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On my homework, I came across a question that asks to compare the capacitances of these two spheres. And answer is $1/2$. Spheres K and L, radii of r and 2r respectively.

I know the formula is 4πε0R for a sphere, but the answer to why it is linear with "$r$" rather than "$r^2$" eludes me. Shouldn't we use the formula for the surface area of a sphere, which is $4πr^2$, to derive the capacitance formula?

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  • $\begingroup$ Assume two concentric spheres of radius $r_1$ and $r_2$. Give one of them charge $+Q$ and the other $-Q$. Calculate the electric field between the two spheres. Get the corresponding expression for V. Use $C=\frac{Q}{V}$ to get the capacitance in terms of $r_1$ and $r_2$. Take $r_2$ to approach $\infty$ and $r_1=R$ to get capacitance of a sphere of radius R. $\endgroup$ Commented Feb 9, 2022 at 14:46
  • $\begingroup$ Why are we taking radius instead of radius squared--isn't capacitance correlated with surface area? Your answer just seems to draw conclusions from other definitions, whereas I am actually questioning the very assumptions made here. $\endgroup$ Commented Feb 9, 2022 at 19:07
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    $\begingroup$ Capacitance has always been linear in the dimensions for many of the simple shapes eg. a parallel plate capacitor with $C = \frac{\epsilon A}{d}$, while there is an area there, it is divided by the separation making it just length not area. $\endgroup$
    – Triatticus
    Commented Feb 9, 2022 at 19:33
  • $\begingroup$ This seems like a great answer. Although, I still can't get my head around the fact that we're utilizing the concept of surface, which is three dimensional, meanwhile using linear distance when it comes to its calculation. Doesn't this contradict the very notion of correlation between the "surface" capacitance and electric potential? $\endgroup$ Commented Feb 14, 2022 at 17:43

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Capacitance is a property of any conductor or system of conductors that what charge must be supplied to the system to raise its potential by 1 Volt, as in the case of a sphere the potential accuired by sphere by gaining a Q charge on its surface is KQ/r (r-radius)enter image description here

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  • $\begingroup$ I am aware of the units' canceling out. As you have mentioned in your answer though, we are talking about the surface. So, why don't we use the sphere's surface area formula, which is r squared? $\endgroup$ Commented Feb 14, 2022 at 17:38
  • $\begingroup$ No,Actually The Potential Acquired by the sphere is not related to the area. If You Didn't know this then you must check out Derivation of Potential Of A Charged Sphere $\endgroup$
    – Kushagra
    Commented Feb 15, 2022 at 5:45

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