Electric Flux Through Closed Surface: Problem Explained

[LucasGB]

The electric flux through a closed surface is the integral of the dot product E.da. Suppose we have a point charge at the center of a sphere. The electric field at the surface of the sphere is constant and can therefore be removed from the integral. Inside the integral we are left with da. But the integral of the area vector for any closed surface is zero! Therefore, the flux is zero, but we know this is not true. What gives?

 

[tiny-tim]

Hi LucasGB!

… Suppose we have a point charge at the center of a sphere. The electric field at the surface of the sphere is constant and can therefore be removed from the integral. Inside the integral we are left with da.

No, if you mean what I think you mean, we are left with ∫da (not ∫da), which is 4πr².

 

[dacruick]

Yes tiny tim is correct. What you are left with when you take E out, is the integral of 1 x da. That means after the integral you have the surface area of a sphere. which is 4πr2

 

[LucasGB]

Oh, I see. So when I have a dot product inside an integral sign, I can't take a vector out and leave a vector in? I must write the dot product in component notation and take a scalar out and leave a scalar in? In that case, this will make sense.

PS.: tiny-tim, where are my mathematical symbols this time?! :D

 

[tiny-tim]

PS.: tiny-tim, where are my mathematical symbols this time?! :D

Take care of them!

I usually only give them out once! ​

 

[Redbelly98]

Another way to explain the problem: the electric field is not constant at the surface of the sphere, as stated in the OP. Only the magnitude is constant -- but since the direction is different everywhere on the surface then the vector is different everywhere on the surface.

p.s. have another dose of math symbols...

 

[LucasGB]

Another way to explain the problem: the electric field is not constant at the surface of the sphere, as stated in the OP. Only the magnitude is constant -- but since the direction is different everywhere on the surface then the vector is different everywhere on the surface.

Oooh, that's true!

But what if I have a situation where the vector is indeed constant throughout the surface? In that case can I take the constant vector out of the integral and leave the vector da inside?

 

[tiny-tim]

But what if I have a situation where the vector is indeed constant throughout the surface? In that case can I take the constant vector out of the integral and leave the vector da inside?

Hi LucasGB!

Yes, if E is constant (magnitude and direction), then

∫ E.da

can be written E.∫ da