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In summary, the conversation discussed the concept of finding the minimum value of q/m for a small particle to be in equilibrium under the action of gravity and electrostatic force when located on the axis of a uniformly charged ring with a negative charge of -Q and a radius of a in a horizontal plane. The solution involved expressing q/m as a function of x using the given variables, finding the position of minimum for the function, and replacing it back into the formula for q/m. Though a difficult process, the correct answer was achieved.
- #1
Jimmy25
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Homework Statement
A uniformly charged ring has a radius a, lies in a horizontal plane, and has a negative charge given by -Q. A small particle of mass m has a positive charge given by q. The small particle is located on the axis of the ring.
What is the minimum value of q/m such that the particle will be in equilibrium under the action of gravity and the electrostatic force? (Use the following variables as necessary: a, k, Q, and g.)?
Homework Equations
I let x = the distance the charge is from the center of the ring.
[tex]E[/tex]=[tex]\frac{kQx}{\sqrt{a^2+x^2}^{3}}[/tex]
The Attempt at a Solution
I started by saying that "mg" must be equal to:
[tex]\frac{kQxq}{\sqrt{a^2+x^2}^{3}}[/tex]
However, I cannot eliminate the x variable from the problem. Am I missing something or do I have the wrong approach altogether?
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- #2
ehild
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You do not need to eliminate x. Express q/m as function of x from the condition qE=mg, and find where this function has its minimum.
ehild
- #3
Jimmy25
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I can express q/m as a function of x however it is also a function of several other variables which have no numerical values. How do I go about finding a minimum for the function algebraically?
- #4
ehild
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"Use the following variables as necessary: a, k, Q, and g".
You need to give a formula, not a number.
ehild
- #5
Jimmy25
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Okay. But how do I find a formula for the minimum using only the variables a, k, Q, and g? I tried to take a derivative (thinking I may be able to then find a local minimum) but it got really nasty.
Am I missing something? - Another way to find the minimum?
- #6
ehild
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[tex]mg = \frac{kQxq}{\sqrt{a^2+x^2}^{3}} \rightarrow q/m= \frac{g}{kQ}\frac{(x^2+a^2)^{3/2}}{x}[/tex]
Find the position of minimum of the function.
[tex]f(x)=\frac{(x^2+a^2)^{3/2}}{x}[/tex]
Replace back this x into the formula for q/m.
ehild
- #7
Jimmy25
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I got it! I had to grunt it out and it turns out the answer was not that pretty but it was correct nonetheless. Thanks for your help!
- #8
ehild
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Congratulation! It was not easy to get the result, but you did it!
1. What is the equation for calculating the force due to a uniformly charged ring?
The force due to a uniformly charged ring can be calculated using the equation F = (k*q*Q*r) / (x^2 + r^2)^1.5, where k is the Coulomb's constant, q is the test charge, Q is the charge of the ring, r is the radius of the ring, and x is the distance between the ring and the test charge.
2. How does the force vary with distance from the ring?
The force due to a uniformly charged ring follows an inverse square law, meaning that as the distance from the ring increases, the force decreases. This relationship is represented by the x^2 in the denominator of the equation.
3. What is the direction of the force due to a uniformly charged ring?
The direction of the force due to a uniformly charged ring is always perpendicular to the plane of the ring, passing through the center of the ring and the test charge. This means that the force will either be pulling the test charge towards the ring or pushing it away, depending on the charges of the ring and the test charge.
4. Can the force due to a uniformly charged ring be repulsive?
Yes, the force due to a uniformly charged ring can be either attractive or repulsive, depending on the charges of the ring and the test charge. If the charges are the same, the force will be repulsive, and if they are opposite, the force will be attractive.
5. What is the relationship between the force due to a uniformly charged ring and the charge and radius of the ring?
The force due to a uniformly charged ring is directly proportional to the product of the charges of the ring and the test charge, and inversely proportional to the distance between them squared. This means that the force will increase with the charges and decrease with the distance between them. The radius of the ring also affects the force, as a larger radius will result in a stronger force due to the increased amount of charge on the ring.
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