http://www.webassign.net/mi3/14-063-Prob3-6.jpg At a particular moment, three sm

Question

http://www.webassign.net/mi3/14-063-Prob3-6.jpg

At a particular moment, three small charged balls, one negative and two positive,are located as shown in the figure. Q1 = 1 nC, Q2= 9 nC, and Q3 = -2 nC.

Remember that you must first convert all quantities to S.I. units. 1 nC = 1 nanocoulomb = 1e-9 C.

At a particular moment, three small charged balls, one negative and two positive,are located as shown in the figure. Q1 = 1 nC, Q2 = 9 nC, and Q3 = -2 nC. Remember that you must first convert all quantities to S.I. units. 1 nC = 1 nanocoulomb = 1e-9 C. (a1) What is the electric field at the location of Q1, due to Q2? E rightarrow 2 = __ N/C (a2) What is the electric field at the location of Q1, due to Q3? E rightarrow 3 = N/C (a3) What is the net electric field at the location of Q1, due to Q2 and Q3? E rightarrow net = N/C (b) Use the electric field you calculated above to find the force on Q1 due to Q2 and Q3. F rightarrow = N (c1) What is the electric field at location A, due to Q1? E rightarrow A, 1 = N/C (c2) What is the electric field at location A, due to Q2? E rightarrow A,2 = N/C (c3) What is the electric field at location A, due to Q3? E rightarrow A,3 = N/C (c4) What is the net electric field at location A? E rightarrow A, net = N/C (d) An alpha particle (He2+, containing two protons and two neutrons) is released from rest at location A. At the instant the particle is released, what is the electric force on the alpha particle, due to Q1, Q2 and Q3? F rightarrow = N

Explanation / Answer

a1) E2 = k q2/r^2 = 9E9*9.0E-9/0.04^2= 50625

so <0,50625>

a2) mag E3 = k q3/r^2 = 9E9*2.0E-9/0.05^2=7200

so breaking into components <7200*3/5, -7200*4/5>

so <4320, -5760>

a3) E = E2 + E3 = <4320, 44865>

b) F = q E = 1.0E-9*<4320,44865> = <4.32E-6, 4.49E-5>

c) E1 = k q1/r^2 = 9E9*1.0E-9/0.03^2=10000

so <10000,0>

c2) E2 = k q2/r^2 = 9E9*9.0E-9/0.05^2= 32400

so break into components

<32400*3/5, 32400*4/5>= <19440, 25920>

c3) E3 = k q3/r^2 = 9E9*2.0E-9/0.04^2=11250
so <0,-11250>

c4> net E = <10000 + 19440+0, 0 + 25920 - 11250>= <29440, 14670>

d) F = q E = 2*1.6E-19*<29440, 14670>= <9.42E-15, 4.69E-15>

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

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