A charge falls from infinity to within r of another charge, find velocity.

In summary, the question asks for the velocity of an electron that falls from infinity to a distance of r=10^-8m from a charge q1=4.8x10^-19C. To solve this, Coulomb's constant k=9x10^-9, electron charge q2=1.602x10^-19C, electron mass m=9.11x10^-31kg, and kinetic energy Ek=½mv² are used. The potential energy change U is calculated using U=k(q1)(q2)/r=6.912x10^-21J. By setting U equal to Ek and solving for v, the velocity is determined to be 1.23x10^5m/s
  • #1
User1247
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0

Homework Statement


Velocity of an electron that falls to r from infinity?
An electron falls from infinity to r=10^-8m from a charge q1=4.8x10^-19C. What is the velocity of the electron?






Homework Equations


U=q1V
V=kq2/r

The Attempt at a Solution


Potential energy change U. coulomb's constant k=9x10^-9, electron charge q2=1.602x10^-19C, electron mass m=9.11x10^-31kg, kinetic energy Ek=½mv²

U=k(q1)(q2)/r=6.912x10^-21J
set U=Ek => v=(2U/m)^0.5=1.23x10^5m/s

Anyone see where I went wrong? Thanks.
 
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  • #2
User1247 said:
U=k(q1)(q2)/r=6.912x10^-21J
Redo this, paying attention to the power of ten.
 
  • #3


Your solution looks correct. However, it would be helpful to see your work and calculations in more detail to ensure that there are no errors. Also, it would be good to mention any assumptions or simplifications you made in your solution. Overall, your approach seems logical and correct.
 

1. What does it mean for a charge to fall from infinity?

When we say a charge falls from infinity, we mean that it starts at a point infinitely far away from another charge. This is a hypothetical starting point that is often used in physics problems to simplify calculations.

2. How is the velocity of the falling charge related to its distance from the other charge?

The velocity of the falling charge is directly related to its distance from the other charge. As the falling charge gets closer to the other charge, the force of attraction between them increases, causing the falling charge to accelerate. This acceleration results in an increase in velocity.

3. What is the equation for finding the velocity of a falling charge?

The equation for finding the velocity of a falling charge is v = √(2qE/m), where v is the velocity, q is the charge of the falling particle, E is the electric field strength, and m is the mass of the falling particle. This equation applies to a charge falling from infinity to within r of another charge.

4. How does the charge of the falling particle affect its velocity?

The charge of the falling particle does not directly affect its velocity. However, it does affect the electric field strength, which in turn affects the velocity through the equation v = √(2qE/m). A higher charge on the falling particle will result in a stronger electric field and a higher velocity.

5. Is the velocity of the falling charge constant?

No, the velocity of the falling charge is not constant. As the falling charge gets closer to the other charge, the force of attraction and acceleration increase, causing the velocity to increase. However, as the falling charge gets even closer and the distance between them decreases, the force of attraction and acceleration decrease, causing the velocity to decrease. This results in a non-constant velocity until the falling charge reaches a stable distance from the other charge.

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