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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.

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  • \$\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 May 12 '17 at 20:12
  • \$\begingroup\$ @jonk Why is it negative? \$\endgroup\$ – stellarhawk 34 May 12 '17 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 May 12 '17 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 May 12 '17 at 20:38
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My old Schaums Outline tells me the sign depends on which plate is your voltage reference.

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