Print I Two charges, Q1 2.30 C\' and Q2= 5.80 C are located at points (0,-3.50 c

Question

Print I Two charges, Q1 2.30 C' and Q2= 5.80 C are located at points (0,-3.50 cm ) and (0,+3.50 cm), as shown in the figure. What is the magnitude of the electric field at point P, located at (5.50 cm, 0), due to Qi alone? Submit Answer Tries 0/10 What is the x-component of the total electric field at P? Submit Answer Tries o/10 What is the y-component of the total electric field at P? Submit Answer Tries 0/10 What is the magnitude of the total electric field at P? Submikt Anwer Tries 0/10 Now let Q2-Q1 = 2.30 . Note that the problem now has a symmetry that you should exploitin your solution, what is the magnitude ofthe total electric field at P 15.0x107 NIC Can you say anything about the y-component of the total electric field before you stars calculating? Submt Answer Incorrect. Tries 1/10 Previous Tries Given the svmmetric situation of the previous problem, what is the magnitude of the force on an electron placed at point P? Submi Answer Tries 0/10O This discussion is closed. send Feedback

Explanation / Answer

solution:

Q1 creates a field at an angle of tan-1(3.5/5.5) = 32.47o

Q2 creates a field at an angle of tan-1(-3.5/5.5) = -32.47o

Now E = k*q/r^2

The distance squared for both is r2 = 0.035^2 + 0.055^2 = 4.25*10-3

Ex = k*Q1/r^2*cos(32.47) + k*Q2/r^2*cos(-32.47)

= [9.0x109*2.30x10-6 / 4.25*10-3]*cos(32.47) + [9.0x109*5.80x10-6/ 4.25*10-3]*cos(-32.47)
= 4.11*106 + 1.036*107

=1.447x107 N/C



b) Ey = k*Q1/r^2*sin(32.47) + k*Q2/r^2*sin(-32.47)

= [9.0x109*2.30x10-6 / 4.25*10-3]*sin(32.47) + [9.0x109*5.80x10-6/ 4.25*10-3]*sin(-32.47)
=2.61*106 - 6.59*106
= -3.98*106 N/C

c) mag = sqrt(Ex^2 + Ey^2) = sqrt((1.447x107)^2 + (3.98x10^6)^2) =1.50*107 N/C

d) Now only the y component is non zero due to symmetry

So Ey = 2*k*Q/r^2*sin() = 2*[9.0x109*2.30x10-6 / 4.25*10-3]*sin(32.47) = 5.22*106 N/C

e) F = E*q = 5.22*106 N/C*1.60x10-19 C = 8.352x10-13N

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