Question

A capacitance is made of two parallel conducting plates as shown in the figure and the distance 

between the plates is ‘d’. (i) Write the value of capacitance as a function of all the possible dimensions 

from L,W, h and d (ii) if a dielectric slab of thickness ‘t’ and area of cross-section A (length L, width W)

(t<d), and relative permittivity εr is inserted in between the plates, what will be the value of 

capacitance (iii) If a voltage V is applied between the plates, find the voltage across various sections 

(iv) If a dielectric slab is of length L/2 , width ‘W’, thickness ‘t’ and relative permittivity εr is inserted 

in between the plates from right end , what is the value of capacitance.

Answer 1

i) Capacitance for parallel plate capacitor is given as ,

$c=\frac{E_0A}{d}, A=WL$

or $c=\frac{E_0WL}{d}(F)$

V$=2V_1+V _3$

$=2V_1+\frac{c_1}{c_3}v_1$

$=(2+\frac{2E_0WL}{d-t}*\frac{t}{E_0E_rWL})V_1$ )

$=V_1=\frac{E_r(d-t)}{2[t+E_r(d-t)]}$

$V_3=\frac{2t}{E(d-t)}.\frac{E_r(d-t)}{2[t+E_r(d-t)]}$

Thus, $V_1=V_2=\frac{E_r(d-t)V}{2(t+E_r(d-t)},V_3=\frac{tV}{t+E_r(d-t)}$

ii) $c_1=c_2=\frac{E_0(WL)}{\frac{d-t}{2}}$

$c_1=c_2=\frac{2E_0WL}{d-t}$

$c_3=\frac{E_0E_rWL}{t}$

$\frac{1}{c_9}=\frac{1}{c_1}+\frac{1}{c_2}+\frac{1}{c_3}$

$\frac{d-t}{2E_0E_rWL}+\frac{d-t}{2E_0WL}+\frac{t}{E_0E_rWL}$

$\frac{1}{C_q}=\frac{d-t}{2E_0WL}+\frac{t}{E_0E_rWL}$

$C_q=\frac{E_0E_rWL}{t+E_r(d-t)}$

iii)V$=V_1+V_2+V_3$

$Q=C_2V$

$\therefore C_1=C_2 \space thus \space V_1=V_2$

iv) $C_1=\frac{E_0WL}{2d}$

$C_2=\frac{E_0WL}{d-t}$

$C_3=\frac{E_0E_rWL}{2t}$

$=C_2$ & $C_3$ are in series

$C_{eq2}=\frac{C_2C_3}{C_2+C_3}=\frac{E_0E_rWL}{2(t+E_r(d-t)}$

$C_{eq2}=C_1+C_{eq1}$

$=\frac{E_0WL}{2d}+\frac{E_0E_rWL}{2(t+E_r(d-t)}$

$C_{eq}=\frac{E_0WL}{2}+[\frac{1}{d}+\frac{E_r}{t+E_r(d-T)}]$