ibiscms/mod/ibis/view.php?id-4930847 Jump to... Grade 2/10/2018 1159 PM O 57.9/1
ID: 1870778 • Letter: I
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ibiscms/mod/ibis/view.php?id-4930847 Jump to... Grade 2/10/2018 1159 PM O 57.9/100 uestion 13 of 18 Map Sapling Learning is placed 38.65 cm to the left of a 68.23-mC charge, as shown in the A 19.41-mC both charges are (depicted as a blue sphere) is placed at rest at a distance particle were to be released from rest, it would follow some complicated path around the two stationary charges. stationary. A particle with a charge of 34.78 cm above the right-most charge. If the Calculating the exact path of the particle would be a challenging problem, but even without such a calculation it is possible to make some definite predictions about the future motion of the If the path of the particle were to pass through the gray point labeled A what would be its speed va at that point? 2.551 pC Number m/s | 34.78 +19.41 mC K-11.60 cm +68.23 mC 38.65 cmExplanation / Answer
potential at initial position,
Vi = k q1 /r1 + kq2 / r2
= [ (9 x 10^9 x 68.23 x 10^-3) / 0.3478] + [ (9 x 10^9 x 19.41 x 10^-3) / sqrt(0.3478^2 + 0.3865^2)]
= 2.10156 x 10^9 Volt
potential at A,
Vf = [ (9 x 10^9 x 68.23 x 10^-3) / (0.3865- 0.1160)] + [ (9 x 10^9 x 19.41 x 10^-3) / 0.1160]
Vf = 3.7761 x 10^9 Volt
Work done = q deltaV = m v^2 /2
(2.551 x 10^-6)(3.7761 - 2.10156)(10^9) = (36.31 x 10^-3 ) v^2 / 2
v = 485 m/s .....Ans
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