A positive test charge q is released near a positive fixed charge Q.

In summary, the question about a charged particle being released in a space with or without friction has been causing confusion. However, it seems that the question is most likely referring to a scenario with no friction, as there is no mention of a coefficient of friction. This means that the only force to consider would be the electric force on the particle, leading to a decreasing acceleration.
  • #1
Curious314
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0
I am not sure if am suppose to consider this charge q, released in the space where there is no friction, or if it is released where there is friction... am I too far from the answer? Because I would assume that it will be in decreasing acceleration. Very confuse with all this topic.

Thanks!
 

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  • #2
I would assume there is no friction in this question. Usually intro physics questions will specify EVERYTHING you need to solve it. Since there is no [itex]\mu[/itex], I would guess that you should only consider the electric force on the test particle.
 
  • #3
that's what i thought. Thank you very much!
 

What is the force experienced by the test charge?

The force experienced by the test charge is given by Coulomb's Law, which states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the test charge will experience a repulsive force from the fixed charge, since both charges are positive.

How does the distance between the two charges affect the force?

The force between the two charges is inversely proportional to the square of the distance between them. This means that as the distance between the charges increases, the force decreases. Conversely, as the distance decreases, the force increases.

What happens to the velocity of the test charge as it approaches the fixed charge?

The velocity of the test charge will decrease as it approaches the fixed charge due to the repulsive force between the two charges. This decrease in velocity is a result of the work done by the electric field in moving the test charge.

Will the test charge ever come in contact with the fixed charge?

No, the test charge will never come in contact with the fixed charge. This is because the electrostatic force between the two charges will continue to repel the test charge, preventing it from reaching the fixed charge. However, the force may become strong enough to significantly slow down the test charge's motion as it approaches the fixed charge.

How can the electric field between the two charges be calculated?

The electric field between the two charges can be calculated by dividing the force on the test charge by the magnitude of the test charge. This will give the electric field strength at any given point between the two charges. Alternatively, the electric field can also be calculated using the formula E = kQ/r^2, where k is the Coulomb's constant, Q is the fixed charge, and r is the distance between the two charges.

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