How Does Electric Flux Differ Across a Hemispherical Surface?

[mdf730]

Homework Statement

A Gaussian surface in the form of a hemisphere of radius R = 6.08 cm lies in a uniform electric field of magnitude E = 9.35 N/C. The surface encloses no net charge. At the (flat) base of the surface, the field is perpendicular to the surface and directed into the surface. What is the flux through (a) the base and (b) the curved portion of the surface?

Homework Equations

\Phi = EA

The Attempt at a Solution

I assumed a would equal b, that is, both would be pi(.0608)^2 * 9.35, or .108 Nm^2/C.
That is indeed the correct answer for b, but a is still incorrect.

 

[pgardn]

So the e field is penetrating just a circular area pi*r^2. And half the surface area of a sphere 4*pi*r^2 divided by 2?

Have I read the question correctly that the base is just the area of a circle?

 

[vanhees71]

According to Gauß's Law, the total flux must be 0. To evaluate this explicitly for the given closed (!) surface, you have to use vectors:
\Phi=\int_A \mathrm{d} \vec{A} \cdot \vec{E}.
Here, \mathrm{d} \vec{A} is the surface-element vector, whose length is the area of that surface element and which points perpendicularly out of the volume enclosed by the surface (by convention).

To calculate it, parametrize the surface (for the spherical part, use spherical coordinates) and evaluate the corresponding usual double integral.

 

[rude man]

You don't need to evaluate any integrals.

Hint: the total flux leaving the closed hemisphere is what? What is the total flux entering from the bottom? Then, what is the total flux leaving the curved surface?

 

[pgardn]

The Op indicates the answer says there is net flux. Yet it also says a Gaussian surface that does not enclose charge. So maybe the question involves two surfaces that are not closed and have e field penetrating them?

 

[rude man]

The Op indicates the answer says there is net flux. Yet it also says a Gaussian surface that does not enclose charge. So maybe the question involves two surfaces that are not closed and have e field penetrating them?

No. There is just one hemispherical closed surface with the E field perpendicular to the flat side and directed into the surface (hint, hint). There is no net charge inside, therefore no net efflux or influx.

The OP is very close to the right answer for (a) also. I can't give more of a hint without flat-out divulging the answer.