Two identical 30 g particles each carry 5.0 ? C of charge. One is held fixed, an

Question

Two identical 30g  particles each carry 5.0?C  of charge. One is held fixed, and the other is placed 1.0 mm away and released.

Find the speed of the moving charge when it's 1.0 cm from the fixed charge answer must be to 2 significant digits

Need some clarification

When I perform the following calculations it says incorrect.

known values: m =.03kg,   q=5*10^-6 C,  r1=.001m,   r2=.01m,  k=8.99*10^9

KE=PE

1/2mv^2 = kq^2[(1/r1)-(1/r2)]

rearranged equation

v^2=2kq^2[(1/r1)-(1/r2)] / m

substitute values

v^2={2(8.99*10^9)(5*10^-6)^2[(1/.001)-(1/.01)]} / .03

v^2=(0.4495 x 900) / .03

v^2=404.55 / .03

v^2= 13485

v= sqrt 13485

v= 23.8m/s

Explanation / Answer

you have done it perfectly correct but only mistake is on taking the square root the correct answer is 116.12m/s

v= sqrt 13485

v= 116.12m/s

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