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Homework Statement:

Two charges +q and +4q are separated a distance of 2r. Use Gauss' Law to find the electric field E at a distance:
a) r around the charge +q
b) 3r around charge +q
Homework Equations:

##A = 4πr^2##
##E.A=\frac {q}{ε_°}##
I tried to work out both a) and b), but I am not sure if I am correct. I drew a picture with a sphere around q first with radius r and then with radius 3r.
For a) ##E.A=\frac {q}{ε_°}## (when using Gauss' Law)
Since ##A=4πr^2##, I substituted this in the equation and solved for E giving me ##E_q= \frac {kq}{r^2}##
My problem is that I am not sure if this is correct and the direction of the field is confusing me. I thought it should be negative.....##E_q= \frac {kq}{r^2}##
I then solved for the electric field using the +4q charge. This I did in the same way as the +q charge. ##E_{4q} = \frac {4q}{4πr^2ε_°}## which gives ##E_{4q}= \frac {4kq}{r^2}## I thought this would be positive.
To get the electric field E: ##E=E_{4q}  E_q## which gives ##E= \frac {3kq}{r^2}##
I am not really sure if I have done this correctly. The directions of the fields are really confusing me.
For b) I basically did the same as for a). The only difference was the distance which increased from r to 3r.
I started by saying ##E_q = \frac {q}{4π(9r^2)ε_°}## and simplifying this gives me ##E_q = \frac {kq}{9r^2}##. I thought the direction would be to the left, making it negative: ##E_q = \frac {kq}{9r^2}##
For the other charge ##E_{4q} = \frac {4q}{4πr^2ε_°}## which gives ##E_{4q} = \frac {4kq}{r^2}##. I reckon the direction would be to the left, also making it negative: ##E_{4q} = \frac {4kq}{r^2}##
Getting E:
##E=E_q+E_{4q}## which gives ##E= \frac {kq}{9r^2} \frac {4kq}{r^2}##
Upon simplification I get:
##E=\frac {37kq}{9r^2}##
I am confused about the directions of the electric fields. Can someone maybe help me by explaining how to get the directions?
For a) ##E.A=\frac {q}{ε_°}## (when using Gauss' Law)
Since ##A=4πr^2##, I substituted this in the equation and solved for E giving me ##E_q= \frac {kq}{r^2}##
My problem is that I am not sure if this is correct and the direction of the field is confusing me. I thought it should be negative.....##E_q= \frac {kq}{r^2}##
I then solved for the electric field using the +4q charge. This I did in the same way as the +q charge. ##E_{4q} = \frac {4q}{4πr^2ε_°}## which gives ##E_{4q}= \frac {4kq}{r^2}## I thought this would be positive.
To get the electric field E: ##E=E_{4q}  E_q## which gives ##E= \frac {3kq}{r^2}##
I am not really sure if I have done this correctly. The directions of the fields are really confusing me.
For b) I basically did the same as for a). The only difference was the distance which increased from r to 3r.
I started by saying ##E_q = \frac {q}{4π(9r^2)ε_°}## and simplifying this gives me ##E_q = \frac {kq}{9r^2}##. I thought the direction would be to the left, making it negative: ##E_q = \frac {kq}{9r^2}##
For the other charge ##E_{4q} = \frac {4q}{4πr^2ε_°}## which gives ##E_{4q} = \frac {4kq}{r^2}##. I reckon the direction would be to the left, also making it negative: ##E_{4q} = \frac {4kq}{r^2}##
Getting E:
##E=E_q+E_{4q}## which gives ##E= \frac {kq}{9r^2} \frac {4kq}{r^2}##
Upon simplification I get:
##E=\frac {37kq}{9r^2}##
I am confused about the directions of the electric fields. Can someone maybe help me by explaining how to get the directions?