Electric Potential of Two Positive Charges at Point A on y-Axis

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SUMMARY

The discussion focuses on calculating the electric potential at point A on the y-axis due to two positive charges, each with a charge of Q = +11 µC, positioned at x = +2 m and x = -2 m. The electric potential at point A, located at (0, 6), is derived using the formula V = ΣKQ/R, resulting in a potential energy of U_initial = 0.1565 J. Additionally, the conservation of mechanical energy is applied to find the speed of a particle with charge q = -5 µC and mass m = 2.4 x 10-4 kg released from rest at point A, leading to a final potential energy of U_final = 0.495 J.

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https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2112/spring/homework/Ch-24-E-Potential/2_pt_chrgs/set3p02.gif" 1.
Homework Statement


Two positive charges, each with Q = +11 µC, are fixed to the x-axis at x = +a and x = -a, where a = 2 m.
(a) Find the electric potential at point A on the y-axis where (xA, yA) = (0, b) and b = 6 m. Take the zero of potential to be at infinity.
(b) A particle with charge q = -5 µC and mass m = 2.4 x 10-4 kg is released from rest at point A. Find its speed at the origin.

Homework Equations


V=[tex]\sum[/tex]KQ/R

U=KQ1Q2/R

The Attempt at a Solution


Part (a) can be solved by using potential equation
For part (b) the potential energy of the particle is conserved, so I've found it. after that i dont' know how to get to the velocity.
Thanks in advance.
 
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Total energy may be conserved, but potential energy alone is not.

What values did you calculate for the potential energy of the particle at point A and at the origin?
 
Conservation of mechanical energy between point A and origin makes sense.
KEinitial +Uinital = KEfinal +Ufinal
K
KE initial is zero, since the particle is released from rest.
KE final = 1/2mv^2
Uinitial=.1565
Ufinal(at origin)=.495

Thanks a lot
 

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