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

In summary, the problem involves two positive charges fixed at points on the x-axis, with a given distance between them. The electric potential is calculated at a point on the y-axis and the speed of a charged particle released from rest at that point is also requested. Using the potential equation, the potential energy of the particle is conserved, but the total energy is not. Through the conservation of mechanical energy, the final kinetic energy can be calculated and used to find the speed of the particle at the origin.
  • #1
lcam2
28
0
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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  • #2
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?
 
  • #3
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
 

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

1. What is the electric potential at point A on the y-axis?

The electric potential at point A on the y-axis is the amount of electric potential energy per unit charge at that point. It is measured in volts (V) and is calculated by taking the sum of the electric potential contributions from two positive charges at that point.

2. How do you calculate the electric potential at point A on the y-axis?

The electric potential at point A on the y-axis can be calculated using the formula V = k(q1/r1 + q2/r2), where k is the Coulomb's constant, q1 and q2 are the charges of the two positive charges, and r1 and r2 are the distances from each charge to point A.

3. What factors affect the electric potential at point A on the y-axis?

The electric potential at point A on the y-axis is affected by the magnitudes of the two positive charges, the distance between them, and the distance from each charge to point A. These factors are all taken into account in the calculation of the electric potential.

4. What is the relationship between electric potential and electric field?

Electric potential and electric field are closely related. The electric field is the force per unit charge exerted on a test charge at a given point, while the electric potential is the amount of electric potential energy per unit charge at a given point. The electric field is the negative gradient of the electric potential, meaning that it is the direction in which the electric potential decreases the most rapidly.

5. How does the electric potential at point A on the y-axis change as the distance between the two positive charges changes?

The electric potential at point A on the y-axis is inversely proportional to the distance between the two positive charges. This means that as the distance between the charges increases, the electric potential decreases. Conversely, as the distance between the charges decreases, the electric potential increases.

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