# Find the Electric Field E using Gauss' Law

## 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?

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berkeman
Mentor
kuruman
The electric field is a consequence of the presence of both charges, therefore you should take $4q+$ into account as well. Superposition theorem.