5. A charge moves a distance of 2.0 cm in the direction of a uniform electric field having a magnitude of 215 N/C. The electrical potential

Question

5. A charge moves a distance of 2.0 cm in the direction of a uniform electric field having a magnitude of 215 N/C. The electrical potential energy of the charge decreases by 6.9 x10-19 J as it moves. Find the magnitude of the charge on the moving particle. (Hint: The electrical potential energy depends on the distance moved in the direction of the field.)

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Vodka 4 years 2021-07-28T20:54:12+00:00 1 Answers 461 views 0

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  1. mit
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    2021-07-28T20:55:18+00:00
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    Answer:

    The magnitude of charge on the particle is 1.6\times 10^{-19}\ C.        

    Explanation:

    Given that,

    The magnitude of electric field, E = 215 N/C

    The electrical potential energy of the charge decreases by, -6.9\times 10^{-19}\ J as it moves.

    We need to find the magnitude of the charge on the moving particle. The change in electric potential energy is given by :

    \Delta U=q\Delta V\\\\q=\dfrac{\Delta U}{\Delta V}………(1)

    The electric potential in terms of electric field is given by :

    \Delta V=-Ed\\\\\Delta V=-215\times 0.02=-4.3\ V

    Equation (1) becomes :

    q=\dfrac{-6.9\times 10^{-19}}{-4.3}\\\\q=1.6\times 10^{-19}\ C

    So, the magnitude of charge on the particle is 1.6\times 10^{-19}\ C.                                    

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