2019-03-01_Physics_Times

The given system can be assumed as shown in

the figure.

4 0  4 0

41 42 43

/3 2

/2 3

K A K

C C C

d d

 

   

1 0  1 0

1

/3 2

,

/2 3

K A K A

C

d d

 

 

2 0 3 0

2 3

2 2

,

3 3

K A K A

C C

d d

 

 

0

eq

K A

C

d





0 1 4 2 4 3 4

1 4 2 4 3 4

2

3

A K K K K K K

d K K K K K K

  

    

    

1 4 2 4 3 4

1 4 2 4 3 4

2

3

K K K K K K

K

K K K K K K

 

    

    

Wrong solution seems to be correct as per the

options given. The circuit is assumed as

(^0)  1 2 3 
2
A 3
A
C K K K
d

  
(^2) 0 4
B
AK
C
d


' ''
1 1 1
Ceq C C
 
0 0  1 2 3  0 4
3
2 2
d d d
K A A K K K AK  
 
 
 1 2 3  4
1 3 1
K K K K K 2 2
 
 
The charge on the capacitor is given by
Q = CV
The energy stored in the capacitor is^2
1
2
 CV
When a dielectric slab of dielectric constant K is
inserted in it, the charge Q is conserved. The
capacitance becomes K times the original
capacitance.
The voltage becomes
1
K
times the initial value.
The change in energy stored is
(^2221)
1
2 2 2
Q Q Q
KC C C K
    
  ^
1 1 (^21)
2
CV
K
 
  
 
Outside the dielectrics the field remains
constant.
Inside the dielectrics the field decrease by k times.
As k k 2  1 the field in the second slab is less than
the field in the first slab.
The energy stored in the capacitor
U u Ad  
u- Energy per unit volume
 
2
0
1
2
U E Ad 
When the capacitors are joined in series,
1 2
1
1
(4 )
series 2
C
U V
n

When the capacitors are joined in parallel,
2
2 2
1
( )
parallel 2
U  n C V
Given,U Useries parallel
3.Sol:
Alter:
4.Sol:
5.Sol:
6.Sol:

7. Sol:


8. Sol:

1 2 2

2 2

1

1

(4 ) ( )

2 2

C

V n C V

n

 2 1

2 1

16 C

C

n n

 

In a series arrangement the potential difference

is

V V V V  1 2 3[V V V V1 2 3   ]

Here, V 3Vnet

The equivalent capacitance CS is given by

3

S

C

C 