Charge is placed 33.65 cm to the left of a 83.73-mC charge, as shown in the figu
ID: 1525628 • Letter: C
Question
Charge is placed 33.65 cm to the left of a 83.73-mC charge, as shown in the figure, and Doth charges are held stationary. A particle with a charge of -2.051 pC and a mass of 16.81 g (depicted as a blue sphere) is placed at rest at a distance 30.29 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
Given q= -2.051 microC
V = 1/4pi0(q1/r1 +q2/r2)
By law of conservation
K1 + U1 = K2 + U2
0 + qV1 = 1/2 mVA2 +qVA
1/4pi0(qq1/r1i +qq2/r2i) = 1/2mVA2 +1/4pi0(qq1/r1A +qq2/r2A)
-9e9 [(14.41e-3*2.051e-6)/0.4527) + (2.051e-6*83.73e-3/0.3029)] = 1/2 * 16.81VA2 + 9e9[(14.41e-3*2.051e-6/0.1010)+(2.051e-6*83.73e-3/0.3029)/0.2355]
90(6.527 +56.695 - 29.257 - 56.695 ) =8.405 VA2
VA2 = 3220.56/8.405 = 383.172
VA = 19.5747 m/s