Two charges, Q1= 3.80 ?C, and Q2= 6.90 ?C are located at points (0,-2.50 cm ) an

Question

Two charges, Q1= 3.80 ?C, and Q2= 6.90 ?C are located at points (0,-2.50 cm ) and (0,+2.50 cm), as shown in the figure.

part 1) What is the magnitude of the electric field at point P, located at (6.00 cm, 0), due to Q1 alone?

part 2) What is the x-component of the total electric field at P?

part 3) What is the y-component of the total electric field at P?

part 4) What is the magnitude of the total electric field at P?

part 5) Now let Q2 = Q1 = 3.80 ?C. 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?

part 6)Given the symmetric situation of the previous problem, what is the magnitude of the force on an electron placed at point P?

Explanation / Answer

(a) magnitude of the electric field at point P, located at (6.00 cm, 0), due to Q1 alone

E = 8.98*109*3.8*10-6/0.065*0.065

     = 8076.68 *103 N/C

(b) the x-component of the total electric field at P

Ex = 8.98*109*3.8*10-6*cos22.62/(0.065)2 + 8.98*109*6.9*10-6*cos22.62/(0.065)2

      = 7455.39*103 + 13537.38 *103

       = 20992.77 *103 N/C

(c) y-component of the total electric field at P

Ey = 8.98*109*3.8*10-6*sin22.62/(0.065)2 - 8.98*109*6.9*10-6*sin22.62/(0.065)2

      = 3106.43*103 - 5640.63 *103

       = -2534.2 *103 N/C

(d) magnitude of the total electric field at P

Enet = sqrt((20992.77 *103)2+(2534.2 *103)2)

         = 21145.178*103 N/C

(e) Here, vertical fields cancel out

so, net electric field = 8.98*109*3.8*10-6*cos22.62/(0.065)2 + 8.98*109*3.8*10-6*cos22.62/(0.065)2

      = 7455.39*103 + 7455.39*103

       = 14910.78 *103 N/C

(f) force on electron placed at P = -1.6*10-19 * 14910.78 *103

                                                    = -2.385*10-12 N    towards -x direction

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