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Physics/Voltage and capacitance


Hi Stephen. I am studying an online course in basic electronics and while the lecturer seems pretty good there's one thing he does not seemed to have explained properly. It is the fact that the capacitance of a capacitor decreases as the voltage increases and vice-versa, i.e:  C = Q/V. I find this very puzzling as I would have thought the opposite case to be true. Any explanation in very basic terms would be appreciated. Thank you.

Actually, you've got the interpretation of this slightly wrong.  Capacitance is a constant that only depends on geometry and any intervening dielectric material, it does not go down as voltage goes up.  As voltage goes up, so does charge.  Q=CV is a way to re-write that equation.  C is constant.  Q/V is a constant.  Double V and you double Q, for example, and the factor of 2 would look like this: C=2Q/2V=Q/V.  Hope that helped.


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Dr. Stephen O. Nelson


I can answer most basic physics questions, physics questions about science fiction and everyday observations of physics, etc. I'm also usually good for science fair advice (I'm the regional science fair director). I do not answer homework problems. I will occasionally point out where a homework solution went wrong, though. I'm usually good at explaining odd observations that seem counterintuitive, energy science, nuclear physics, nuclear astrophysics, and alternative theories of physics are my specialties.


I was a physics professor at the University of Texas of the Permian Basin, research in nuclear technology and nuclear astrophysics. My travelling science show saw over 20,000 students of all ages. I taught physics, nuclear chemistry, radiation safety, vacuum technology, and answer tons of questions as I tour schools encouraging students to consider careers in science. I moved on to a non-academic job with more research just recently.

Ph. D. from Duke University in physics, research in nuclear astrophysics reactions, gamma-ray astronomy technology, and advanced nuclear reactors.

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