# 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
I am confused about the directions of the electric fields. Can someone maybe help me by explaining how to get the directions?
To get directions, you need to use the vector form of Gauss' Law. Are you familiar with how to use the (2-D) vector form?

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html#c1 kuruman
Homework Helper
Gold Member
The electric field depends on the position of the point where you want to know the field not just the distance from one of the charges. Without that information you cannot answer this question. This is a bad problem as stated. Are you sure you are looking for the electric field and not the electric flux through spheres centered at charge +q?

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