Answer to Question #181426 in Electric Circuits for Abdulmajid

Question

show that for two capacitor of area A seperated by a distance D apart with a di electric material of tickness T in between them. a capacitance is given by the expression C= permitivity of A (1/d-1 +permitivity of r/t

Answer 1

$c=\dfrac{Q}{A}$ (1)

$\sigma=$ surface charge density

$\sigma=\dfrac{Q}{A}$

$Q=\sigma A$

$c=\dfrac{\sigma A}{V}$ from (1)

Distance of plates from each other = $D$

Thickness of slab = $T$

$t_1+t_2+T=D$

$t_1+t_2=D-T$

Potential difference

$V=\dfrac{\sigma D}{\epsilon_o}$

$\Rightarrow V=\dfrac{\sigma}{\epsilon_o}(t_1+t_2+\dfrac{T}{K})$

$\Rightarrow V=\dfrac{\sigma}{\epsilon_o}(D-T+\dfrac{T}{K})$

$\Rightarrow \dfrac{Q}{c}=\dfrac{\sigma}{\epsilon_o}(D-T+\dfrac{T}{K})$

$\Rightarrow \dfrac{\sigma A}{c}=\dfrac{\sigma}{\epsilon_o}(D-T+\dfrac{T}{K})$

$\Rightarrow c=\dfrac{\epsilon_o A}{(D-T)+\dfrac{T}{K}}$