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mixer [17]
4 years ago
9

A positive point charge Q is fixed on a very large horizontal frictionless tabletop. A second positive point charge q is release

d from rest near the stationary charge and is free to move. Which statement best describes the motion of q after it is released? A) Its speed will be greatest just after it is released. B) As it moves farther and farther from Q, its speed will decrease. C) As it moves farther and farther from Q, its speed will keep increasing. D) Its acceleration is zero just after it is released. E) As it moves farther and farther from Q, its acceleration will keep increasing.
Physics
1 answer:
andrezito [222]4 years ago
4 0

Explanation:

A positive point charge Q is fixed on a very large horizontal friction less tabletop. A second positive point charge q is released from rest near the stationary charge and is free to move.

As per the law of conservation of energy the change in kinetic energy is equal to the change in potential energy.

The mathematical expression for the conservation of energy is given by :

\dfrac{1}{2}mv^2=-k\dfrac{q_1q_2}{r_2^2}-(-k\dfrac{q_1q_2}{r_1^2})

On solving the above equation, we get the value of v as :

v=\sqrt{\dfrac{2kq_1q_2}{m}(\dfrac{1}{r_1^2}-\dfrac{1}{r_2^2})}

In the above expression, as the term (\dfrac{1}{r_1^2}-\dfrac{1}{r_2^2}) increase, as r_2 increases, the value of \dfrac{1}{r_2} decreases. So, the correct option is (B) "As it moves farther and farther from Q, its speed will increase".

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b).

tan(theta) = 6400 km/(4.0*10^6 km)

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c).

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