Fifty parallel plate capacitors are connected in series. The distance between the plates is \(d\) for the first capacitor, \(2 d\) for the second capacitor, \(3 d\) for the third capacitor, and so on. The area of the plates is the same for all the capacitors. Express the equivalent capacitance of the whole set in terms of \(C_{1}\) (the capacitance of the first capacitor).

When multiple capacitors are connected in series, the total capacitance can be calculated using the formula: \[\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} + ... + \frac{1}{C_{n}}\] Where: - \(C_{eq}\) is the equivalent capacitance - \(C_{1}, C_{2},..., C_{n}\) are the capacitances of the capacitors connected in series.

It is given that the area of the plates is the same for all capacitors, and the distance between the plates increases in a linear manner (\({d, 2d, 3d,\dots}\)). So, the capacitance of a capacitor can be defined using the formula: \[C = \frac{A \epsilon_0}{d}\] Where: - \(C\) is the capacitance - \(A\) is the area of the plates - \(\epsilon_0\) is the vacuum permittivity - \(d\) is the distance between the plates Since the area and vacuum permittivity are constant for each capacitor, we can observe that the capacitance is inversely proportional to the distance between the plates. So, for the \(i^{th}\) capacitor with a plate distance of \(id\), we can write: \[C_i = \frac{C_1}{i}\]

Now, we have the relationship between each capacitor's capacitance and the first capacitor's capacitance. We can plug this relationship into the series capacitance formula from step 1: \[\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{\frac{C_1}{2}} + \frac{1}{\frac{C_1}{3}} + ... + \frac{1}{\frac{C_1}{50}}\] We can write this as a summation: \[\frac{1}{C_{eq}} = \sum_{i=1}^{50} \frac{1}{\frac{C_1}{i}}\] Now, we can simplify this expression by multiplying both sides by \(C_1\): \[C_1 = \sum_{i=1}^{50} i\] This sum can be evaluated using the arithmetic series sum formula: \[C_1 = \frac{n(n+1)}{2}\] Where \(n = 50\): \[C_1 = \frac{50 \cdot (50+1)}{2}\] Now, we can solve for the equivalent capacitance, \(C_{eq}\): \[C_{eq} = \frac{C_1}{1275}\] And finally, we have the expression for the equivalent capacitance of the whole set of capacitors in terms of \(C_{1}\): \[C_{eq} = \frac{C_1}{1275}\]