Two charges, Q1= 3.40 C, and Q2= 5.60 C are located at points (0,-3.50 cm ) and

Question

Two charges, Q1= 3.40 C, and Q2= 5.60 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 (6.00 cm, 0), due to Q1 alone? Tries 0/99 What is the x-component of the total electric field at P? Tries 0/99 What is the y-component of the total electric field at P? Tries 0/99 What is the magnitude of the total electric field at P? Tries 0/99 Now let Q2 = Q1 = 3.40 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? Tries 0/99 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(3.5/6.0) =30.25o

Q2 creates a field that is down and to the right at angle of arctan(-3.5/4.5) =30.25o

Now E = k*q/r^2 The distance squared for both is 0.035^2 + 0.0^2 = 0.004825

Ex = k*Q1/r^2*cos(30.25) + k*Q2/r^2*cos(-30.25)

= 9.0x10^9*2.00x10^-6/0.004825*cos(30.25) + 9.0x10^9*5.60x10^-6/0.004825*cos(-30.25)

=1.94x10^8N/C

and

b) Ey = k*Q1/r^2*sin(30.25) + k*Q2/r^2*sin(-30.25)

= 9.0x10^9*3.40.00x10^-6/0.004825*sin(30.25) + 9.0x10^9*5.60x10^-6/0.004825*sin(-30.25)

= -8.1x10^7N/C


c) mag = sqrt(Ex^2 + Ey^2) = sqrt((1.94x10^8N/C)^2 + (-8.1x10^7)^2) =2.10x10^9N/C

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

So Ey = 2*k*Q/r^2*sin() = 3.40*9.0x10^9*3.40.0x10^-6/0.004825*sin(30.25) = 4.28x10^6N/C

e) F = E*q =4.28x10^6N/C*1.60x10^-19C = 6.48x10^-13N

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