0
\$\begingroup\$

How to calculate flux density, using Gauss' Law, between the plates of a perfect parallel-plate capacitor with +7C and -7C on the top and bottom plates, and a plate area of 1m^2, with a separation of 1m.

The answer is -7C/m^2, but i cannot understand why or how.

\$\endgroup\$
4
  • \$\begingroup\$ Um. There's 7 Coulombs spread across a square meter. The flux density actually is the same regardless of the distance between the plates (ignoring fringing.) This density figure isn't often a concern to designers. On the other hand, the electric field strength does depend on the distance between the plates and is measured in volts per meter. It does matter to designers more, I think. These concerns are different for inductors, where the field density (tesla) does matter a lot regarding materials used and where the strength (amps/meter) is less of a focus. What don't you understand? Need math? \$\endgroup\$
    – jonk
    Commented May 12, 2017 at 20:12
  • \$\begingroup\$ @jonk Why is it negative? \$\endgroup\$ Commented May 12, 2017 at 20:24
  • \$\begingroup\$ I've never seen a need where a signed flux density was important to know about capacitors. And I can't think of one, off-hand. \$\endgroup\$
    – jonk
    Commented May 12, 2017 at 20:27
  • \$\begingroup\$ I also can't think of a reason to estimate a negative magnetic field density in negative-Tesla, either. (Perhaps for some esoteric faster than light drive using negative matter?) I think the minus sign is a mystery that isn't about the physics itself. Look to human error or wayward imagination, I'd say. \$\endgroup\$
    – jonk
    Commented May 12, 2017 at 20:38

2 Answers 2

1
\$\begingroup\$

The flux density is "created" by the charged particles just as our friend Maxwell tells us: \$\nabla \cdot \vec{D} = \rho \$. The other way to look at this relationship is to think that the net charge contained in any volume can be calculated by taking a surface integral of \$\vec{D}\$ over the surface of this volume. Note that \$\vec{D}\$ is a vector quantity.

See the three steps in the drawing below, red marks positive charge, blue negative, and black is the flux density \$\vec{D}\$:

  1. For a point charge, the \$\vec{D}\$ can be calculated just by drawing a sphere around the charge and saying that at distance \$r\$ the surface density is \$ D = Q / 4\pi r^2\$.

  2. For a single (very large) plate that has a charge density of \$ 7~C/m^2\$, the \$\vec{D}\$ can be calculated by taking a small (square meter) section of the plate. We can see that the charge contained in the section is 7 C and the surface area over which the 2 square meters (one surface on both sides of the plate). So \$D = \frac{7~C/m^2}{2} = 3.5~C/m^2\$.

  3. In phase three, we take a second plate that has a charge of \$-7~C/m^2\$. This second plate creates a \$\vec{D}\$ around itself just like the first one, just flux going in the opposite direction, that is, towards the plate. The vector sum of these two fields is \$\vec{D} = 7~C/m^2\$ towards the negatively charged plate. The sign of the answer is just a matter of how you set your coordinate system. Normally we say that z-axis points upwards and in your question, the top plate is the positively charged one. As a conclusion, the \$\vec{D} = - 7 C/m^2 \vec{\hat{z}}\$, where \$\vec{\hat{z}}\$ is the unit vector.

Flux around point charge, single plate, and two parallel plates.

\$\endgroup\$
0
\$\begingroup\$

My old Schaums Outline tells me the sign depends on which plate is your voltage reference.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.