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rewona [7]
3 years ago
9

Two charges, qA and qB, are separated by a distance, d, and exert a force, F, on each other. Analyze Coulomb's law and answer th

e following questions.
(a) What new force will exist if qA is doubled?
(b) What new force will exist if qA and qB are cut in half?
(c) What new force will exist if d is tripled?
(d) What new force will exist if d is cut in half?
(e) What new force will exist if qA is tripled and d is doubled?
Physics
1 answer:
maria [59]3 years ago
6 0

Answer: a. F doubled

b. F reduced by one-quarter i.e

1/4*(F)

c. 1/9*(F)

d. F increased by a factor of 4 i.e 4*F

e. F reduces 3/4*(F)

Explanation: Coulombs law states the force F of attraction/repulsion experience by two charges qA and qB is directly proportional to thier product and inversely proportional to the square of distance d between them. That is

F = k*(qA*qB)/d²

a. If qA is doubled therefore the force is doubled since they are directly proportional.

b. If qA and qB are half, that means thier new product would be qA/2)*qB/2 =qA*qB/4

Which means the product of charge is divided by 4 so the force would be divided by 4 too since they are directly proportional.

c. If d is tripped that is multiplied by 3. From the formula new d would be (3*d)²=9d² but force is inversely proportional to d² so instead of multiplying by 9 the force will be divided by 9

d. If d is cut into half that is divided by 2. The new d would be (d/2)²=d²/4. So d² is divided by 4 so the force would be multiplied by 4

e. If qA is tripled that is multiplied by 3. F would be multiplied by 3 also, if at the same time d is doubled (2*d)²= 4*d² . Force would be divided by 4 at same time. So we have,

3/4*F

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Explanation:

Let KE₁ represent the initial kinetic energy.

Let v₁ represent the initial velocity.

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Let v₂ represent the final velocity.

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Initial kinetic Energy (KE₁) = E

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Final velocity (v₂) =?

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