Consider two points in an electric field. The potential at point 1, V1, is 44 V. The potential at point 2, V2, is 175 V. An electron at rest

Question

Consider two points in an electric field. The potential at point 1, V1, is 44 V. The potential at point 2, V2, is 175 V. An electron at rest at point 1 is accelerated by the electric field to point 2.
(a) Write an equation for the change of electric potential energy ΔU of the electron in terms of the symbols given.
(b) Find the numerical value of the change of the electric potential energy in electron volts (eV).

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Mít Mít 4 years 2021-08-20T09:23:29+00:00 1 Answers 36 views 0

Answers ( )

  1. thaiduong
    0
    2021-08-20T09:25:23+00:00
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    Answer:

    Ans 1 : The change of electric potential energy \Delta U= q(V_{2} -V_{1} ).

    Ans 2 : Numerical value of the \Delta U = 131 eV

    Explanation:

    Given :

    The potential of one end (V_{1} ) = 44 V

    The potential of other end (V_{2} ) = 175 V

    We know that electric potential is the electric potential energy per unit charge.

    Ans 1 :

    So we write that, \Delta V = \frac{\Delta U}{q}

    ∴ \Delta U = q \Delta U

    Here, electron moves from higher potential to lower potential.

    ∴  \Delta U = q (V_{2} -V_{1} )

    Ans 2 :

    For numerical value we have to only subtract two potential,

    ⇒  \Delta U = (175 -44) eV

        \Delta U = 131eV

    Where e = 1.6 \times 10^{-19} charge of electron.

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