Suppose that the electric potential outside a living cell is higher than that inside the cell by 0.090 V. How much work is done by the electric force when a sodium ion (charge = +e) moves from the outside to the inside?

A particle has a charge of +3.1 μC and moves from point A to point B, a distance of 0.16 m. The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at A and B is EPEA - EPEB = +7.5 x 10-4 J. (a) Find the magnitude of the electric force that acts on the particle. (b) Find the magnitude of the electric field that the particle experiences.

In a television picture tube, electrons strike the screen after being accelerated from rest through a potential difference of 23000 V. The speeds of the electrons are quite large, and for accurate calculations of the speeds, the effects of special relativity must be taken into account. Ignoring such effects, find the electron speed just before the electron strikes the screen.

1 Answer

  • 2 months ago
    Favorite Answer

    1) By W = qV

    =>W = 1.6 x 10^-19 x 0.090 = 1.44 x 10^-20 J

    2)(a) As ΔW = FΔs

    =>7.5 x 10^-4 = F x 0.16

    =>F = 4.69 x 10^-3 N

    (b) By F = qE

    =>4.69 x 10^-3 = 3.1 x 10^-6 x E

    =>E = 1.51 x 10^3 N/C

    3) by work energy relation

    =>W = ΔKE

    =>qV = 1/2mv^2

    =>v = √[2qV/m]

    =>v = √[{2 x 1.6 x 10^-19 x 23000}/{9.1 x 10^-31}]

    =>v = 8.99 x 10^7 m/s

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