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Two small charged objects repel each other with a force F when separated by a distance d. If the charge on each object is reduced to 1/ 14 of its original value and the distance between them is reduced to d/ 24 the force becomes Group of answer choices F F * 24 / 196 F * 576 / 14 F * 576 / 196 F * 196 / 576

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cjrodriguez89

Answer:

Ff = F₀ *(576/196)

Explanation:

  • Assuming that both charges are equal each other, we can express the repulsion force between the charges (assuming that we can treat them as point charges) using Coulomb's Law, as follows:

       [tex]F_{o} = \frac{k*q^{2} }{d^{2}} (1)[/tex]

  • Now, if q reduces to q/14, and d is reduced to d/24, the new value of the force will be:

       [tex]F_{f} = \frac{k*(q/14)^{2} }{(d/24)^{2}} = \frac{k*q^{2}}{d^{2}} * \frac{(24)^{2}}{(14)^{2}} (2)[/tex]

       ⇒[tex]F_{f} =\frac{k*q^{2}}{d^{2}} * \frac{(24)^{2}}{(14)^{2}} = F_{o} * \frac{576}{196} (3)[/tex]

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