Chapter 21 015 The charges and coordinates of two charged particdes held fixed i

Question

Chapter 21 015 The charges and coordinates of two charged particdes held fixed in an xy plane are 01-3.33 uC, X-3.07am, -0.604 cm and oz -5.53 uC, - 2.28 cm, 2-1.08 cm. Find the (a) magnitude and (b) direction (with respect to +x-axis in the range (-180:180°]) of the electrostatic force on partide 2 due to particle 1. At what (c) x and (d) y coordinates should a third partide of charge q-4.53 LC be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero? (a) Number (b) Number (c) Number (d) Number Click if you would like to Show Work for this question: Qpen Show Work Units Units

Explanation / Answer

Distance between q1 and q2 is r= ((3.07-(-2.28))^2+(0.604-1.08)^2)^1/2 = 5.37cm= 0.0537m

So, magnitude of electrostatic force on particle 2 due to particle 1 is

F = k*q1*q2 / r^2 = 9*10^9*3.33*10^-6*5.53*10^-6 / 0.0537^2

or, (a) F = 57.4 N

Direction of this force is directly towards F1 { because they are unlke charges }, and is given by tan^-1 (y2-y1 / x2-x1)

So, (b) direction is -5.08 degrees.

Now, magnitude of force on 2 due to 1= magnitude of force on 2 due to 3

Or, k*q1*q2/ r2^1 = k*q2*q3/ r3^2 { where r1= distance between q1 and q2, r3= distance between q3 and q2 }

Or, q1/r1^2 = q3/r3^2

So, q1/q3= (r1/r3)^2

So, 3.33*10^-6/ 4.53*10^-6 =  (r1/r3)^2

So,  (r1/r3)^2= 0.735

Or, ( ((x1-x2)^2+(y1-y2)^2) = 0.735 * ((x3-x2)^2+(y3-y2)^2) )

Or, 0.094+1.4+5.2+0.365-1.3+1.166 = 0.735 ( x3^2+4.56x3+5.2+y3^2-2.16y3+1.166 )

Or, 6.9 = 0.735x3^2+3.35x3+3.8+0.735y3-1.58y3+0.857

Or, 2.243=  0.735x3^2+3.35x3+0.735y3^2-1.58y3 ___________(A)

Also, the position of q3 should be along the line joining q1 and q2, so, from the equation of the line joining q1 and q2, which is y= 0.877 - 0.89x

So, using this value of y in (A),

2.243=  0.735x3^2+3.35x3+0.735(0.877-0.89x3)^2-1.58(0.877 - 0.89x3)

2.243=  0.735x3^2+3.35x3+0.64-0.11x3+0.58x^2 -1.38+1.4x3

2.98 = 1.3x3^2 + 4.64x3

Or, 1.3x3^2 + 4.64x3- 2.98 = 0

Solving this quadratic, x3= 0.55 or -4.12. The charge q3 should be nearer to q2 than q1,

so, (c) x= -4.12 cm

y3= 0.877 - 0.89x3

So, (d) y3= 4.54 cm

if unclear then ask, please upvote to appreciate

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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