Suppose two point charges, Q1 = +3.37 x 10^-6 C and Q2 = -8.21 x 10^-6 C, are separated by a distance d = 1.99 m. Using

Question

Suppose two point charges, Q1 = +3.37 x 10^-6 C and Q2 = -8.21 x 10^-6 C, are separated by a distance d = 1.99 m.

Using Coulomb’s Law, find the attractive force between the charges.

A)
F = 0.0626 N

B)
F = 0.0313 N

C)
F = -0.0626 N

D)
F = -0.0313 N

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Adela 5 months 2021-08-09T14:13:27+00:00 1 Answers 6 views 0

Answers ( )

  1. ngocdiep
    0
    2021-08-09T14:15:23+00:00
    Reply

    Option (C) is the correct one.

    Explanation:
    Coulmbs law here is,

    \frac{9 \times {10}^{9} \times ( - 8.21 \times {10}^{ - 6} )(3.37 \times {10}^{ - 6} )}{ {(1.99)}^{2} } \\ \\ = - 0.0628 \\

    Which is almost equal to optionC.

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