A 15.41-mC charge is placed 33.15 cm to the left of a 51.23-mC charge, as shown

Question

A 15.41-mC charge is placed 33.15 cm to the left of a 51.23-mC charge, as shown in the figure, and both charges are held stationary. A particle with a charge of -2.051 mu C and a mass of 29.31 g (depicted as a blue sphere) is placed at rest at a distance 29.84 cm above the right-most charge. If the particle were to be released from rest, it would follow some complicated path around the two stationary charges. Calculating the exact path of the particle would be a challenging problem, but even without performing such a calculation it is possible to make some definite predictions about the future motion of the particle. If the path of the particle were to pass through the gray point labeled A what would be its speed v_A at that point?

Explanation / Answer

Change i potential energy of particle = Change in kinetic energy

Kq1q2/r = 0.5mv^2

[9*10^9*51.23 * 2.051*10^-9 / 0.2984 + 9*10^9*15.41 * 2.051*10^-9 / (0.2984^2 + 0.3315^2)]-[9*10^9*51.23 * 2.051*10^-9 / 0.23205 - 9*10^9*15.41 * 2.051*10^-9/ 0.09945 ]= 0.5*(0.02931)(v^2)

4599.71 - 1214.955 = 0.5 *0.02931v^2

v = 480.58 m/s

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