Solution:
If the plates are connected in series, the total capacitance \( C_{\text{eq}} \) of \( n-1 \) capacitors is given by \( \frac{1}{C_{\text{eq}}} = \frac{n-1}{C} \), which yields \( C_{\text{eq}} = \frac{C}{n-1} \).
A parallel plate capacitor is made by stacking \( n \) similar metallic plates equally spaced from one another. The capacitance of the capacitor formed by any two neighbouring plates is \( C \). The total capacitance of the combination will be
Leave a Reply