Charge moving toward a conductor. Solution makes no physical sense

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SUMMARY

The discussion centers on the dynamics of a point charge q with rest mass m released from a distance d from an infinite grounded conducting plane, as described in Griffiths' "Electromagnetism" textbook. The derived solution for the time it takes for the charge to reach the plane is given by (πd/q) √(2πϵ_0 md). However, the mathematical model suggests that the distance r increases over time, contradicting physical expectations that the charge should accelerate towards the plane. The issue arises from the application of the method of images, where a missing minus sign in the force equation leads to incorrect interpretations of velocity and acceleration.

PREREQUISITES
  • Understanding of electrostatics and point charges
  • Familiarity with the method of images in electrostatics
  • Knowledge of differential equations and their applications in physics
  • Basic concepts of force and motion in classical mechanics
NEXT STEPS
  • Study the method of images in electrostatics for grounded conductors
  • Learn about the implications of force equations in dynamic systems
  • Explore differential equations related to motion under electrostatic forces
  • Investigate the physical interpretations of mathematical solutions in physics
USEFUL FOR

This discussion is beneficial for physics students, educators, and researchers interested in electrostatics, particularly those studying the dynamics of charges near conductors and the application of mathematical models in physical scenarios.

thatsunpossibl
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This is not an assigned homework problem, just something I came across. It's in Griffiths E&M book.
Also I apologize if my equations don't look right. They are showing up all weird on my computer screen.

Homework Statement



A point charge q of rest mass mass m is released from rest a distance d from an infinite grounded conducting plane. How long will it take for the charge to hit the plane?

Homework Equations



F=k (q_1 q_2)/r^2 =m[r]\ddot{}[/itex]


The Attempt at a Solution


The solution is given as

(πd/q) √(2πϵ_0 md)

Griffth implies that we should use the method of images, comparing the problem to two oppositely charged particles moving toward each other. I can get the solution by solving the differential equation above. In fact, mathematically the only solution to that is that r as a function of time comes out to r being some constants times t to the 2/3 power. The problem is that it doesn't seem to be a real physical solution! It implies that r gets bigger as time goes forward, which is not the case. also, if you then differentiate that equation to find v as a function of t, we'll get v as some constants times t to the -1/3 power. This implies that v decays as time goes forward, which can't be true. The charge has to be gaining speed as it falls right? Same with acceleration. Shouldn't it be gaining acceleration as it gets closer to the sheet?



 
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thatsunpossibl said:
This is not an assigned homework problem, just something I came across. It's in Griffiths E&M book.
Also I apologize if my equations don't look right. They are showing up all weird on my computer screen.

Homework Statement



A point charge q of rest mass mass m is released from rest a distance d from an infinite grounded conducting plane. How long will it take for the charge to hit the plane?

Homework Equations



F=k (q_1 q_2)/r^2 =m[itex]\ddot{r}[/itex]

ehild


The Attempt at a Solution


The solution is given as

(πd/q) √(2πϵ_0 md)

Griffth implies that we should use the method of images, comparing the problem to two oppositely charged particles moving toward each other. I can get the solution by solving the differential equation above. In fact, mathematically the only solution to that is that r as a function of time comes out to r being some constants times t to the 2/3 power. The problem is that it doesn't seem to be a real physical solution! It implies that r gets bigger as time goes forward, which is not the case. also, if you then differentiate that equation to find v as a function of t, we'll get v as some constants times t to the -1/3 power. This implies that v decays as time goes forward, which can't be true. The charge has to be gaining speed as it falls right? Same with acceleration. Shouldn't it be gaining acceleration as it gets closer to the sheet?

A minus sign is missing from the equation: The mirror charge is opposite to the real one, so the force points inward, towards the plane.


ehild
 

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