Does Earth's Inward Electric Field Result in Outward Electric Flux?

AI Thread Summary
The Earth has an inward electric field of 150 N/C at its surface, leading to confusion about the net electric flux through a surrounding spherical surface. The flux is calculated using the equation Φ = EAcos(theta), where the electric field lines point inward, suggesting a negative flux. This indicates a net flux into the volume rather than outward, leading to the conclusion that the net flux is -150*(4*pi*r^2), where r is the Earth's radius. The discussion clarifies that while the term "net" can imply a non-zero value, in this context, it signifies inward flux despite the outward orientation of the surface vectors. Ultimately, the net flux is considered negative, representing an inward flow of electric field lines.
hitemup
Messages
81
Reaction score
2

Homework Statement



The Earth possesses an electric field of (average) magnitude 150 N/C near its surface. The field points radially inward. Calculate the net electric flux outward through a spherical surface surrounding, and just beyond, the Earth's surface.

Homework Equations



Flux = EAcos(theta)

The Attempt at a Solution



I know that flux is proportional to the number of lines passing through a surface. Thus, we only need to calculate the the flux with respect to earth.

What confuses me is that, is there any net flux outward? Since the field lines point inward, and infinitesimally small surface vectors point outward, the sign of the flux will be negative. If the flux is negative, then there is a net flux into the volume. So I don't know if the answer is zero or simply (+-?)150*(4*pi*r^2), where r is the radius of the Earth.
 
Physics news on Phys.org
all the flux is outward on a closed surface. do you know gauss theorem ?
 
Xsnac said:
all the flux is outward on a closed surface. do you know gauss theorem ?

\Phi = \oint \vec{E}*d\vec{A}

d\vec{A} always points outward from the enclosed surface.
So depending on \vec{E}, the flux can either be positive or negative?
 
hitemup said:
\Phi = \oint \vec{E}*d\vec{A}

d\vec{A} always points outward from the enclosed surface.
So depending on \vec{E}, the flux can either be positive or negative?
I'm not sure but I think that flux can eighter be 0 or positive . not negative.
 
What confuses me is that, is there any net flux outward? ##\qquad##correct
Since the field lines point inward, ##\qquad## ##\qquad####\qquad####\qquad##correct
and infinitesimally small surface vectors point outward, ##\qquad##correct
the sign of the flux will be negative. ##\qquad## ##\qquad## ##\qquad## ##\qquad##correct
If the flux is negative, then there is a net flux into the volume. ##\qquad##correct

So I don't know if the answer is zero or simply (+-?)150*(4*pi*r^2), where r is the radius of the Earth.
##\qquad## no reason to shy away from what you deduced: ##-## 150*(4*pi*r^2)

This does represent a net outward flux, with a negative sign. Net in the scientific sense of non-zero(*). Like a net force can have a negative sign if the positive axis points the other way. Common language doesn't like that and claims that net outward has to be positive definite. Maybe that explains our hesitation.

(*) And even that isn't very scientific: net is net and doesn't say anything about the value; it can be zero just as well.
 
Last edited:
  • Like
Likes hitemup
Xsnac said:
I'm not sure but I think that flux can eighter be 0 or positive . not negative.
The flux can be and in this case is negative (outward). That's equivalent to saying that the net flux is inward.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top