A proton moves 10.0 cm on a path parallel to the direction of a uniform electric field of strength 3.0 N/C. What is the change in electrical

Question

A proton moves 10.0 cm on a path parallel to the direction of a uniform electric field of strength 3.0 N/C. What is the change in electrical potential energy?

in progress 0
Thiên Di 3 years 2021-09-02T13:23:22+00:00 2 Answers 88 views 0

Answers ( )

  1. lethu
    0
    2021-09-02T13:24:23+00:00
    Reply

    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

    =-(1.60*10^{-19}C )(3.0N/C)(0.10m)\\=-4.8*10^{-20}J

    ΔPE= -4.8×10⁻²⁰J

  2. Neala
    0
    2021-09-02T13:25:14+00:00
    Reply

    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.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )