CAPACITA

Vidyamandir Classes

Solutions to Home Work Test - 4/Physics Capacitors 1 q2 1 q2 ; Ui  2 10 2 2

1.(D)

Uf 

2.(C)

q V B  qE t

3.(C) 5.(D) 6.(A)

HWT - 1



V 

E   B 0 B

  0 B  V 

1  q12  q22    2  c  The plates of the capacitor have equal and opposite charge. The net charge enclosed is zero, hence the net flux will be zero. 1 dV 10 7.(D) 8.(B) Use U  CV 2 E     50 Vm 1  2 2 dr 20  10 C

K 0 A d



C A

4.(A)

Use U 

E0 where E0 = electric field in free space k E = electric field in medium 1 2 10.(C) Use U  cv 12V 2 Q 9.(A)

E

11.(C) The circuit reduces to

Q  3F  12V  36 C Q

12V

Now

4F 6F

Q1 C  1 Q2 C2

3F

2F

Q1

Q

Q2

 Q1  2Q2

Q = Q1 + Q2 Solve for Q1 & Q2 12.(C) When battery remains connected C’ = KC Q’ = KQ E’=E U ’ = KU U and Q both increases 13.(B)

1 1 Use U  K q1q2     r2 r1  Where r2 = 0.06 m r1 = 0.1 m

VMC/Capacitors

70

Solutions to HWT-4/Physics

Vidyamandir Classes 14.(A) In steady state, the capacitor behaves like as an open circuit i.e., capacitor will not pass the current Use q  CV  20  106  45  9  1014 C . 15.(D)

1 C1V12  450  106 J 2

E1 

q  q2  10 Volt Common Potential V  1 c1  c2 E2 

1  C1  C2 V 2  150  106 J . 2

Capacitors 1.(C)

4 3 4 R  64  r 3 3 3



HWT - 2

R  4r

Qtotal  64 q

C'   4 0  4r  4C . 2.(B)

C' 

2 0  2 L  1 Q2 b ; E log   2 2 0 2 b a log   a

…(i)

1  2Q  1  2Q  b  log   2 C' 2 2 0  2 L  a

…(ii)

2

E' 

From (i) & (ii) 3.(B)

E'  2 E

2 K 0 A 2 0 A ; C2  d d

C1 

1 1 1   C' C1 C2

 4.(C)

2

 A 0  2 2 E   Ed   A 0 E d d  

5.(B)

total charge 5  103  sec .  50 sec . charge rate 100  106

6.(D)

t

7.(C)

C0 

A 0 d

C 15  5 C0 3

…(i)



KA 0 d

;

C

9.(D)

W  U  0 .

…(ii)

k  5.

r  0 k . 8.(D) 10.(D)

1 qV 1  Required Ratio = 2 qV 2

C

K 0    b  d

VMC/Capacitors



  33.9 m

11.(D) Energy loss =

71

1 C1C2 V1  V2 2 2  C1  C2 

Solutions to HWT-4/Physics

Vidyamandir Classes 12.(B)

 A C  0 and C'  d

0 A 1  d '  t t   k 

Solve for C = C'

13.(A)

4F

2F

2F

4F

x B

y

6V

q q  4  106 and  2  106 Vx  V A VA  V y Here Vx = 6V and Vy = 0. We get VA = 4V Similarly for lower side branch We get VB = 2V.



VA  VB  4  2  2V .

14.(C) Net emf in circuit E  16  6  10V

C

23 6  F 23 5

Charge on each capacitor q 

6  10  12 C 5

Potential difference across 2F capacitor = 15.(B)

12  6V . 2

Done earlier similar type of question.

Capacitors

1.(B)

1 Fx Ratio = 2 1 1 CV 2 2

3.(D)

C

K 0 A d

…(i)

From (i) and (ii) : C0  4.(B)

HWT - 3

 n 1 K 0

2.(D)

Use C 

and

 A C0  0 d

A

d

…(ii)

C C  . K 2

0 A

We have, C 

t K Here K   and t  0 d t 

 A C  0  C0 . d

VMC/Capacitors

72

Solutions to HWT-4/Physics

Vidyamandir Classes 1 q2 2 C

5.(D)

Use w 

7.(A)

Use conservation of energy

8.(A)

1 1 1 2    C' C C C

6.(C)

C' 

A0 KA 0 C   1  K  2d 2d 2

meV 2  0  e V2  V1  2



C' 

C 2

C 2 When filled with dielectric C1  4C1C2  C Q  15

4C 5 Q  CV C

Get V = 10 V. 9.(A) 12.(A)

10.(D) Basic Question

1 E  CV 2 2

…(i)

From both equation ms T 

H  msT

and

1 CV 2  V  2

11.(A) …(ii)

2ms T C

13.(D) Done earlier similar type of questions. 14.(A)

1 1 1   C C3 C1  C2 q  CV  Now



 C1  C2  C3

C1  C2  C3 C'  C1  C2

C

 C1  C2  C3 C1  C2  C3

.V

q'  C' V   C1  C2 V

Change in charge q  q1 15.(A) Similar type of question is done earlier.

Capacitors 1.(C)

 A 0 A Use 0  t d d' t  K

4.(D)

V  V1  V2  V3

5.(D)

Resultant field =

HWT - 4 2.(B)



3.(A)

1 1 1  11  q     1 2 3  6 V1   6V . 1



Ed

q  6C

Q1  Q2 2 0 A

 Q  Q2  Q1  Q2 V  1 d  2C  2 0 A 

VMC/Capacitors

73

Solutions to HWT-4/Physics

Vidyamandir Classes 6.(C)

Done similar type of Question.

7.(A)

E1 

1 Q2  C0V02 2 C0 / 2

E2 

1 C0 2 1 · V0  C0V02 2 2 4

E1 4  E2 1 8.(A)

q  q2 C V  C2V2 V  1  11 C1  C2 C1  C2

9.(C)

10.(C)

16 C 3 5 16 C 5



C  48 F

11.(C) Similar type of question done in DC circuit 12.(D) Three capacitor in parallel so Ceq = 3C. 13.(B)

U initial 

1 C1V02 2

U final 

 C1V0  1  C1V0  1 C1    C2   2  C1  C2  2  C1  C2 

2

=

1  C1V0    2  C1  C2 

2

2

 C1  C2 

U initial C  C2 C  C  1  2 U final C1 C



U final 

U initial v  2 2

14.(C) Stored charge on capacitor becomes zero only when it is discharged through resistance or when two capacitors of equal capacitances are charged and then connected to opposite terminals. Here the capacitances are efficient. 4F

4F

4F

15.(A) A

A

8F



A

A

4F

8F



A

B 6F

VMC/Capacitors

74

Solutions to HWT-4/Physics

Vidyamandir Classes

Capacitors A

A

4F

1F

1.(D)

HWT - 5



1F

1F



A

B 2F

B

B 2F 3F

2.(A)

3F

A

B

 9F

3F

3.(A)

3C  C  3.75 3C  C

5.(C)

6.(C)

U 

7.(A)

A



2d/3

d/3

K2

K1

C  5F

4.(B)

C Where

Done Earlier

C1C2 C1  C2

C1 

K1 0 A d/3

…(i)

C2 

K1 0 A 2d / 3

…(ii)

1 2  4   2   8 J . 2 C B



 A  A 2 0 A Ceq  0  0  d d d

C 8.(D)

9.(B)

10.(B)

Ceq 

0 A 4 0 A .  d1 d 3d  2 K1 K 2

 A C  0  0.5 pF d For 8 capacitor Cnet = 8 × 0.5 = 4pF.

Use C' 

A A K 2 0 2  2 d d

K1 0

13.(A) 15.(C)

11.(A)

12.(A)

Cnet 

3 0 A 0 AK  d 4d

14.(B)

Cseries 

4 F 7

C parallel  7 F

VMC/Capacitors

75

Solutions to HWT-4/Physics

Vidyamandir Classes

Capacitors 1.(A) 5.(D) 6.(B)

2.(A) 0 A 0 A 2 0 A C   d d d

CBC 

10  10  10  15 F 10  10

C AC 

5  15 75  F 5  15 20

HWT - 6 3.(A)

4.(B)

12.(D)

75  2000 20 = 7500 C

Charge on capacitor of 5F 

VA  VB 

7500 C  1500 volt 5F

VA  2000 volt VB  2000  1500  500 volt 7.(A) 8.(D)

Flux shared by 8 cubes = Flux shared by 1 cube =

q 0 q . 8 0

9.(A)

10.(C)

11.(A)

13.(C)

14.(B)

15.(A)

VMC/Capacitors

76

q

 2  1  6

0

Solutions to HWT-4/Physics