Find min velocity so that particle grazes the shell

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Homework Help Overview

The problem involves a particle with mass and charge projected towards a non-conducting spherical shell with the same charge. The objective is to determine the minimum initial velocity required for the particle to just graze the shell.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the conservation of energy and angular momentum in the context of the problem. There is a question about whether the velocity of the particle can be assumed to be zero at the surface of the shell when determining the minimum velocity.

Discussion Status

Some participants have offered insights into the conservation of angular momentum and energy, suggesting that the velocity does not need to be zero at the surface. There is an ongoing exploration of the implications of these conservation laws on the initial velocity required.

Contextual Notes

Participants express uncertainty about the concepts of angular momentum and central forces, indicating a need for clarification on these topics as they relate to the problem.

Saitama
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Homework Statement


A particle of mass 1 kg and charge 1/3 μC is projected towards a non conducting fixed spherical shell having the same charge uniformly distributed on its surface. Find the minimum initial velocity of projection required if the particle just grazes the shell.
1zv7vnr.jpg

a)[itex]\sqrt{\frac{2}{3}}[/itex]
b)[itex]2\sqrt{\frac{2}{3}}[/itex]
c)[itex]\frac{2}{3}[/itex]
d)none

Homework Equations


The Attempt at a Solution


Since the charge is uniformly distributed over the shell, the potential at any point on the surface or inside the shell is KQ/R, where R is the radius. Therefore,
[tex]\frac{1}{2}mu^2=\frac{KQq}{R}[/tex]
where u is the initial velocity and Q and q are the charges of shell and particle respectively. Plugging the values, i get
[tex]u=\sqrt{2}[/tex]

But this is wrong, please help me proceed in the right direction.
 
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The velocity of the charged particle (strange particle it is, with 1 kg mass:smile:) need not be zero at the surface of the sphere: remember that not only the energy conserves in a central field but the angular momentum, too.

ehild
 
ehild said:
The velocity of the charged particle (strange particle it is, with 1 kg mass:smile:) need not be zero at the surface of the sphere: remember that not only the energy conserves in a central field but the angular momentum, too.

ehild

Why can't we assume that velocity becomes zero at the surface as we need to find the minimum velocity?
About what point we will conserve angular momentum?
 
Pranav-Arora said:
Why can't we assume that velocity becomes zero at the surface as we need to find the minimum velocity?

As the angular momentum must be conserved.

Pranav-Arora said:
About what point we will conserve angular momentum?

About the centre of the sphere, of course. It is a central force field! ehild
 
ehild said:
As the angular momentum must be conserved.
I think i need to clear my concepts about angular momentum and central forces.

About the centre of the sphere, of course. It is a central force field!

Thanks, i think i have got the answer. :smile:
Conserving angular momentum, i get:
Vf=V/2

Vf is final velocity and V is the initial velocity.
Using conservation of Energy
[tex]\frac{1}{2}mV^2=\frac{KQq}{R}+\frac{1}{8}mV^2[/tex]
Solving, i get:
[tex]V=2\sqrt{\frac{2}{3}}m/s[/tex]

Thanks for the help!
 
Excellent! :cool: (although you omitted the "f" subscript from the energy equation)

ehild
 
ehild said:
Excellent! :cool: (although you omitted the "f" subscript from the energy equation)

ehild

Thanks! :smile:
Yes, i skipped the step and wrote 1/8mV^2.
 
Pranav-Arora said:
Thanks! :smile:
Yes, i skipped the step and wrote 1/8mV^2.

I see. Think of the supervisors, their mind is not so quick as yours: Do not skip the first step.

ehild
 
ehild said:
I see. Think of the supervisors, their mind is not so quick as yours: Do not skip the first step.

ehild

Haha, i can do it here because the supervisor is you. :smile:
 

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