10-19 × 0.3
= 4.806 × 10^-20 J.
