Chapter 21, Problem 032 Your answer is partially correct. Try again. Figure (a)

Question

Chapter 21, Problem 032 Your answer is partially correct. Try again. Figure (a) shows charged particles 1 and 2 that are fixed in place on an x axis. Particle 1 has a charge with a magnitude of 19.1 = 12e. Particle 3 of charge q3 = +12e is initially on the x axis near particle 2.Then particle 3 is gradually moved in the positive direction of the x axis. As a result, the magnitude of the net electrostatic force on particle 2 due to particles 1 and 3 changes. Figure (b) gives the x component of that net force as a function of the position x of particle 3. The scale of the x axis is set by x 1.60 m. The plot has an asymptote of F2,net = 1.168 × 10-25 N as x oo. As a multiple of e and including the sign, what is the charge q2 of particle 2? c (m 0 Number Units This answer has no units the tolerance is +/-2%

Explanation / Answer

This problem can be solved by two ways.

CASE 1 when   q1 &  q2 are both positive

q1 = +12e , q3 = +12e and q2 = ?

The net force on q2 is zero from fig.b when q3 is at 2xs (2 units) distance, which means force due to q1 on q2 is equal and opposite to force due to q3 on q2.

Therefore.   r32 = 2xs = 3.20m   where, r32 is distance between q3 and q2

Also, F12 = F32 at r32 = 2xs

(k.q1.q2) / (r12)2 = (k.q3.q2) / (r32)2 where, r32 is distance between q3 and q2 and similarly r12

q1 / (r12)2 = q3 / (r32)2

r12 = r32 since q1 = q3 = 12e

or r12 = r32 = 2xs = 3.20 m

When q3 is taken to infinity then the force experienced by q2 is only due to q1,

Therefore, F2,net = (k.q1.q2) / (r12)2 = 1.168 x 10-25 N where, k is constant = 8.99 x109 Nm2C-2

e = 1.6 x 10-19 C is SI unit

F2,net = (k.q1.q2) / (r12)2

1.168 x 10-25 = [ ( 8.99 x109 ) x (12 x 1.6 x 10-19) x q2 ] / 3.202

q2 = [ 1.168 x 10-25 x 3.202 ] / [ ( 8.99 x109 ) x (12 x 1.6 x 10-19) ]

q2 = 6.93 x 10-17 C

or   q2 = [ 6.93 x 10-17 ] / [ 1.6 x 10-19 ] e

.q2 = + 433.0737 e

CASE 2 when   q1 &  q2 are both negative

q1 = -12e , q3 = +12e and q2 = ?

The net force on q2 is zero from fig.b when q3 is at 2xs (2 units) distance, which means force due to q1 on q2 is equal and opposite to force due to q3 on q2.

Therefore.   r32 = 2xs = 3.20m   where, r32 is distance between q3 and q2

Also, F12 = F32 at r32 = 2xs

(k.q1.q2) / (r12)2 = (k.q3.q2) / (r32)2 where, r32 is distance between q3 and q2 and similarly r12

q1 / (r12)2 = q3 / (r32)2   direction of forces are already taken into acount we are just maching its magnitude.

r12 = r32 since - q1 = q3 = 12e

or r12 = r32 = 2xs = 3.20 m

When q3 is taken to infinity then the force experienced by q2 is only due to q1,

Therefore, F2,net = (k.q1.q2) / (r12)2 = 1.168 x 10-25 N where, k is constant = 8.99 x109 Nm2C-2

e = 1.6 x 10-19 C is SI unit

F2,net = (k.q1.q2) / (r12)2

1.168 x 10-25 = [ ( 8.99 x109 ) x (- 12 x 1.6 x 10-19) x q2 ] / 3.202

q2 = [ 1.168 x 10-25 x 3.202 ] / [ ( 8.99 x109 ) x (- 12 x 1.6 x 10-19) ]

q2 = - 6.93 x 10-17 C

or   q2 = [ - 6.93 x 10-17 ] / [ 1.6 x 10-19 ] e

.q2 = - 433.0737 e

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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