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  4. Two capacitors of capacitances C1 and C2 are connected in series, assume that C1< C2. The equivalent capacitance of this arrangement is C, where

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Q. Two capacitors of capacitances $C_1$ and $C_2$ are connected in series, assume that $C_1< C_2$. The equivalent capacitance of this arrangement is $C$, where

Electrostatic Potential and Capacitance

Solution:

$C_1< C_2$
$\therefore \frac{C_1}{C_1+C_2}< \frac{1}{2} \text { and } \frac{C_2}{C_1+C_2} >\frac{1}{2}$
$C=\frac{C_1 C_2}{C_1+C_2}=C_1 \cdot \frac{C_2}{C_1+C_2} >\frac{C_1}{2}$
Similarly, $C< \frac{C_2}{2}$