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Q. A capacitance of an air capacitor is $25\, \mu F$ the separation between the parallel plates is $8\, mm$. An aluminum plate of thickness $2 mm$ thickness is introduced symmetrically between the plates. The new capacitance will be

Electrostatic Potential and Capacitance

Solution:

$C _{ m }=\frac{\varepsilon_{ 0 } A }{ d - t +\frac{ t }{ K }}$
$C_{0}=\frac{\varepsilon_{0} A}{d-t}(K=\infty$ for metals $)$
$C_{0}=\frac{\varepsilon_{0} A}{d}$
$\frac{ C _{ m }}{ C _{0}}=\frac{\varepsilon_{ 0 } A }{ d - t } \times \frac{ d }{\varepsilon_{ 0 } A }=\frac{ d }{ d - t }$
$\frac{C_{m}}{C_{0}}=\frac{8}{8-2}=\frac{8}{6}$
$C_{m}=25 \times \frac{4}{3}$
$=33.3 \mu F$