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Electric Potential: Velocity of a particle

  • Thread starter Bryon
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  • #1
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Homework Statement


Two positive charges, each with Q = +12 µC, are fixed to the x-axis at x = +a and x = -a, where a = 2 m.

Part A was not a poblem to find.
(a) Find the electric potential at point A on the y-axis where (xA, yA) = (0, b) and b = 5 m. Take the zero of potential to be at infinity.

V(A) = 40065.62617V

(b) A particle with charge q = -6 µC and mass m = 1.5 x 10-4 kg is released from rest at point A. Find its speed at the origin.

V at the origin = ???


Homework Equations


F= ma

V(a)-V(b) = ∫E•dl = kQ/r

w = 0.5*m*(v1^2 - v0^2) = F•d

potential energy = kq1q2/r

3. The Attempt at a Solution

I think I have the equations I need for this problem. For this problem do I need to find a way to relate work to the potential energy?
 

Answers and Replies

  • #2
gneill
Mentor
20,793
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The electrical potential is specified in Volts. The Volt unit is equivalent to

[tex] [Volt] = \frac{[Joule]}{[Coulomb]}[/tex]

So if your charged particle falls through a potential difference ∆V, it changes its kinetic energy accordingly.
 

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