- #1
lluke9
- 27
- 0
Okay, so I know
[tex]C=Q/ΔV[/tex]
And ΔV is the sum of the electric fields multiplied by the distance between the charges, so if the first charge has a charge of Q and the other has -Q with R distance between, the electric potential/voltage is:
[tex]ΔV = [(kQ/r^2) + (kQ/r^2)]R[/tex]
so
[tex]ΔV = 2kQ/R[/tex].
And [tex]C = Q/ΔV[/tex]
so...
[tex]C = Q/(kQ/R)[/tex]
and...
[tex]C = R/2k[/tex]
and...
[tex]C = R/2[1/(4Πε_0)][/tex]
and
[tex]C = 2ΠRε_0[/tex]
What...?
It would make some semblance of sense if R were inversely proportional to capacitance, but it isnt...
[tex]C=Q/ΔV[/tex]
And ΔV is the sum of the electric fields multiplied by the distance between the charges, so if the first charge has a charge of Q and the other has -Q with R distance between, the electric potential/voltage is:
[tex]ΔV = [(kQ/r^2) + (kQ/r^2)]R[/tex]
so
[tex]ΔV = 2kQ/R[/tex].
And [tex]C = Q/ΔV[/tex]
so...
[tex]C = Q/(kQ/R)[/tex]
and...
[tex]C = R/2k[/tex]
and...
[tex]C = R/2[1/(4Πε_0)][/tex]
and
[tex]C = 2ΠRε_0[/tex]
What...?
It would make some semblance of sense if R were inversely proportional to capacitance, but it isnt...
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