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
Expert's answer
2021-05-31T06:20:22-0400

i) Capacitance for parallel plate capacitor is given as ,

c=E0Ad,A=WLc=\frac{E_0A}{d}, A=WL

or c=E0WLd(F)c=\frac{E_0WL}{d}(F)

V=2V1+V3=2V_1+V _3

=2V1+c1c3v1=2V_1+\frac{c_1}{c_3}v_1

=(2+2E0WLdttE0ErWL)V1=(2+\frac{2E_0WL}{d-t}*\frac{t}{E_0E_rWL})V_1 )

=V1=Er(dt)2[t+Er(dt)]=V_1=\frac{E_r(d-t)}{2[t+E_r(d-t)]}

V3=2tE(dt).Er(dt)2[t+Er(dt)]V_3=\frac{2t}{E(d-t)}.\frac{E_r(d-t)}{2[t+E_r(d-t)]}

Thus, V1=V2=Er(dt)V2(t+Er(dt),V3=tVt+Er(dt)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) c1=c2=E0(WL)dt2c_1=c_2=\frac{E_0(WL)}{\frac{d-t}{2}}

c1=c2=2E0WLdtc_1=c_2=\frac{2E_0WL}{d-t}

c3=E0ErWLtc_3=\frac{E_0E_rWL}{t}

1c9=1c1+1c2+1c3\frac{1}{c_9}=\frac{1}{c_1}+\frac{1}{c_2}+\frac{1}{c_3}

dt2E0ErWL+dt2E0WL+tE0ErWL\frac{d-t}{2E_0E_rWL}+\frac{d-t}{2E_0WL}+\frac{t}{E_0E_rWL}

1Cq=dt2E0WL+tE0ErWL\frac{1}{C_q}=\frac{d-t}{2E_0WL}+\frac{t}{E_0E_rWL}

Cq=E0ErWLt+Er(dt)C_q=\frac{E_0E_rWL}{t+E_r(d-t)}


iii)V=V1+V2+V3=V_1+V_2+V_3

Q=C2VQ=C_2V

C1=C2 thus V1=V2\therefore C_1=C_2 \space thus \space V_1=V_2


iv) C1=E0WL2dC_1=\frac{E_0WL}{2d}

C2=E0WLdtC_2=\frac{E_0WL}{d-t}

C3=E0ErWL2tC_3=\frac{E_0E_rWL}{2t}

=C2=C_2 & C3C_3 are in series

Ceq2=C2C3C2+C3=E0ErWL2(t+Er(dt)C_{eq2}=\frac{C_2C_3}{C_2+C_3}=\frac{E_0E_rWL}{2(t+E_r(d-t)}


Ceq2=C1+Ceq1C_{eq2}=C_1+C_{eq1}

=E0WL2d+E0ErWL2(t+Er(dt)=\frac{E_0WL}{2d}+\frac{E_0E_rWL}{2(t+E_r(d-t)}

Ceq=E0WL2+[1d+Ert+Er(dT)]C_{eq}=\frac{E_0WL}{2}+[\frac{1}{d}+\frac{E_r}{t+E_r(d-T)}]

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