A charge of 2.93 is held fixed at the origin. A second charge of 2.93 is release

Question

A charge of 2.93 is held fixed at the origin. A second charge of 2.93 is released from rest at the position (1.19 m, 0.553 m). If the mass of the second charge is 2.76 g, what is its speed when it moves infinitely far from the origin? Ive already solved for this answer as 6.53 m/s A) At what distance from the origin does the 2.93 charge attain half the speed it will have at infinity?

Explanation / Answer

In general , the total energy of the charge is : Eo = Q^2/4[(pi)(8.854(10)^(-12)r] + 1/2mv^2 . The fourth person is right the charge must be in microcoulombs --- Q= 2.93(10)^-6 coulombs . We also have the initial distance between charges is --- r = sqrt (1.19^2+.553^2)=1.31221530246meters , and the initial velocity is zero because the charge is released from rest . r = infinity . At r = infinity , we can figure out its speed by conservation of energy : E = [ (2.93(10)^(-6)]^2/[4(pi)8.854(10)^-12(1.31221530246)] + 1/2m(0)^2 = 0 + 1/2(2.76(10)^-3)Vf^2 Eo= .05880046507 = 1.38(10)^-3Vf^2 Vf^2 = .05880046507/(1.38(10)^-3)=42.6090326591 ------> Vf = 6.52755947189meters/second . We want r when V=1/2Vf = 3.26377973594meters/second . Plug this velocity into --- Eo = Q^2/4[(pi)(8.854(10)^(-12)r] + 1/2mv^2 : Eo= .05880046507= Q^2/4[(pi)(8.854(10)^(-12)r] + 1/2(2.76(10)^-3)(3.26377973594)^2 .05880046507= [ (2.93(10)^(-6)]^2/[4(pi)8.854(10)^-12(r)] + .01470011627 .07715887006/r = .05880046507-.01470011627= .04410034880 --- solve this equation for r : ------------------------------------------------------------------------------------------ r = (.07715887006)/(.04410034880) = 1.7496 meters ---- solution -------------------------------------------------------------------------------------------

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