Electrostatics Mtg 5x3o5k

)

is proportional to

approximately x2

(c) ; Energyofacapacitor U =

or lll -= K Kr+K2

x F--+l

= Ur- U,

\Zrcna"

ZAesK3

a

Finally

/\ or AU =t 4lx2

- fftotsingle ddd +" KAeo Aeo{&+F) 2AesK3

a

q

2Oax2 or LU= - " z 4Teoa

* rS

x:+a

0

lnitiaily

t I I uz= Qql a^l"aa* "-*l #

2AesK3

But C

e

x:_a

2Qq

or AU _

tdd .'---=++..C Aes(K1+ K)

an

equipotential line.

AeoK,

C, and Crarcinparallel

,.

Any line perpendicular to

.x-axis

{,

,,=.,(+)";2= tf

..cz:*,(9"#=

to

.x-axis.

.'. u;:

or Q(J|+\: -qE or Q= (2 ,^ +,'n^,. 23.

Potential decreases

runs parallel

and 53 closed.

+. ffi:,

:

The fieldEisuniform and

(b) : Net electrostatic energY = 0

.. 9!. *

switch S. Option (a) represents the correct answer"

as we move along the direction of electric field.

occur unless the two capacitors C, and C, are either connected in series or in parallel. Charges would be maintained as they are unless S,, S, and E, all are closed. This condition is given in none of the options. Hence the potentials of,C, and C, will be maintained asVr=30 volt and

22.

(a) : Due to attraction ofpositive charge, the negative charge is bound and so it will not flow to capacitor B through the

(b)

No change of any potential across any capacitor will

Vr=20 volt, with

E

--

LCV' 2

fnitiuffy, Totalenersv 1

:

ur: )c1vl+

Lcr? * lcvr"

2'2"

vly

After , common potential =

....,(i) lz

\

216 t/

ct4 -

+ c"vz

Ct+

-lFc, l/. --.tr

Cz

. r/_cvt+cv2 =vt+ vz 2 C+C

,/\

Final total energy = U,

u,: )
f,{v,

produce any electric field in between the two cylinders. The-charge distributions given in the alternatives (a), (c), (d) produce a radial electric field between the cylinders.

C,V'

35.

)lc'v

vector

û

=t

i

from the centre of the given sphere. Let the

Let the position vector ofthe centre ofcavity from the centre

* vr)'

ofthe sphere be {. Clearly { is constant. Consider the uniformly charged sphere without cavity. Electric field at I can-be obtained using Gauss law.

....(ii)

Ex4nr2=ox!nr3 '

.'. Loss in energ5r

.'.

is

position vector of this point from the centre of cavity be 72.

\/,

: Ur- U, r?* jcvi-

(b) ; Consider a point in the emptied space whose position

3

fv, + vz\2

The direction

of E, vector is along /,

?$

f,v?

* v'r-

ar,vr1 =

f,vr-

J

Now consider a negative charge of the same density in the region of cavity. Due to this charge, electric field at I is given by

rr),.

(c) : Field lines do not enter the metallic portion of the shell as field is zero inside the metal. Hence options (a) and (d) are

; LZ=

incorrect. The field lines are normal to metal surface. Hence option (b) is incorrect. Both the conditions are satisfied in option (c). Hence (c) represents the correct answer.

30.

(c) : According to option (c), the electric fiElds at O due to (i) P and S nullify each other G\

(ii) pand

Znullifyeach

31.

O-

-;r2 J

Supbrposition of the two charges forms the emptied space. To get the net electric

field due to actual charge distribution superfunpose the above two fields.

P(-)

u

other (iii) R and U addupto2E z'(+) Hence option (c) represents the answer.

so we can write

q=1r.

=f,rar?+zv2r- v?- v7- 2v|lzj

:

or Er=I, ' 3 [[ "=tl "0,,

R

E=Er+E,

(+)

o_

7rt' J

=

This implies that the electric field in the emptied region is non-zero and uniform. :

s(-)

v

(d) : At any point over the spherical Gaussian surface, net electric field is the vector sum of electric fields due to all the three charges + 4u-grand?z(c) :All the three charge sheets will produce electric field at P. The field will be along negative Z-axis. Hence

*2o * 2ro

"=1" l2to

= or L: 33.

(d)

34.

(a) ;

2c:

a]c-;r It can be seen that potential both atA and. B are zero. When the charge is moved from I to.B, work done by the electric field on the charge will be zero.

EoK.

will

appear a pgtential difference between the two cylinders if there is an electric field in the space between them, since the poterrtial difference is related to T_here

the electric field

dv

r

as

=-E.dv,

and E=

,L2nesr'

is distance from the axis of-cylindrical charge distribution. The charge distribution given in alternative (b) does not where

36. (c):

37.

(c): The electric field at O due to the two charges ql3 will get cancelled. Electric field due to

( -2"\

|.-iJ

will

be directed in

-ve

r-axts,

Lo=

I Qql3) 4lteo-T'- =*Rt q

Potential energy ofthe system

Electrostatics

217 or

-4me, l(+)' +(-+) ' "d-n)1 - "' L 2R

2Rsin60o

or fh = Z"

2Rcos60"l

The magnitude of,the forie between B and C

-1

l(

or

zn\( q\1

q' l\31\3/l_= ,n' =t54;;p' 4nes f lznsinoo';2 r. 3- ;"r"Rt=

.

l( n *t-'n\f Theelectriotential ato.lt=rl l\: ? l/l=0. 4nesL .'t l-"' (a) : Time constant r: RC.

t*".;=f

-

r/t,.:r,=

Aeo d-X

--;-----: +.

