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Q.
Find the capacitance of the infinite ladder between point $A$ and $B$
Electrostatic Potential and Capacitance
Solution:
Let $C$ be the capacitance of infinite ladder
$\frac{1}{C'}=\frac{1}{C}+\frac{1}{5}=\frac{5+C}{5 C}$
$C'=\frac{5 C}{5+C}$
$C''=2+C'$
$C''=2+\frac{5 C}{5+C}$
$C''=C$ (Addition of one or more $5 \mu F$ and $2 \mu F$ do not change the total capacitance. So, overall capacitance should be $C$ )
$C=\frac{10+2 C+5 C}{5+C}$
$5 C+C^{2}=10+7 C$
$C^{2}=10+2 C$
$C^{2}-2 C-10=0$
$D=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\therefore C=\frac{2 \pm \sqrt{44}}{2}$
$C =\frac{2+6.6}{2}, C =\frac{2-6.6}{2}$
$C =\frac{8.6}{2}, C =\frac{-4.6}{2} $
$C =-2.3, \,C =4.3$
-ve value neglected
So, total capacitance is $4.3\, \mu F$
