# Conducting shell and proton with KE

• origamipro
In summary, Q net is +.04E-16C. Both charge sources are modeled from the very center. How does the proton come to rest?

## Homework Statement

http://imageshack.us/photo/my-images/543/15jd.jpg/

[IMG=[PLAIN]http://img543.imageshack.us/img543/8674/15jd.jpg][/PLAIN] [Broken]

## Homework Equations

Qnet outside sphere = Q1+Q2
KEi + PEi = KEf + PEf

q = 1.6E-19
mass of q = 1.672E-27

## The Attempt at a Solution

+(q ∫ KQ1 / R^2 ] 3R to R) + (q ∫ K(-Q2) / R^2 ] 3R to R) - 1/2 m v^2

Last edited by a moderator:
I don't know if it's just me, but I cannot see the picture that you posted.

Bump. Q net is +.04E-16C. Both charge sources are modeled from the very center. How does the proton come to rest?

Your Q net does not look right to me. Show your work.

Presumably the proton comes to rest only momentarily before being accelerated away from the net positive charge on the shell.

Q1 = 4E-16 Q2 = ?
q = 1.6E-19 qm = 1.67E-27

at the surface of the shell kinetic energy must equal zero. The potential energy at the surface does not need to be accounted for since it only wants value of Q2 when proton gets to the surface. Is this assumption correct, otherwise calculating E field at its surface becomes an approximation since the integral does not converge.

so proton traveling in negative x direction, with two electric potentials both modeled from the same source point.

(1/2)m(v^2) -(q ∫ kQ1/r^2 ] 3R to R) + (q ∫ KQ2/r^2 ] 3R to R) = 0
(4.0915x10^-24) -2.74286E-24 + (1.6E-19)Q2(4.28571E10) = 0
solving for Q2 = -1.9667E-16

origamipro said:
Q1 = 4E-16 Q2 = ?
q = 1.6E-19 qm = 1.67E-27

at the surface of the shell kinetic energy must equal zero. The potential energy at the surface does not need to be accounted for since it only wants value of Q2 when proton gets to the surface. Is this assumption correct, otherwise calculating E field at its surface becomes an approximation since the integral does not converge.
The potential energy is key. The change in potential energy between the staring and ending locations will equal the magnitude of the change in kinetic energy of the particle. The potential for a charge q very near the surface of a spherical conductor with charge Q and radius r is kQq/r. The spherical charged surface behaves like a point charge located at its center.

so proton traveling in negative x direction, with two electric potentials both modeled from the same source point.

(1/2)m(v^2) -(q ∫ kQ1/r^2 ] 3R to R) + (q ∫ KQ2/r^2 ] 3R to R) = 0
(4.0915x10^-24) -2.74286E-24 + (1.6E-19)Q2(4.28571E10) = 0
solving for Q2 = -1.9667E-16
No need to integrate. Use conservation of energy via initial KE and change in PE for the two locations.

Right that is what i showed.

Ke = 1/2 m v^2

the integration is the delta V(voltage) = q(kQ1/3R - KQ2/R)

Ah. My mistake. When I read the problem I misread the key phrase "...from a distance of 2R from the surface of the sphere." That puts the initial distance at 3R from the center. I suspect that whoever provided the answer for the problem did the same thing.

So I agree with your result that Q2 should be ~ -1.97 x 10-16 C and not the given value.

## 1. What is a conducting shell?

A conducting shell is a hollow metallic object that has a uniform charge distribution on its surface. It is an ideal conductor, meaning that charges can freely move around the surface of the shell.

## 2. How does a conducting shell affect a proton with kinetic energy (KE)?

A conducting shell has a strong electrostatic field on its surface, which can interact with a charged particle like a proton with KE. This interaction can cause the proton's path to bend or change direction.

## 3. What is the difference between a conductor and an insulator?

A conductor is a material that allows charges to move freely, while an insulator does not allow charges to move easily. Conductors are typically metals, while insulators are non-metal materials like plastics or rubber.

## 4. Can a conducting shell have a net charge?

Yes, a conducting shell can have a net charge. However, the charge will only reside on the surface of the shell, and the interior will remain neutral. This is because charges repel each other and will spread out evenly on the surface of the shell.

## 5. How can the electrostatic field of a conducting shell be manipulated?

The electrostatic field of a conducting shell can be manipulated by changing the amount of charge on its surface or by changing the shape or size of the shell. This can be done using an external source of charge or by physically altering the shell itself.