# Determine the electric flux through each of the sides

1. May 21, 2007

### Rasine

A 2.10- mC charge is placed at the center of a cube of sides 6.40 cm. Determine the electric flux through each of the sides.

ok so electric flux=E(A)

so i figure that the electic flux will be the same for all six sides of the cube so i am just calculating the flux on one side then X6.

so that would be =(2.10x10^-6)(.064)(.064)=8.60x10^-9...for one side

and that would be 5.16x10^-8 N m^2/C for all 6 sides

but this isn't right...please tell me what i am doing wrong

2. May 21, 2007

### hage567

So what is the charge? It looks like you wrote 2.10 mC, in the question, and then used 2.10x10^-6 C in the calculation.

You never actually calculated the electric field due to the point charge. The E in EA is electric field, in N/C. Check your units, they don't work out to the units of flux the way you have it.

3. May 21, 2007

### Rasine

ohh thats right...ok so now if i calcuate what E is

E=Kq/r^2

now r is at the the center of the cube so r=.064/2

E=(8.99x10^9)(2.1x10^-6)/.032^2=1.84x10^7

A=.064*.064=.004096

now 1/6flux=(1.84x10^7)(.004096)

flux=4.53x10^5....but that still isn't the right answer

4. May 21, 2007

### hage567

$$\Phi=\oint{\bf E} \cdot{d}{\bf A}=\frac{q}{\epsilon_o}$$

Look closely at this. What does it say?

5. May 21, 2007

### Rasine

that means that the flux= the charge on the inside of the soild/ the constant e which is 8.85 x10^-12

6. May 21, 2007

### hage567

Yes, so the net electric flux through any closed surface is equal to the net charge inside the surface divided by $$\epsilon_o$$.

How would you go about solving the problem knowing this? This makes it MUCH easier!

7. May 21, 2007

### Rasine

well the net charge inside is just 2.1x10^-6 C/e is that right?

8. May 21, 2007

### hage567

That's right. As long as you're sure it's 2.10 $$\mu C$$, and not 2.10 mC.

9. May 21, 2007

### Rasine

right it is the first value...i don't have the notation for that in my keypad....

but the when i do that operation =2.37x10^5 N m^2/C which is wrong

10. May 21, 2007

### hage567

OK, but that is for the flux through the entire surface.
I think you want the flux through one side, so divide by six.