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My teacher told that,

Consider a parallel plate capacitor of capacitance C, if we isolate one of the plate, then capacitance of that plate becomes double i.e 2C.

Can anyone help me out in finding out the reason? enter image description here

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    \$\begingroup\$ Add a diagram. How would you isolate one plate? \$\endgroup\$
    – Transistor
    Oct 3, 2021 at 13:49
  • \$\begingroup\$ Ya, I have done it @ transistor \$\endgroup\$ Oct 3, 2021 at 13:54

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This is incorrect.

Capacitance is \$C=\large\epsilon\frac{A}{d}\$, where \$\large\epsilon\$ = the permittivity of the medium (usually vacuum or air, \$\epsilon = 8.85*10^{-12}\$). \$C\$ can also be defined at \$C=\frac{Q}{V}\$, where \$Q\$ is the charge on one plate.

In a parallel plate capacitor, \$A\$ and \$d\$ are clearly obvious. Imagine that \$d\$ increases by separating the plates -- the capacitance decreases.

In the limit, as you separate the plates, the capacitance will not decrease to 0, but to a limiting value - see http://www.vias.org/matsch_capmag/matsch_caps_magnetics_chap2_10.html

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  • \$\begingroup\$ Sir/madam ,I didn't understand that how your answer relates to my question, can you please tell me then what happens when we isolate a plate? What is the capacitance of the plate? \$\endgroup\$ Oct 4, 2021 at 11:29
  • \$\begingroup\$ From the reference above, a sphere (with a radius, and in an isolated universe) would have a capacitance of 4.π.e0.radius. An isolated parallel plate would have no more capacitance than a sphere of the same size. \$\endgroup\$
    – jp314
    Oct 4, 2021 at 18:11
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Unless I misunderstand the question, the capacitance will not double. The self capacitance of the earth is measured in uF. There are parallel plate capacitors with mutual capacitance in the Farads. If removing one of the plates doubled the capacitance, then one of the plates would have greater capacitance than the earth. That doesn't seem correct.

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At first the capacitace between 2 parallel plates:

enter image description here

(the formulas are captured from Wikipedia)

Then the self capacitance (see NOTE1) of a single disc:

enter image description here

Let the parallel plates also be discs with radius =a. Their areas are A=Pi*(a^2). If you say that the self capacitance of the disc is twice the capacitance of the capacitance between 2 discs you get condition

d=(Pi/4)*a

That's nothing special. I guess many of us would draw just the right capacitor for the wanted doubling if they must draw a sketch of a capacitor which has circular plates and air insulation.

NOTE1: The capacitance of the imagined capacitor where one of the plates is the disc and the other plate is the infinitely large sphere at infinity around the disc.

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Essentially proportional to the surface area of the capacitor plate. In mutual capacitance, the capacitive surface is the surface directed at the other parallel plate. In self-capacitance, the entire surface area of both sides of the plate are counted. See Wikipedia for self-capacitance.

https://en.m.wikipedia.org/wiki/Capacitance#Self_capacitance

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