Find the equivalent capacitance between AB:

CONCEPT:

Capacitor:

• The capacitor is a device in which electrical energy can be stored.
  • In a capacitor, two conducting plates are connected parallel to each other and separated by an insulating medium carrying charges of equal magnitudes and opposite signs.
  • The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or a semi-conductor called a dielectric.​

​1. Capacitors in series

• When two or more capacitors are connected one after another such that the same charge gets generated on all of them, then it is called capacitors in series.
• The net capacitance/equivalent capacitance (C) of capacitors in series is given by,

\(⇒\frac{1}{C} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}+...+ \frac{1}{{{C_n}}}\)

2. Capacitors in parallel

• When the plates of two or more capacitors are connected at the same two points and the potential difference across them is equal, then it is called capacitors in parallel.
• The net capacitance/equivalent capacitance (C) of capacitors in parallel is given by,

\(⇒ C = C_1+ C_2+...+ C_n\)

CALCULATION:

The given diagram is,

     -----(1)

• In figure 1 the infinite capacitors are connected in series.

From figure 1 the equivalent capacitance between AB is given as,

\(⇒\frac{1}{C_{AB}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}+...+ \frac{1}{{{C_∞}}}\)

\(⇒\frac{1}{C_{AB}} = \frac{1}{{{C}}} + \frac{1}{{{2C}}}+\frac{1}{{{4C}}}...\)     -----(1)

We know that equation 1 is a case of infinite geometric progression and the sum of the infinite geometric progression is given as,

\(⇒ S=\frac{a}{1-r}\)     -----(2)

Where a = First term and r = Common ratio

In equation 1,

\(⇒ a=\frac{1}{C}\)

\(⇒ r=\frac{1}{2}\)

So by equation 1 and equation 2,

\(⇒\frac{1}{C_{AB}} =\frac{\frac{1}{C}}{1-\frac{1}{2}}\)

\(⇒\frac{1}{C_{AB}} ={\frac{2}{C}}\)

\(⇒C_{AB} =\frac{C}{2}\)

• Hence, option 2 is correct.