What was the electron's initial speed?

  • Thread starter rayhan619
  • Start date
  • #1
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Homework Statement



A 2.4 mm diameter sphere is charged to -4.4 nC. An electron fired directly at the sphere from far away comes to within 0.31 mm of the surface of the target before being reflected.
a) What was the electron's initial speed?
b) At what distance from the surface of the sphere is the electron's speed half of its initial value?
c) What is the acceleration of the electron at its turning point?

Homework Equations





The Attempt at a Solution


how do i start this problem?
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
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When the electron is fired at the sphere it will have kinetic energy right?

How much work will you need to do to slow it down and reverse?

Isn't Work = q*ΔV ?

V at ∞ is = 0. So what is the Voltage Potential at .31 mm?
 
  • #3
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so the kinetic energy = 1/2 mv^2
what is the kinetic energy i use to slove for v?
 
  • #4
LowlyPion
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so the kinetic energy = 1/2 mv^2
what is the kinetic energy i use to slove for v?
That's the wrong question.

What is the V you have at .31mm to let you solve for the ½mv²?
 
  • #5
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can you plz explain it a little more.
its due at 11 and im totally lost
 
  • #6
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how do you do part b if you get a?
 
  • #7
LowlyPion
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can you plz explain it a little more.
What is the formula for determining Voltage?

V = kq/r perhaps?

q is your charged ball. And r = .31 mm.

That helps you determine the Voltage and then you can get the kinetic energy.

That gives you the Velocity at ∞.

So what voltage will it be moving through when the Velocity is half the Velocity at ∞ ?

The acceleration back out at .31mm is a little easier.

You can determine that from F = q*E = m*a
 

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