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Two Charges Energy

  • Thread starter salman213
  • Start date
302
1
1. Two charges are arranged on the x axis. q1 = +8.00 μC, and
q2 = +5.00 μC. a = 25.0 cm. q1 is held fixed and q2 is released from rest.
Both q1 and q2 have a mass of 20.0 g.
Calculate x2, the position of q2, when it reaches a speed of 8.00 m/s.
a is the distance between q1 and q2. q1 is at the origin and q2 is along the x axis at a distance "a"


2. Ek= 1/2mv^2
V = kq/r



3.
first i converted m to kg = 20 x 10^-3

Ek = 1/2mv^2
V = kq1/r + kq2/r

1/2mv^2 = kq1/r + kq2/r
1/2(20x10^-3)(8)^2 = k(8x10^-6)/x + k(5x10^-6)/(x-0.25)

solving for x gives me wrong answer
 

Answers and Replies

Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,474
20
Ek = 1/2mv^2
V = kq1/r + kq2/r
Why are you adding the potential energies? You are supposed to be equating the energy of the initial state (which is purely potential) to the energy of the final state (which is a combination of kinetic + potential).
 
302
1
so initially

Ei = kq1/r2

and final

Ef = 1/2mv^2 + kq1/r

?
 
85
0
so initially

Ei = kq1/r2

and final

Ef = 1/2mv^2 + kq1/r

?
That sounds almost right. But also remember, voltage is not the energy of the charge. It is the energy a 1C of charge would have, but you do not have 1C charges now.
 
302
1
but doesnt the forumula incorporate teh charge

kq/r
with that "q"
 
85
0
but doesnt the forumula incorporate teh charge

kq/r
with that "q"
Nope, the q1 is the voltage DUE TO the charge q1, not the energy OF q2. :)
 
302
1
so does the answer of it involve the formula

W = q(vf-vi)

:S:S:S
 
85
0
so does the answer of it involve the formula

W = q(vf-vi)

:S:S:S
Yes, the energy of the charge is given by qV, where V is the voltage at a point.
 
302
1
ok but how would i incorporate the energy due to charge 1 that is held fixed..

so far Ek = 1/2m1v1^2
V = qV2

is that all i need or in Ek I need to add the potential of the fixed charge
 
302
1
kq1q2/.25 = 1/2mv^2 + kq1q2/(x-0.25)
 

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