A positive charge of Q = 5.40 uC is fixed in place. From a distance of ri = 4.20

Question

A positive charge of Q = 5.40 uC is fixed in place. From a distance of ri = 4.20 cm a particle of mass m = 5.50 g and charge q = +4.00 uC is fired with an initial speed of v = 68.0 m/s directly toward the fixed charge. What is the minumum distance rf between the two charges?

I had it as a previous question and i tried using the input equation that someone answered with and still got 1.86. Im on my last try and if someone can get the correct answer and show me how they got it, it would be greatly appreciated!!!

Explanation / Answer

initial energy= kqQ/r +1/2*mv^2 final energy =kqQ/R kqQ/r +1/2*mv^2 =kqQ/R kqQ(1/0.042 - 1/R) +1/2*0.0055*68*68=0 0.1944(23.80 -1/R) + 12.716 =0 R= 1.12cm

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