# Gauss Theory Question !

1. Jul 15, 2014

### VU2

Two concentric imaginary spherical surface of radius R and 3R, respectively, surrounds a point charge -Q, located at the center of the surface. When compared to the electric flux I1 through the surface of radius R, the electric flux I2 through the surface 3R is.

I know the answer is that I1=I2 based on Gauss's Theorem, I=q/ε, where q is the charge enclosed. But wouldn't the two flux's change if we instead use, I=E(dA), because of different radius in the dA portion? Can someone explain to me how will we get the same, I, if we use the latter equation? Thanks.

2. Jul 15, 2014

### Tanya Sharma

The surface area changes , but so does the electric field . The electric field at distance R and 3R are different .

Both surfaces give same result.

3. Jul 15, 2014

### VU2

Thanks Tanya for replying. But is the magnitude of E the same for both radius's?

4. Jul 15, 2014

### Tanya Sharma

What is the formula of electric field at distance 'x' due to a point charge ?

5. Jul 15, 2014

### VU2

E=kq/x^2

6. Jul 15, 2014

### Tanya Sharma

Now for x = R and x=3R ,do you get same values of E or different ?

7. Jul 15, 2014

### VU2

So at the magnitude of the electric field is actually 4 times more for radius R than 2R?

8. Jul 15, 2014

### VU2

Im sorry, I meant 9 times more.

9. Jul 15, 2014

### VU2

Yeah, its much different. Thanks!