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It takes longer for the charge in an RC circuit to fall down to half its initial value than for the stored energy. Why?

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The energy stored in the electric field of a capacitor is given by

$$E=\frac{CV^2}{2}$$

The relationship between voltage, charge and capacitance is

$$C=\frac{Q}{V}$$

Substituting for V is the first equation

$$E=\frac{Q^2}{2C}$$

Therefore when the charge $Q$ is one half the original, the stored energy in capacitor is one quarter. The capacitor loses energy at a faster rate than the loss of charge. So it takes longer for the charge to be half than the energy to be half.

Hope this helps.

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Because energy is proportional to the square of the charge, than if charge fall down to half its initial value than energy fall down to quarter of its initial value

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The current answers are correct in the mathematical descriptions, but let's look at the underlying physics. Your mistake is in assuming that the charge on the capacitor alone determines the energy stored in the capacitor, and that it is a linear relationship between $E$ and $Q$.

But what do we really mean by the energy stored in the capacitor? Well, it is the work needed to move the charge from one plate to the other plate (and hence achieving the charge separation the capacitor was designed to do). What does this work depend on. Well, it depends on

  1. How much charge you move, and
  2. The force you need to apply to each charge to get it to move to the other plate

Since the force is directly related to the potential between the plates, we essentially have shown that the energy needs to depend on both the charge $Q$ and the voltage difference $V$ between the plates.

Through dimensional analysis, the energy must depend on the product of $Q$ and $V$. Hence if $Q$ and $V$ each fall exponentially, then their product must fall twice as quickly.

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