Solutions Hwt Electrostatics

Vidyamandir Classes

Solutions to Home Work Test - 4/Physics Electrostatics 1.(D)

HWT - 1

At any instant T cos   mg

T sin  

Kq 2

x2 From Equation (i) and (ii) Kq 2 x2

q2 

mg x · · x2 2 K  2

…(i) …(ii)

 mg tan 

x  tan   2   

mg 3 x 2K  Differential w.r.t. 't' q2 

We get Vx1 / 2  constant

V  x 1 / 2

 2.(A)

Using conservation of energy Ki = 0

Ui 

1 2qQ 4 0 a

1   1   5 

At centre Uf = 0. Gain in KE = Loss in PE 

K 

3.(A)

v Q(fixed) q Let r be the closest distance when velocity is v and r1 be the closest distance when velocity is 2V Using conservation of energy.

1 2 KQq mv  2 r ' We get r = r/4 4.(A)

1 2qQ  1  1   4 0 a  5

KQ1 KQ2   3 1

KQ1

3  102

…(i)

1 KQq 2 m  2v   2 r'

…(ii)

Q1  3Q2

109 . 3

 10  Q1 

Calculate force using coulombs Law. 5.(D)

W  U a 



K 5  106

VMC/Electrostatics

 10  106   9  101 J .

0.5

59

Solutions to HWT-4/Physics

Vidyamandir Classes 6.

Kq1q2 r



2

mv 2 r



Kq1 q2 r

v

T 

2r  v

42 r 3 m Kq1q2

7.(C)

K 

d 16   2. d   16  8

7.(B)

Mass should be same as force of repulsion is same on both masses.

7.(B)

Equipotential surface is perpendicular to the electric field.

10.(B)

Let us consider 1 Q and 2 Q

 

K 2Q 2 F = Force before touching = F' = Force after touchup =

r2

K 9Q 2 4r 2

F ' > F. 11.(A)

a

F me

and a' 

m a'  e a mp

m a'   e  mp 



12.(D) Use Q = ne

F mp a

   1.6  1019 m / s 2  

u = 2 × 1012

Mass  2  1012  9.1  1031  2  1018 . 75

 1.6  1019 C   1.32  1013 C

13.(C)

Q 

14.(B)

Only the linear momentum of charge '  q' is constant.

9.1  10

15.(A)

31

Fnet on Q 

2 Qq Ka 2

Q2

 K



2a



2

0

Q  Q 2q    0 2  2 Ka

2q   Q / 2 Q  2 2 . q

VMC/Electrostatics

60

Solutions to HWT-4/Physics

Vidyamandir Classes

Electrostatics

HWT - 2

2 Qqx

1.(C)

F 

3.(D)

Principle of Quantization ; Q = ne. where n = 1, 2, 3,….. .





3/ 2 4 0  2  x 2

2.

Incomplete Question.

4.

Incomplete Question.

Charge on inner surface is '  Q' due to induced charges and no charge on surface

5.(B)

r (a < r < R) and + Q on the outer surface.

6.(C)

Due to induction of charges.

7.(C)

F 

KQ 2 r2

FC  FA  FB F 8.

Done earlier.

9.(A)

Charges will be induced but equal and opposite, so the net charge is zero.





10.(D) Use coulomb’s Law: 10  103  10 

9  109  q 2

 0.6  F 

11.(D)



2

F1 

q  2  106 C  2C .

KQ 2 r2

K Q  3Q 2

r 8



3F 8

A 1C

Fnet 

12.(D)

FA2  FB2  2 FBC FAC cos 60 C

C

= 3.12 N. B 1C

60

FBC

2C

VMC/Electrostatics

61

Solutions to HWT-4/Physics

Vidyamandir Classes 13.(D) The new force is reduced by the value of dielectric constant i.e., 4.

F F  ; where k = dielectric constant. K 4 14.(C) However there is force of repulsion on both the bobs but of same magnitude. So no effect on time period. P' 

 . g

So T  2

15.(C) Net force on string is zero due to charges. Only force of tension in the string acts. So no change in time period. +q Fe

Fe

q

Electrostatics 1.(B) 3.(A) 6.(B)

|F|

K 8q 2

and | F ' | 

Kq 2



HWT - 3

F . 8

2.(B) r r Because in case of sphere, charge distributes uniformly on whole surface quickly. 2

2

1c 1c 1c 1c



O 1

2

4

8

1 106 1 106 1 106  F  k    .......    12   2 2  3 2   F  9  10 10 9

7.(D)

6

 1  = 9  10    12000 N . 1  1   4 

Since electrons from earth flow, it means the body is positively charged.

T sin  

8.(C) 30



10

T

T cos 

9.(C)



3

1 1 1  12  2  2  ........   2 3  

  E0



T cos 

Fe



K  10  106



2

12

T  9  109  1010  2 T = 1.8 N

mg 2a a cos 45  E0 a 2 .



10.(A) Here electric field provides the constant force. In SHM frequency depends on variable force and not on constant force. Constant force only shifts the mean position. 11.(AD) |Q1| > |Q2| because number of field lines are greater at this charge. Since the magnitude |Q2| is less. The null point exists near the Q2. Note:

Null point exists near the charge with smaller magnitude.

VMC/Electrostatics

62

Solutions to HWT-4/Physics

Vidyamandir Classes    ˆj    2 q 2 ˆj 2 0 r 2 0 r

13.(B)

E

14.

dq  4r 2 dr . 





5 r3  5 r  = 4r 2 dr 0    = 40  r 2   dr R  4 R  4





r  5 r3 5 r3  r4  q   dq  40   r 2    dr  40  .  Now, 4 R  4 3 R.4     0

15.

t

2h .me eE

2h  ae

E

r

zero



0 r  5 r    . 4 0  3 R 

HWT - 4

Fnet = 0

  4 r

R 2

2.(A)

2

Put values to get t  2.9  109 s .

Electrostatics

1.(A)

Kq

a E2 r 0  As E  1 2 R/2 2 E r2 r 1



2

  Kr a  dr

 4r   Kr  dr 2

a



 R / 2 2  R 2

8

0

a = 2. 3.(C)

Electric field inside the shell is zero and outside the shell E 

4.(B)

K.E. = work done against field = qEy. +Q

5.

R –Q

E E

i.e. E  r 2

for 0  r  R

 2 0 r

r R and put values to get E  2  106 Nc 1

Use E 

8.(C)

For Non-conducting sphere ; + + + + + + +

r2

0 for r > R.

7.(B)

9.(D)

Q 4 0 r

Kq

+q

– – – – – – –

a

E

O

R

qE m

g eff 

g 2  a2

T  2

  qE  g2     m 

2

10.(A) Electric field lines are always perpendicular to the equipotential surface

VMC/Electrostatics

r

63

11.(B)

Solutions to HWT-4/Physics

Vidyamandir Classes 12.(B)





     2 0 2 0 0 zero



K  4R 2



 E   and E  r 2 r r Electric field inside the cavity is uniform and Non-zero.

13.(CD) E  14.(B)

KQ 2



2

a a    15.(C) Potential at  a, 0 , 0  and  0 , 0 , 0  is zero due to charges q and  q at  0 , 0 ,  and  0 , 0 ,  , So W = 0. 2 2    

Electrostatics

HWT - 5

1.(D)

Direction of E changes since it is a vector quantity but magnitude remains same and V remains same as it is scalar Quality.

2.(D)

E

y ˆ y ˆ y ˆ i  j k x y z

E

 d  20  40 x   2 dx  x 2  4  x2  4



E at x  4 m  3.(C)

E 



10 V in + x-direction. 9 m





d 3x 2  5   6 x dx

E at  2 , 1, 0    6   2   12 V m 1 4.(B) 5.(D)

(B) is not possible because number of field lines should be proportional to magnitude of charge which is not obeyed in this Figure. (C) is not possible because field lines can never intersect.

