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 k ( n ) = k ( 1 + α n )   where '  x ' is he distance measured from one of the plates.

If ( α d ) < < 1  , the total capacitance of the system is best given by the expression:

Option 1 -

A K ε 0 d 1 + α d 2

Option 2 -

A ε 0 K d 1 + α 2 d 2 2

Option 3 -

A K ε 0 d ( 1 + α d )

Option 4 -

A ε 0 K d 1 + α d 2 2

0 2 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    a month ago
    Correct Option - 1


    Detailed Solution:

    Capacitance of element

    Capacitance of element, C ' = K ( 1 + α x ) ε 0 A d x

    1 C ' = 0 d ? d x K ε 0 A ( 1 + α x )

    1 C = 1 K ε 0 A α l n ? ( 1 + α d )

    Given: α d ? 1

    1 C = 1 K ε 0 A α α d - α 2 d 2 2 ; 1 C = d K ε 0 A 1 - α d 2

    C = K ε 0 A d 1 + α d 2

     

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R
Raj Pandey

Please find the answer below

V
Vishal Baghel

Now, using junction analysis
We can say, q? + q? + q? = 0
2 (x - 6) + 4 (x - 6) + 5 (x) = 0
x = 36/11, q? = 36 (5)/11 = 180/11
q? = 16.36µC

R
Raj Pandey

dC? = (ε? + kx)A / dx [For 0 < x < d/2]
1/C? = ∫ dx / (ε? + kx)A) from 0 to d/2
= (1/Ak) [ln (ε? + kx)] from 0 to d/2
= (1/kA) ln (1 + kd/ (2ε? )
C? = kA / ln (1 + kd/ (2ε? )

Similarly dC? = (ε? + k (d-x)A / dx [For d/2 ≤ x ≤ d]
C? = kA / ln (1 + kd/ (2ε? )
Clearly, C? = C? = C
For series combination:
C_eq = C? / (C? + C? ) = C/2 = kA / (2ln (2ε? + kd)/2ε? )

A
alok kumar singh

Charge remains same

 

A
alok kumar singh

Charge on C1 = CV

            Charge on C2 = 4CV

When connected in parallel

V C = 5 V 3  

Q ' 1 = 5 3 C V Q ' 2 = 1 0 3 C V

? E = 1 2 C V 2

E ' = 2 5 1 8 C V 2 + 2 5 9 C V 2  

2 5 6 C V 2 = 5 0 E 6

x = 50

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