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  • A parallel plate capacitor has plates of area A separated by distance'd' between them. It is filled with a dielectric which has a dielectric constant that varies as

A parallel plate capacitor has plates of area A separated by distance'd' between them. It is filled with a dielectric which has a
dielectric constant that varies as \mathrm{k}(x)=\mathrm{K}(1+\alpha x) where ^{\prime} x^{\prime} is the distance measured from one of the plates. If (\alpha \mathrm{d})< < 1, the total capacitance of the system is best given by the expression:    


Option: 1 \frac{A\epsilon _{0}K}{d}\left ( 1+\left ( \frac{\alpha d}{2} \right )^{2} \right )
 

 
Option: 2 \frac{AK\epsilon _{0}}{d}\left ( 1+\frac{\alpha d}{2} \right )

Option: 3 \frac{A\epsilon _{0}K}{d}\left ( 1+\frac{\alpha^{2} d^{2}}{2} \right )  

Option: 4  \frac{AK\epsilon _{0}}{d}\left ( 1+\alpha d\right )
 

Answers (1)

best_answer

 

 

Dielectrics -

Dielectric:  A dielectric is an insulating material in which all the electrons are tightly bounded to the nuclei of the atoms and no free electrons are available for the conduction of current. They are non-conducting materials. They do not have free charged particles like conductors have. They are of two types.

1. Polar : The centre of +ve and –ve charges do not coincide. Example HCl, H_2O, They have their own dipole moment

2. Non-Polar : The centers of +ve and –ve charges coincide. Example CO_2 , C_6H_6 . They do not have their own dipole moment.

 

When a dielectric slab is exposed to an electric field, the two charges experience force in opposite directions. The molecules get elongated and develops a  surface charge density \sigma _{p} . This leads to development of an induced electric field Ep , which is in opposition direction of external electric field Eo . Then net electric field E is given by  E = E_{o} - E_{i} .

This indicates that net electric field is decreased when dielectric is introduced.

The ratio  \frac{E_0}{E}=K  is called dielectric constant of the dielectric. Hence, Electric field inside a dielectric is   E_i=\frac{E_0}{K}.

E=E_{0}-E_{i} \ and \ E=\frac{E_0}{k}\\ So,\ E_{0}-E_{i}=\frac{E_{0}}{K}

\begin{array}{ll}{\text { or } E_{0} K-E_{i} K=E_{0}} \\ {\text{ or } E_{0} K-E_{0}=E_{i} K} \\ {\text { or } \quad E_{i}=\frac{K-1}{K} E_{0}} & {} \\ {\text { or } \frac{\sigma_{i}}{\varepsilon_{0}}=\frac{K-1}{K} \frac{\sigma}{\varepsilon_{0}}} \\ or\ {\sigma_{i}=\frac{K-1}{K} \sigma} \\ {\text { or } \quad \frac{Q}{A}=\frac{K-1}{K} \frac{Q}{A}} \\ {\text { or } \quad Q_{i}=Q\left(1-\frac{1}{K}\right)} \end{array}$                                                             

 

This is irrespective of the thickness of the dielectric slab,i.e., whether it fills up the entire space between the charged plates or any part of it.

 

 

{C}'=\frac{\epsilon _{0}A}{d-t+\frac{t}{k}}

 

-

 

 

If K filled between the plates -

{C}'=K\frac{\epsilon _{0}A}{d}={C}'=Ck

 Given K=K_{0}+K_0\alpha x

V= -\int_{0}^{d} Edr= V=\int_{0}^{d}\frac{\sigma }{K\epsilon _{0}} dx

V= \frac{\sigma}{\varepsilon _{0}} \int_{0}^{d}\frac{1}{K+K \alpha x} dx= \frac{\sigma }{K \alpha \varepsilon _{0} }[ln(K+K \alpha d)-lnK]

V= \frac{\sigma }{K \alpha \varepsilon _{0} }ln (1+ \alpha d)

For 

ln (1+ \alpha d)=\alpha d-\frac{(\alpha d)^2}{2}+\frac{(\alpha d)^3}{3}.......

V= \frac{\sigma }{K \alpha \varepsilon _{0} }ln (1+ \alpha d)

  C= \frac{Q}{V}=\frac{\sigma A}{V}= \frac{\sigma A}{\frac{\sigma }{K\varepsilon _{0} \alpha }ln (1+ \alpha d)}      

      here, C_0=\frac{\varepsilon _{0}A}{d}

C=\frac{Kd\alpha }{ln\left ( 1+ \alpha d \right )}C_{0}=\frac{KdC_0\alpha }{\alpha d-\frac{(\alpha d)^2}{2}+\frac{(\alpha d)^3}{3}.....}=\frac{KdC_0\alpha }{(\alpha d)*(1-\frac{\alpha d}{2}+\frac{(\alpha d)^2}{3}...)}

since \alpha d << 1

So

   \\ C=\frac{KdC_0\alpha }{(\alpha d)*(1-\frac{\alpha d}{2} )}= \frac{KdC_0\alpha }{(\alpha d)}*(1-\frac{\alpha d}{2} )^{-1}=\frac{KdC_0\alpha }{(\alpha d)}*(1+\frac{\alpha d}{2} ) \\ \\ \\ =\frac{AK\varepsilon _0}{d}(1+\frac{\alpha d}{2} )

So, option (2) is correct.

Posted by

Ritika Jonwal

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