In Figure (a) three positively charged particles are fixed on an x axis. Particl

Question

In Figure (a) three positively charged particles are fixed on an x axis. Particles B and C are so close to each other that they can be considered to be at the same distance from particle A. The net force on particle A due to particles B and C is 3.22 × 10-23 N in the negative direction of the x axis. In Figure (b), particle B has been moved to the opposite side of A but is still at the same distance from it. The net force on A is now 4.64 x 1024 N in the negative direction of the x axis. What is the ratio qdB? BC Number Units

Explanation / Answer

For configuration a)

Net force on A due to B and C :

Fnet = k*(qA)*(qB + qC)/r^2 = 3.22*10^-23 N

here, r = distance between A and B

Now, for configuration b)

Fnet = k*qA*qC/r^2 - k*qA*qB/r^2 = k*qA*(qC - qB) = 4.64*10^-24 N

So, taking the ratio :

[ k*(qA)*(qB + qC)/r^2 ] / [ k*qA*(qC - qB) ] = 3.22*10^-23 / 4.64*10^-24

So, (qB + qC) / (qC - qB) = 6.94

So, (qC/ qB + 1) / (qC/ qB- 1) = 6.94

So, qC/ qB = 1.337 <---------- answer

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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