Animated Solution for Physics - Physics and Measurement: Two capacitors with capacitance values $C_1=2000\pm 10\text{ pF}$ and $C_2=3000\pm 15\text{ pF}$ are connected in series. The voltage applied across this combination is $V=5.00\pm 0.02\text{ V}$. The percentage error in the calculation of the energy stored in this combination of capacitors is _______.

Imagine you are building a delicate electronic circuit. You have two capacitors, $C_1$ and $C_2$, and you connect them in series across a battery. In the real world, no component is perfect. Your capacitors have slight manufacturing variations, and your voltmeter isn't infinitely precise. This problem asks us to find out how these tiny uncertainties "snowball" or propagate into the final calculation of the energy stored in the system.

The energy stored in a capacitor system is given by the elegant formula:

To find the percentage error in $U$, we first need to figure out the percentage error in the equivalent capacitance $C_{eq}$ and the voltage $V$.

For capacitors in series, the equivalent capacitance is found using the reciprocal rule:

Here is where many students fall into a trap. You cannot simply add the percentage errors of $C_1$ and $C_2$ because they are not being multiplied or divided in a simple linear way. Instead, we must use the power of calculus. By differentiating the entire equation, we can see exactly how a small change (or error) in $C_1$ and $C_2$ affects $C_{eq}$.

Differentiating both sides gives:

Since errors represent the maximum possible uncertainty, we always take the absolute values and add them up. Rearranging this to find the fractional error in $C_{eq}$, we get:

First, let's find the actual value of $C_{eq}$:

Now, we substitute all our known values into our beautiful error equation:

Let's break down the math inside the parentheses. It looks intimidating, but it simplifies wonderfully:

Converting this fraction to a percentage, we find that the error in the equivalent capacitance is exactly $0.5\%$.

The Energy Error

Now we return to our energy formula, $U=\frac{1}{2}{C}_{eq}{V}^2$. According to the standard rules of error propagation, when quantities are multiplied, their fractional errors add up. And when a quantity is raised to a power, its fractional error is multiplied by that power.

Therefore, the percentage error in energy is:

We already know the first part is $0.5\%$. Let's calculate the percentage error for the voltage:

Plugging this back into our total error equation:

The final percentage error in the calculation of the stored energy is $1.30$.