How Close Can a Proton Get to a Charged Sphere?

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SUMMARY

The discussion focuses on calculating the closest approach of a proton to a charged conducting sphere and determining the voltage at the sphere's surface. The proton, with an initial velocity of 6.28 x 105 m/s, approaches a sphere with a radius of 85.2 cm and a charge of 4.83 μC. The relevant equation used is V = (1/(4πε0))(q/r), leading to an electric potential of approximately 50964.4V at the sphere's surface. The participants emphasize the need to understand the relationship between electric potential and kinetic energy to solve the problem effectively.

PREREQUISITES
  • Understanding of electric potential and kinetic energy concepts
  • Familiarity with Coulomb's law and the equation for electric potential
  • Basic knowledge of protons and their properties, including mass and charge
  • Ability to perform calculations involving scientific notation
NEXT STEPS
  • Study the derivation and application of Coulomb's law in electrostatics
  • Learn about the concept of electric potential energy and its relation to kinetic energy
  • Explore the principles of conservation of energy in electrostatic systems
  • Investigate the behavior of charged particles in electric fields
USEFUL FOR

Students in physics, particularly those studying electromagnetism, as well as educators and anyone preparing for exams involving electrostatic concepts and calculations.

AnthroMecha
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Homework Statement



A proton is moving with an initial velocity of 6.28x10^5 m/s straight towards a large conducting
sphere of radius 85.2 cm. The sphere has a charge of 4.83 μC on it. Protons have a mass of
1.69x10^-27 kg.

a) Determine how close the proton can come to the surface of the sphere.
b) What is the voltage on the surface of the sphere?


Homework Equations



possibly 1/2mv^2=-qΔV

The Attempt at a Solution



Don't even know where to begin on this one
 
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AnthroMecha said:

Homework Statement



A proton is moving with an initial velocity of 6.28x10^5 m/s straight towards a large conducting
sphere of radius 85.2 cm. The sphere has a charge of 4.83 μC on it. Protons have a mass of
1.69x10^-27 kg.

a) Determine how close the proton can come to the surface of the sphere.
b) What is the voltage on the surface of the sphere?

Homework Equations



possibly 1/2mv^2=-qΔV

The Attempt at a Solution



Don't even know where to begin on this one
That's a good relevant equation, since at closest approach, the proton's velocity will be zero.

What is the electric potential at the location of the proton, due to the sphere?
 
SammyS said:
That's a good relevant equation, since at closest approach, the proton's velocity will be zero.

What is the electric potential at the location of the proton, due to the sphere?

I'm coming up with 50964.4V electric potential (I may be using the wrong equation).

I used V=(1/(4πε0))(q/r)

q=4.83μC, r=85.2cm
 
AnthroMecha said:
I'm coming up with 50964.4V electric potential (I may be using the wrong equation).

I used V=(1/(4πε0))(q/r)

q=4.83μC, r=85.2cm
That's good for part (b).

To find the potential due to the sphere at the location of the proton at the time its velocity is 6.28×105 m/s, you have to know how far the proton is from the center of the sphere at that time.

Did they give that quantity ?
 
SammyS said:
That's good for part (b).

To find the potential due to the sphere at the location of the proton at the time its velocity is 6.28×105 m/s, you have to know how far the proton is from the center of the sphere at that time.

Did they give that quantity ?

no. I'm wondering if it involves some sort of integral. I hope there's nothing like this one on the test, because I have to go take it in an hour, lol! I kind of procrastinated a bit on this one. :(
 

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