Calculate the electric force between two point charges of -54.0 μC and +36.0 μC when they are 2.00 cm apart. **note — the metr

Question

Calculate the electric force between two point charges of -54.0 μC and +36.0 μC when they are 2.00 cm apart.

**note — the metric prefix “micro” = μ = x10 – 6 **

Is this attractive or repulsive?

Can someone help me please? Thank you!!!!

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Khang Minh 4 years 2021-08-01T05:12:28+00:00 1 Answers 32 views 0

Answers ( )

    0
    2021-08-01T05:14:02+00:00

    Answer:

    43740 N and attractive force

    Explanation:

    Given that,

    Charge 1, q₁ = -54.0 μC

    Charge 2, q₂ = +36.0 μC

    The distance between charges, r = 2 cm = 0.02 m

    We need to find the force between charges. The formula for the force between charges is given by :

    F=k\dfrac{q_1q_2}{r^2}

    Put all the values,

    F=9\times 10^9\times \dfrac{54\times 10^{-6}\times 36\times 10^{-6}}{(0.02)^2}\\\\F=43740\ N

    As the charges are opposite, the force between them is attractive. Hence, the required force is 43740 N.

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )