calculate the magnitude of the electric field intensity in vacuum at a distance of 20 cm from a charge of 5 * 10 raise to power – 3 column​

Question

calculate the magnitude of the electric field intensity in vacuum at a distance of 20 cm from a charge of 5 * 10 raise to power – 3 column​

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Philomena 5 years 2021-08-11T17:43:33+00:00 1 Answers 18 views 0

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  1. Farah
    0
    2021-08-11T17:45:12+00:00
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    Answer:

    1.1259*10^9 Newton per Columb

    Explanation:

    the magnitude of the electric field intensity can be calculated using the expresion below;

    E=Kq/r^2

    Where k= constant

    q= electric charge

    r=distance= 2cm= 20*10^-2m( we convert to m for unit consistency

    :,K=59*10^9 Columb

    If we substitute the value into above formula we have

    E=(9*10^9)*(5*10^-3)/(20*10^-2)^2

    =1.1259*10^9 Newton per Columb

    Therefore,the magnitude of the electric field intensity in vacuum at a distance of 20 cm is 1.1259*10^9 Newton per Columb

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