Sapling Learning Map.e A 37.91-mC charge is placed 38.15 cm to the left of a 55.

Question

Sapling Learning Map.e A 37.91-mC charge is placed 38.15 cm to the left of a 55.23-mC charge, as shown in the figure, and both charges are held stationary. A particle with a charge of-3.551 and a mass of 3631 g (depicted as a blue sphere) is placed at rest at a distance 34.34 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 va at that point? Number 3.551 pC mis 34.34 cm +37.91 mC k-11.45 cm +55.23 mC 38.15 cm

Explanation / Answer

We need to know where point A is w/r/t the first two charges.

The solution method is as follows:

initial E = EPE = charge * potential

E = -3.551e-6C * 8.99e9N·m²/C² * [55.23e-3C/0.3434m + 37.91e-3C/(0.3434² + 0.3815²)m]

E = -7,492.12 J

The final E is the sum of the EPE at point A, which requires the distances from each of the first two charges, and the KE of the third particle (which is why we're given the mass). It's a short step from there to the velocity.

PER YOUR COMMENT:

final E = -3.551e-6C * 8.99e9N·m²/C² * [55.23e-3C/(0.3815 – 0.1145) + 37.91e-3C/0.1145m]

so final E = -17,173.10 J

KE = initial E - final E = 9680.98 J = ½mv² = ½ * 0.03631kg * v²

solves to

v = 730.23 m/s

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