6.(D)

Work done is zero an equipotential surface.

7.(A)

A is + ve because field lines are away from the charge and B is ve be cause field lines are towards the charge. Since intensity at A is more than B, So | A | > | B |.

8.(C)

Basic formula

11.(B)

Module Question.

12.(D)

9.(A)

Basic concept

+ + 

T



T cos 

T sin   QE  F = QE

T sin 

mg

VMC/Electrostatics

|E| 

v r

 r 

600  3m 200

T cos   mg

+ +

10.(D)

tan  

Q 0

Q . 0 mg

  tan 

64

Solutions to HWT-4/Physics

Vidyamandir Classes y 13.(C)

K . 8q

–2q

+ 8q L

 L  a

x

a

2



K 2q a2

Solution to get. At x = 2L. 

Kqr

14.(A) Use E 

R3

to get E  2  106 NC 1

15.(D)

Electrostatics

HWT - 6

1.(B)

Electric field inside cavity is non-zero, on conductor electric field is zero and outside conductor non-zero.

2.(B)

Use   2rE E



3.(C) 4.(B)





    90 Solve to get   tan 1



 (Q1 + Q2 + Q3)

5.(B)

Q1

(Q1 + Q2) Q1

4R

2



Q  Q2  Q3 Q1  Q2  1 2 2 4  2 R  4  3R 

Solve to get:

6.(A)

 2

Q : Q2 : Q3 = 1 : 3 : 5.

  P  E  PE sin 

max  PE at   90 7.(C) 9.(C)

8.(C)

E 

dV dr

VC  VB  



a

V  10.(C)

E  J 

I 2

3  3R. 2

dV    E .dr



ab



BC  AB sin 60   2 R 

I dr  2 r 2

1 1     a a  b 

I 2r 2

VMC/Electrostatics

65

Solutions to HWT-4/Physics

Vidyamandir Classes 11.(D)

U  PE cos 

=

  · E cos    24.5  104 J . E sin  tan 

12.(C) Use   P E sin  . 13.(C)

W 

2



 PE sin  . d    PE cos 12

1

14.(B)

For   60

PE = 2W

For   180

W1 = 4W 15.(D) Use max  PE to get max  1.5  103 N  m .

From negative to positive charge

Electrostatics

1.(A)

HWT - 7

Enet E is parallel to the axis of the dipole but opposite in direction.

q

–q

p

+q

2.(C) F1

F2 Torque is experienced by the dipole. Non-zero Translational force is also experienced as F1  F2 in Non-uniform electric field.

q A



3.(C)

P



q 4.(B)



(  0) X-axis

+q

Eaxis  K .

2P r

3

and E  K

P r3

Eaxis 2  E1 1

d  d net  q    E  qE   2 2

q

5.(C) F=qE

 O

net  qd  E

F   qE

  P E sin  and   I 

d +q



I   PE sin 

I 

f 

VMC/Electrostatics

PE .   sin    

  

I / PE

66

Solutions to HWT-4/Physics

Vidyamandir Classes QA QB  RA RB

VA = VB 

6.(ABCD)



QA  QB .

Also  A   B and E A  EB . 7.(A)

Potential inside the hollow sphere is same at all points and same as that on the surface.

8.(B)

Since they are at same potential and of identical shapes V 

kq r

So, q is same on both. 9.(B)

Linear momentum of electron Le  Linear momentum of proton L p 

Le  Lp



10.(D)

W  q  V   100 1.6  1019

11.(D)

E   6 x   6 x 2   12 vm 1

2me eV

2 m p eV me . mp

  4  10  2.24  1016 J 12.(C)

V 

1 r

13.(C) Electric field is non-zero but potential is zero. 14.(D) Let r be radius of small drop and r' be radius of bigger drop and V & V ' be the respective potential

n.

