Two charges, Q1= 2.80 C, and Q2= 5.90 pC are located at points (0,-4.00 cm ) and
ID: 1661532 • Letter: T
Question
Two charges, Q1= 2.80 C, and Q2= 5.90 pC are located at points (0,-4.00 cm ) and (0,+4.00 cm), as shown in the figure 2, What is the magnitude of the electric field at point P, located at (5.00 cm, o), due to Q1 alone? Submit Answer Tries 0/20 What is the x-component of the total electric field at P? Submit Answer Tries 0/20 What is the y-component of the total electric field at P? Submit Answer Tries 0/20 What is the magnitude of the total electric field at P? Submit Answer Tries 0/20 Now let Q2 = Q1 = 2.80 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? Submit Answer Tries 0/20 Given the symmetric situation of the previous problem, what is the magnitude of the force on an electron placed at point P? Submit Answer Tries 0/20Explanation / Answer
r = (42+52) = 6.4 cm 2= 6.4 *10--2 cm
r2 = 41 cm2
A) Electric field at P by Q1 alone as E1 = k.Q1/r2 = 9x109.(2.80x10-6)/(41 x10-4 )
E1 = 6.14 x 106 N/C
B) from right triangle we know that cos = 5/6.4 = 0.78
sin = 4/6.4 = 0.625
Electric field at P by Q2 alone as E2 = k.Q2/r2 = 9x109.(5.90x10-6)/(41x10-4)
E2 = 12.95 x 106 N/C
X componen as Ex = E1 cos + E2 cos = 6.14x106.(0.78) + 12.95x106.(0.78) = 14.89 x106N/C
C) Y component as Ey = E1 sin - E2 sin = 6.14x106(0.625) - 12.95x106.(0.625) = -4.256 x 106N/C
D) Etot = (Ex2 + Ey2) = [(14.89 x106)2+(-4.256 x106)2] =15.48 x106 N/C
E) If Q1 =Q2 = 2.80C that E1 = E2 and the y component of E1 and E2 cancel out so Etot = 2E1 cos
Etot = 2.(6.14 x106).(0.78) = 9.578 x 106 N/C
F) Electric force F =q.Etot = (1.6x10-19).(9.578 x106) = 1.532 x10-12 N
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