A positive test charge of 5.0 x 10^-4 C is in an electric field that exerts a force of 2.5 x 10^-4 N on it. What is the magnitude of the el

Question

A positive test charge of 5.0 x 10^-4 C is in an electric field that exerts a force of 2.5 x 10^-4 N on it. What is the magnitude of the electric field at the location of the test charge?

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Neala 4 years 2021-09-03T05:23:38+00:00 1 Answers 27 views 0

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  1. thachthao
    0
    2021-09-03T05:25:26+00:00
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    Answer:

    E = 0.5N/C

    Explanation:

    In order to calculate the magnitude of the electric field you use the following formula:

    E=\frac{F}{q}

    q: charge = 5.0*10^-4 C

    F: force on the charge = 2.5*10^-4N

    You replace the values of q and F in the equation for E:

    E=\frac{2.5*10^{-4}N}{5.0*10^{-4}C}=0.5\frac{N}{C}

    hence, the magnitude of the electric field at the position of the carge is 0.5N/C

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