At a particular moment, one negative and two positive charges are located as sho

Question

At a particular moment, one negative and two positive charges are located as shown in the figure. Your answers to each part of this problem should be vectors. It helps a great deal to make a diagram with arrows representing the various electric field contributions, and then check the signs of your components against these arrows.

An alpha particle (He2+, containing two protons and two neutrons) is released from rest at location A. Determine the initial acceleration of the alpha particle. Mass of the alphaalpha-particle is 6.646×1027 kg

Choose correct answers:

2.38×10^15,6.48×10^14,0 m/s^2

6.9 m/s^2,87.9

5.34×10^15,2.16×10^14,0 m/s^2

6.9 m/s^2 , 272.1

Explanation / Answer

= tan^-1(4/3) = 53.13 deg

F1 =F1x = kqq1/r1^2 = (9*10^9*3*10^-6*2*1.6*10^-19)]/(0.03^2) = 9.6*10^-12 N to the right

F2 = kqq2/r2^2 = (9*10^9*9*10^-6*2*1.6*10^-19)]/(0.05^2) = 1.04*10^-11 N away from q

F2x= F2cos = 1.04*10^-11cos53.13 = 6.24*10^-12 N

F2y = F2sin = 1.04*10^-11sin53.13 = 8.3*10^-12 N

F3 = F3y= kqq3/r3^2 = (9*10^9*-7*10^-6*2*1.6*10^-19)]/(0.04^2) = -1.26*10^-11 to the below

Fnetx = F1x + F2x + F3x = 9.6*10^-12 +6.24*10^-12 + 0 = 1.6*10^-11 N

Fnety = F1y + F2y + F3y = 0 + 8.3*10^-12 -1.26*10^-11 = -4.3*10^-12 N

ax = Fnetx/m = (1.6*10^-11)/( 6.646*10^27) = 2.4*10^15 m/s^2

ay = Fnety/m = (-4.3*10^-12)/( 6.646*10^27) = -6.48*10^14 m/s^2

= tan^-1(ay/ax) = tan^-1[(-6.48*10^14)/( 2.4*10^15)] = -87.9 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.