How Is Work Calculated When Changing Orbital Radius in an Electric Field?

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SUMMARY

The work W required to change the orbital radius of a negatively charged particle -q in the electric field of a positively charged particle Q is derived as W = (Qq/8πε₀)(1/r₁ - 1/r₂). This expression utilizes the formula for electric potential energy and the relationship between kinetic energy and potential energy (TE = KE + PE). The factor of 8 arises from the integration of the electric field over the change in radius, which is crucial for accurate calculations in electrostatics.

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jperk980
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A particle of positive charge Q is fixed at point P. A second particle of mass m and negative charge -q moves at constant speed in a circle of radius r1, centered at P. Derive an expression for the work W that must be done by an external agent on the second particle to increase the radius of the circle of motion to r2. (Use epsilon_0 for 0, r1 for r1, r2 for r2, and Q, m, and q as necessary.)
I know the answer to be (Qq/8pi*episolon_0)(1/r1-1/r2)
I know that i use the equation w=k*q1q2/r and i believe r to be (1/r1-1/r2). I do not understand how to get 8 though could someone please help!
 
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TE= KE+PE
you forgot calculate the PE
increase the radius = change of TE
 
jperk980 said:
I know that i use the equation w=k*q1q2/r and i believe r to be (1/r1-1/r2). I do not understand how to get 8 though could someone please help!

Show your calculations briefly so that we can take it up from there.
 

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