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19 February, 06:06

If a particle with a charge of + 4.3 * 10-18 C is attracted to another particle by a force of 6.5 * 10-8 N, what is the magnitude of the electric field at this location?

2.2 * 1026 NC

2.8 * 10-25 NC

1.5 * 1010 N/C

6.6 * 10-11 N/C

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Answers (1)
  1. 19 February, 09:39
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    Answer: 1.5*10^10 N/C

    Explanation:

    E = F/q

    Where E = magnitude of the electric field

    F = force of attraction

    q = charge of the given body

    Given F = 6.5*10^-8 N

    q = 4.3 * 10^-18 C

    Therefore, E = 6.5*10 ^-8 / 4.3*10^-18

    E = 1.5*10^10 N/C
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