4 3 4 r   r' 3 3 3

V'  K .

nq 1/ 3

n 15.(B)



 r



r'  n1 / 3 r

Kq 2 / 3 .n  n2 / 3 V . r

KQ0 q Kq 2 KqQ0   0   2



q

q 2 2

.

Electrostatics

HWT - 8

W [By definition of electric potential] Q

1.(C)

V 

2.(C)

Potential inside a conductor is constant. Note: Surface of a conductor acts as equipotential surface.

3.(A)





Use W  Q  E . r   Q e1 ˆi  e2 ˆj  e3kˆ .  aiˆ  bjˆ 

W  Q  e1a  e2b  4.(D)

+q

Initially: A

Ui 

q

+Q L

C

L

B

KQq KQq Kq 2   L L 2L +q

Q

–q

Finally: A

VMC/Electrostatics

2L

B

L

D

67

Solutions to HWT-4/Physics

Vidyamandir Classes Uf 

 Kq 2 KQq KQq   2L 3L L

Work done = U  U f  Vi   5.(C)

As done earlier

V ' = n2/3V 40 = (N)2/3 × 10

qQ 6 0 L



.

N = 8.

6.(C)

1.6  1019   1.6  1019   U K J   14.4 eV

7.(A)

Use U  U f  U i .

8.(B)

1010

–q

q/2 a

A

q/2

B

r

P

C

 q/2 q q/2  YP  K     r  a r r  a 9.(C)

Since both spheres have same ' ' but are of different radii; So they have different value of charges.



K 4 0

V 

1   R  r   0 0

10.(A) Use V0 

11.(C)

1  QR qr  Q q    R  r    2   4r 2  0  4R

V 

1 4 0 r

 qi



R  r .

9  109  3  106  27  103V . 1

1 1  Kq t1 VA  VB  Kq      rA rB  rA rB



t1  rA rB Similarly: t2  rB rC Since

rA  rB  rC



t1 < t2.

12.(B)

Potential decreases in the direction of electric field and remains same in perpendicular direction.

13.(B)

Discussed in class

14.(C)

V 

KQ depends on both charge and size r

15.(A) Since proton is moving against the field so work done by proton is negative and hence work done by field is negative. Proton is moving from low potential to high potential so energy increases.

VMC/Electrostatics

68

Solutions to HWT-4/Physics

Vidyamandir Classes

Electrostatics 1.(B)

HWT - 9

V Slope = 0 Slope = + ve Slope = 0

Slope = – ve 3

2

1

4 x

dV E   slope of V- x graph. dx



E 2 > E 4 > E 1 = E 3.

K q1 q2 r

2.(A)

Use U 

3.(B)

Since no electric field present, so no work is done in all three cases i.e., WA  WB  WC  0 .

4.(C)

In uniform field a straight line perpendicular to field is equipotential

6.(A)

Charge on Nucleus = 47  1.6  1019 C

V' 

9  109  47 1.6  1019 3.4  10

14

5.(C)

 1.99  106 V

7.(D)

Surface of a conductor is equipotential. So VA = VB = VC

8.(A)

Use V ' = n2/3 V Where V ' = Potential of bigger drop V = Potential of small drop N = no. of droplets

9.(B)

Potential at any point inside sphere is same and equal to the value of potential at the surface of sphere.

10.(CD) 13.(A)

11.(C)

qenclosed 

6C 2 



12.(B)

8C 2 

Use:

 E.ds 

 . 0

 7C (due to charge at left equal).

due to due to  half  half disc  rod 

= 2C

net  14.(A)

2C 0

  20 Vm as flux linked with the closed body is independent of the geometry of the surface.

15.(A) Area Vector = 20ˆi

  E .ds  5  20  100 unit

VMC/Electrostatics

69

Solutions to HWT-4/Physics