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pishuonlain [190]
3 years ago
9

Two small objects each with a net charge of Q (positive) exert a force of magnitude F on each other. We replace one of the objec

ts with another whose net charge is 4Q. The original magnitude of the force on the Q charge was F; what is the magnitude of the force on the Q now?
A. 16F
B. 4F
C. F
D. F/4
Physics
1 answer:
erik [133]3 years ago
5 0

Answer:

option B

Explanation:

we know,

Force between two charge is calculated by

F = \dfrac{kq_1q_2}{r^2}

r is the distance between the two charges

k is Coulomb constant.

give two small object with charge Q

now, Force

F = \dfrac{kQQ}{r^2}......(1)

now,

New force if another charge is equal to 4 Q

F' = \dfrac{kQ(4Q)}{r^2}

F' = 4\dfrac{kQQ}{r^2}

now from equation 1

F' = 4 F

Hence, the correct answer is option B

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Explanation:

The statement is incomplete. We proceed to present the complete statement: <em>A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if </em><em>a. </em><em>The mass is doubled? </em><em>b.</em><em> The mass is halved? </em><em>c.</em><em> The amplitude is doubled? </em><em>d.</em><em> The spring constant is doubled? </em>

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c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

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