A charge q 1 = +3.00 nC is placed at the origin of an xy -coordinate system, and

Question

A charge q1 = +3.00 nC is placed at the origin of anxy-coordinate system, and a charge q2= -1.00 nC is placed on the positivex-axis at x = 4.00 cm. (a) If a third charge q3 =+6.00 nC is now placed at the pointx = 4.00 cm, y = 3.00 cm, find the x-and y-components of the total force exerted on this chargeby the other two.
1 N (x-component)
2 N (y-component)

(b) Find the magnitude and direction of this force.
3 N
4° (counterclockwise from thex-axis)
(a) If a third charge q3 =+6.00 nC is now placed at the pointx = 4.00 cm, y = 3.00 cm, find the x-and y-components of the total force exerted on this chargeby the other two.
1 N (x-component)
2 N (y-component)

(b) Find the magnitude and direction of this force.
3 N
4° (counterclockwise from thex-axis)

Explanation / Answer

charges q1 = 3 * 10^-9 C            q2 = -1 * 10 ^ -9 C            q3 = 6 * 10 ^ -9 C X-component : --------------- force on q3 due to q1 is F= Kq1 * q3 / ( 4cm ) ^ 2                                        = Kq1 * q3 / ( 0.04 m ) ^ 2 where K = coulomb's constant = 8.99 * 10^9 N m^2 / C ^2 Plug the values weget   F = 101.13 * 10^-6 N Force on q3 due to q2 is F' = 0     SInceboth at same point in x-direction So, Net x-component of force F " = F + F ' =101.13*10^-6 N Y-component : --------------- force on q3 due to q1 is F= Kq1 * q3 / ( 3cm ) ^ 2                                        = Kq1 * q3 / ( 0.03 m ) ^ 2 where K = coulomb's constant = 8.99 * 10^9 N m^2 / C ^2 Plug the values weget   F = 179.8 * 10^-6N It is along positive y-axis force on q3 due to q2 is F'= Kq2 * q3 / ( 3cm ) ^2                                        = Kq2 * q3 / ( 0.03 m ) ^ 2 where K = coulomb's constant = 8.99 * 10^9 N m^2 / C ^2 Plug the values weget   F' = 59.93 * 10^-6N It is along negative y-axis So, Net y-component F "' = F-F'                                         = 119.87 * 10 ^ -6 N (b). Magnitude of the force = [ F " ^ 2 + F "'^2]                                           = 156.83 * 10 ^ 3 N Let the force vector makes an angle with x-axis thentan = F"'/F"                                                                                        = 49.84 degrees force on q3 due to q1 is F= Kq1 * q3 / ( 3cm ) ^ 2                                        = Kq1 * q3 / ( 0.03 m ) ^ 2 where K = coulomb's constant = 8.99 * 10^9 N m^2 / C ^2 Plug the values weget   F = 179.8 * 10^-6N It is along positive y-axis force on q3 due to q2 is F'= Kq2 * q3 / ( 3cm ) ^2                                        = Kq2 * q3 / ( 0.03 m ) ^ 2 where K = coulomb's constant = 8.99 * 10^9 N m^2 / C ^2 Plug the values weget   F' = 59.93 * 10^-6N It is along negative y-axis So, Net y-component F "' = F-F'                                         = 119.87 * 10 ^ -6 N (b). Magnitude of the force = [ F " ^ 2 + F "'^2]                                           = 156.83 * 10 ^ 3 N Let the force vector makes an angle with x-axis thentan = F"'/F"                                                                                        = 49.84 degrees force on q3 due to q2 is F'= Kq2 * q3 / ( 3cm ) ^2                                        = Kq2 * q3 / ( 0.03 m ) ^ 2 where K = coulomb's constant = 8.99 * 10^9 N m^2 / C ^2 Plug the values weget   F' = 59.93 * 10^-6N It is along negative y-axis So, Net y-component F "' = F-F'                                         = 119.87 * 10 ^ -6 N (b). Magnitude of the force = [ F " ^ 2 + F "'^2]                                           = 156.83 * 10 ^ 3 N Let the force vector makes an angle with x-axis thentan = F"'/F"                                                                                        = 49.84 degrees

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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