es/4710/assignments/84590 Three charges are situated at the comers of a square a

Question

es/4710/assignments/84590 Three charges are situated at the comers of a square as shown. The square has sides d- 0.2 m. The charges in the three corners are qA-40 nC, qp +30 nC, and qc- 40 nC. What is the magnitude of the electric field at the free corner? N/C What is the direction of the electric field vector relative to horizontal? at o Select an answer the horizontal to the Select an answer If an electron were situated at the free corner, what would be the magnitude of the electrostatic force on it? N Use E notation if needed Assuming no other significant forces, what would be the magnitude of the acceleration of the electron? m/s2 What is the direction of the acceleration relative to the horizontal? l?Select an answer. the horizontal to the Select an answer ' at Insert Priso Delete Home End

Explanation / Answer

Given

charges are at three corners of a square of side d = 0.2 m

qA = -40 nC , qB = +30 nC , qC = -40 nC

the distance from the charge qB to free corner is rb = sqrt(2) *d = sqrt(2) *0.2 = 0.28284 m

and rA = 0.2 m , rC = 0.2 m

electric field is E= kq/r^2 , force F = kq1*q2/r^2

the electric field along x axis is Ex =k(-qA/rA^2+qB cos45/rB^2)

= (9*10^9)((-40*10^-9) /0.2^2 + 30*10^-9 cos45 / 0.28284^2)N/C

= -6613.46884 N/C

and y component of the field is  

Ey = k(-qC/rC^2+qB sin45/rB^2)

Ey = (9*10^9)((-40*10^-9) /0.2^2 + 30*10^-9sin45 / 0.28284^2)N/C

Ey = -6613.46884 N /C

the electric field is  

E = Exi + Ey j

E = -6613.46884 N/C i -6613.46884 N/C j

E = sqrt((-6613.46884 )^2+(-6613.46884 )^2) N/C

E = 9352.85732 N/C

the direction is  

theta = arc tan ((-6613.46884)/(-6613.46884)) = 45 degrees

the direction of hte electric field vector relative to the horizontal is 45 degreeesw

if electron is there at the free corner

Fx = k*q(qA/rA^2-qB cos45/rB^2) N

Fx = (9*10^9*1.6*10^-19)((40*10^-9) /0.2^2 - 30*10^-9 cos45 / 0.28284^2)N = 1.0581550*10^-15 N

Fy = k*e(qC/rC^2-qB sin45/rB^2)

Fy = (9*10^9*1.6*10^-19)((40*10^-9) /0.2^2 - 30*10^-9sin45 / 0.28284^2)N = 1.0581550143724*10^-15 N

F = Fx i + Fy j = 1.0581550*10^-15N i + 1.0581550*10^-15N j

magnitude is  

F = sqrt((1.0581550*10^-15)^2+(1.0581550*10^-15)^2) N = 1.4964571*10^-15 N

the direction is theta = arc tan (Fy/Fx) = arc tan ((1.0581550*10^-15)/(1.0581550*10^-15)) = 45 degrees

the acceleration is F = ma

a = F/m

a = 1.4964571*10^-15 / (9.1*10^-31) m/s2 = 1.6444*10^15 m/s2

direction is in the force direction itself

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

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