Electric Flux Comparison: Gauss's Theorem and Enclosed Charge

[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.

 

[Tanya Sharma]

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.

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.

 

[VU2]

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

 

[Tanya Sharma]

But is the magnitude of E the same for both radius's?

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

 

[VU2]

E=kq/x^2

 

[Tanya Sharma]

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

 

[VU2]

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

 

[VU2]

Im sorry, I meant 9 times more.

 

[VU2]

Yeah, its much different. Thanks!