physics electric charges and fields?

An electron a proton fall from rest in a uniform electric field 3×10^4 N/c. electrom takes 3×10^-9 s and proton takes 1×10^-7 compare the distance of the fall

1 Answer

  • NCS
    Lv 7
    1 month ago

    both have charge magnitude

    |q| = 1.6e-19 C

    and so both experience a force of

    |F| = |q| * E = 4.8e-15 N

    electron distance

    d = ½at² = ½(F/m)*t² = ½(4.8e-15N / 9.11e-31kg)*(3e-9s)²

    d = 2.37e-3 m ≈ 2 cm

    proton distance

    D = ½(4.8e-15N / 1.67e-27kg)(3e-9s)²

    D = 1.29e-5 m ≈ 13 µm

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