Two charges, Q1= 3.50 ?C, and Q2= 6.00 ?C are located at points (0,-3.50 cm ) an

Question

Two charges, Q1= 3.50 ?C, and Q2= 6.00 ?C are located at points (0,-3.50 cm ) and (0,+3.50 cm), as shown in the figure:

P (5.0cm, 0)

What is the magnitude of the electric field at point P, located at (5.0 cm, 0), due to Q1 alone?

*I was able to solve this one but am struggling with the rest.

8.45?106N/C

1.) What is the x-component of the total electric field at P?

2.) What is the y-component of the total electric field at P?

3.) What is the magnitude of the total electric field at P?

4.) Now let Q2= Q1= 3.50 uC. Note that the problem now has a symmetry that you should exploit in your solution. What is the magnitude of the total electric field at P?

5.) Given the symmetric situation of the previous problem, what is the magnitude of the force on an electron placed at point P?

Explanation / Answer

Ex= Kq/r^2*cos(theta) E2x= 9*10^9*6*10^-6*(5/6.103)/ (6.103*10^-2)^2 = 11877735.6 N/C = 11.88 *10^6 N/C E1x= 9*10^9*3.5*10^-6*(5/6.103)/ (6.103*10^-2)^2 = 6928679.11 = 6.93*10^6 N/C 1.)Ex net= 18806414.71 = 18.80*10^6 N/C 2.) Ey= Kq/r^2*sin(theta) E1y= 9*10^9*3.5*10^-6*(3.5/6.103)/ (6.103*10^-2)^2 =4850075.377 upwards = 4.85*10^6 N/C upwards E2y= 9*10^9*6*10^-6*(3.5/6.103)/ (6.103*10^-2)^2 = 8314414.93 downwards = 8.31*10^6 N/C downwards so Ey net= 8314414.93- 4850075.377= 3464339.55 downwards = 3.46*10^6 N/C 3.) magnitude of total electric field = sqrt(Ex^2+ Ey^2) = sqrt((18.80*10^6)^2+ (3.46*10^6)^2) = 19115742.203 N/C = 19.12*10^6 N/C 4.) magnitude of the total electric field will be 2*E1x since the Y component will cancel out each other. thus NET MAGNITUDE= 13.86*10^6 N/C 5.) F= q*E = 3.5*10^-6*13.86*10^6 N =48.51 N PLEASE RATE THIS ANSWER A+.. :)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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