# Answer to Question #200847 in Electrical Engineering for Deepak

Question #200847

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.

1
2021-05-31T06:20:22-0400

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\n_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)}]"

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