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 The space between the plates of a parallel plate capacitor is filled with a ‘dielectric’  whose ‘dielectric constant’ varies with  distance as per the relation :        

K(x)=K_{0}+\lambda x(\lambda =a\: constant)

The capacitance C, of this capacitor, would   be related to its ‘vacuum’ capacitance Co as per the relation :                                                                

 

                           

 

  • Option 1)

    C=\frac{\lambda d}{ln\left ( 1+K_{0} \lambda d\right )}C_{0}

  • Option 2)

    C=\frac{\lambda }{d.ln\left ( 1+K_{0} \lambda d\right )}C_{0}

  • Option 3)

    C=\frac{\lambda d }{ln\left ( 1+ \lambda d/K_{0}\right )}C_{0}

  • Option 4)

    C=\frac{\lambda }{d.ln\left ( 1+ K_{0}/\lambda d\right )}C_{0}

 

Answers (2)

best_answer

As we discussed in

The boundary Conditions -

\dpi{100} dV=-\int_{r_{1}}^{r_{2}}\overrightarrow{E}\cdot \vec{d}r=-\int_{r_{1}}^{r_{2}}Edr\cos \theta

-

 

 

Capacitance of Conductor -

Q\propto V

Q=CV

- wherein

C - Capacity or capacitance of conductor 

V - Potential.

 

 Given K=K_{0}+\lambda 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+\lambda x} dx= \frac{\sigma }{\lambda\varepsilon _{0} }[ln(K_{0}+\lambda d)-lnK_{0}]

V= \frac{\sigma }{\lambda\varepsilon _{0} }ln (1+\frac{\lambda d}{K_0})

C= \frac{Q}{V}=\frac{\sigma S}{V}= \frac{\sigma S}{\frac{\sigma }{\lambda }ln(1+\frac{\lambda d}{K_0})} S= surface area of plate.

here, C_0=\frac{\varepsilon _{0}S}{d}

C=\frac{\lambda d }{ln\left ( 1+ \lambda d/K_{0}\right )}C_{0}

 

 


Option 1)

C=\frac{\lambda d}{ln\left ( 1+K_{0} \lambda d\right )}C_{0}

Option 2)

C=\frac{\lambda }{d.ln\left ( 1+K_{0} \lambda d\right )}C_{0}

Option 3)

C=\frac{\lambda d }{ln\left ( 1+ \lambda d/K_{0}\right )}C_{0}

Option 4)

C=\frac{\lambda }{d.ln\left ( 1+ K_{0}/\lambda d\right )}C_{0}

Posted by

Avinash

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