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.

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

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

What is the magnitude of the total electric field at P?

Now let Q2= Q1= 3.50 ?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?

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

Explanation / Answer

Q1 creates a field that is up and to the right at angle of arctan(3.5/4.5) =37.87o Q2 creates a field that is down and to the right at angle of arctan(-3.5/4.5) =37.87o Now E = k*q/r^2 The distance squared for both is 0.035^2 + 0.045^2 = 0.00325 Ex = k*Q1/r^2*cos(37.87) + k*Q2/r^2*cos(-37.87) = 9.0x10^9*2.00x10^-6/0.00325*cos(37.87) + 9.0x10^9*6.30x10^-6/0.00325*cos(-37.87) =1.81x10^7N/C and b) Ey = k*Q1/r^2*sin(37.87) + k*Q2/r^2*sin(-37.87) = 9.0x10^9*2.00x10^-6/0.00325*sin(37.87) + 9.0x10^9*6.30x10^-6/0.00325*sin(-37.87) = -7.31x10^6N/C c) mag = sqrt(Ex^2 + Ey^2) = sqrt((1.81x10^7)^2 + (7.31x10^6)^2) =1.96x10^7N/C d) Now only the y component is non zero due to symmetry So Ey = 2*k*Q/r^2*sin(?) = 2*9.0x10^9*2.0x10^-6/0.00325*sin(37.87) = 6.80x10^6N/C e) F = E*q = 6.80x10^6N/C*1.60x10^-19C = 1.09x10^-12N

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