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I am trying to intuitively understand why equation $Q=CV$ should work. I can understand why $V$ should be proportional to $Q$ but not why $Q$ should be proportional to $V$. In the end, $C$ is constant so $\frac{1}{C}$ will be a constant too. So the equation should work. But intuitively I am not able to understand

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  • $\begingroup$ Please use mathjax to format mathematical expressions. To learn more about mathjax, please read MathJax basic tutorial and quick reference. $\endgroup$ – Yashas Mar 18 '17 at 4:02
  • $\begingroup$ $Q\frac1C=V$ would work just as well as $Q=CV$. Someone just made a choice back in the days and chose one of these versions - and gave the name $C$ to the constant, whatever it was. $\endgroup$ – Steeven Mar 18 '17 at 9:16
  • $\begingroup$ Not clear why $Q \propto V$ means something different from $V \propto Q$. $\endgroup$ – sammy gerbil Aug 25 '18 at 20:01
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A capacitor in an electrical circuit is similar to a spring in a mechanical system, with $Q$ being like the amount of stretching $x$ of the spring, $V$ is like the force $F$ produced by the spring, and $1/C$ is like the spring constant $k$.

Now you say you understand the equation $V=\dfrac{1}{C} Q$, which makes sense because the charge is creating the voltage, so if you have twice the charge, it must make twice the voltage. Now according to what I said in the first paragraph, this is analagous to the equation $F=kx$. This also makes sense: the more you stretch the spring, the more force you get.

Now lets look at the mechanical version of the equation you don't understand. The mechanical version is $x=F/k$. This says that if the spring has a force $F$, then it must have been stretched by a proportional amount. For example if the spring force is ten percent greater, then the spring must have been stretched by ten percent more, if it stretched any farther, the force would be more than ten percent greater, and if it stretched less, the force would be less. Another way of looking at things is to say $k=F/x$. In this equation it is obvious that $F$ and $x$ have a constant proportion.

You can apply this same logic to the equation $Q=CV$. If the only way to increase the voltage by ten percent is to increase the charge by ten percent. Put another way, the ratio of voltage to charge has to be a constant: $\dfrac{V}{Q}=\dfrac{1}{C}$.

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What about: If you double the charge, you double the voltage, thus V is proportional to Q. If you double the voltage, you also double the charge, so Q is also proportional to V.

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Look at the way how capacitance is defined: $C=\frac{Q}{V}$

Capacitance is defined as the number of charges required to raise the potential by $1 Volt$. So, for a capacitor of high capacitance, more charge is required to raise the potential by $1 Volt$.

For example, if the capacitance of a capacitor in $50\mu F$, then it means that $50\mu C$ of charge is required to raise the potential of capacitor by $1 Volt$.

That's just the way how capacitance is defined.

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