-

-.

disc. For rod 8 C only the cube

at

8C

f

(d)

a

/eo

::

;

Charge densify, o =

n -Qt+Qz 4 . A+Qz+Qz 1o =o vz, ---g---l "yt I

Itt

Clttt".

:. 43. '

Area

--

-

6

t,/

:

Lq

a

i, - )cvi

rÛ

,(KC)Vi

LJ rel="nofollow">

o,

%:

K

whereK >

I

Lto.

: For equilibrium, total potential . kQq , kQq , kQ2 a a2a o qnq+t

(b)

energy = zero

Consider Q when it is at a distance x:from origin. Let F represent the force between q and e: Resolve -F into two rectangular components. Resultant force : 2Fcos0

44.

o

p--3___{

o, u= _x.

_Qt

... I =92=Q, r35

:

-a)

F

=0

88t7Qz=gz

=

(0,

Hence options (a) and (d) are correct.

89;3Q!t: 5Q,:'8,,.

(d)

U=

accbrding to Gauss's law.

E"

Enersy,

.:,

Frux=*(+.T- rr)=-+

.'.

41.

(c) E:

7 C is placed inside

:

Fcos0

.'. o&:KwhereK>l "'. QrQo Qn (b) tt= Zo : potential difference across the battery.

aa n . _7 a.

3 C isoutside the surface. .'. Flux = total charge enclosed

(b)

-

Fcos0 (0,0)

and

lnitial capacity:

is inside the cube due to the

is inside.

(0, a)

Final capacity = /(f : -0 .'. Initial charse e-o O^ CVFinal charge Q= (KQV,

(a) : By Gauss's law charges,inside' the enclosed place z^

F

(a), (d) z (a) Q> Qu

I : I and K = 2 in theabove equation 3x2Ret "=d-3Vt +6d -2d -6W:6Reoi.. 5d +3Vt'

contributes to the flux-

FsinO

0). Hence option (d) represents the answer.

4-vt.* x(d*Aru*)

6C

is not proportional to x, the motion is not

maximum at points (2a,0) and (- 2a,

x

Substituting

39.

-!-4"tren

e.}f"

origin

x+ K(d-x)'

KAes

where ,t =

2kOax

force will be zero at

KAes

cr+c2 =

=

simpleharmonic. The charge Q will execute oscillatory motion under the forcc,2,Fcos0. The

l

NowC=ctC'

^

.{rR

-P-lx- + a') "r(xz + 421rtz

Since F*

J

38.

4:2Fcos0

,T

(b), (d) : Charging battery is removed. The charge on the capacitor remains constant. The plates ofcapacitor are moved farther apart. The capacity

(#)

decreases when dincreases.

The options (a) a4d (c) are incorrect.

PotentialV=A. C

Since q is constant and C decreases, Irshould increase. Option

(b) is conect.

218 Enerw

45.

U:Lt 2C

or v.= v'A or

Hence options (b) and (d) are conect.

Again, along similar lines

(a) : Potential inside the shell is equal to the potential on the

or

I

ofa shell.

:.

is a point on the surface of

sphere. It is an internal point

.3)

vn=-"*[*

v,:#1".#1

LV:

Vn

-

Vo

or

on the shell.

r)

Vn-Vu =V

r/2(4zre,,R)

When shell is given a charge (-3Q), the change of potential at A dnd B will be same. Let the potential due to charge

...Workdone=qLV

(- 3Q) be - tt. .'. New potentials

or W=

- It)

at A. and B are (Vn New difference (Vn t/) (VB

:

-

-

-

and

(V"- l/)

Ia:VA

-

Vu

= As

before .'. New difference = V.

(a) : lncombination(a),

I c.tr

11

or

10

c.r:

49.

r11l : C* t* r* rU

|i r*

50.

Since the battery is disconnected, the charge remains constant.

d

option (a) is conect.

(b) Option (b) is incorrect. (c) Afterthe slab is inserted, capacity changes. KeoA ... c,= and v, : O : "oA,v , _4_ = L

dC'dKeoAK

'

t/ tlt\

or tt=

C'd

or

E:

(d)

W

V

*.

22 I 2

I 2

e6AV

KesAd

ORtion (c) is conect.

J

Y,'+lt

g,1y,9

where V, = V/K.

Vr*

V,

charge per unit length of rylinder

(d) : The lines offorce never enter ametallic sphere because intensity is zero inside a conductor/metal. Hence options (b)

Option (d) represents the correct answer.

sl.

(b) (c):

(a) Magnitudes of electric fields in the slab and outside the slab are

p *ano. Refer ro Eor

f.

op,ion 1";

+(8.)(+)'

*]

trd

i,

incorrect.

(b) The direction o{ --(En/h electric field remains'the' "

(b)

is

therefore correct.

(c) Electric field lines alway flow from higher potential to lower potential.

option (d) is conect.

(b) z Let A and B denote the centres of two rings. Vn=

(c) : Let ?', : ... E =

same. Option

= Change in energy

= ! cv,

Jzgneonl

figures (i) and (ii)

-'. Intensitv" E:l!-l \d),

_o E:'

r)

(a) is inconect

eÂ

fu+

Q)di -

and (c) are incorrect. The lines of force are at normal to the surface. Hence option

(a)

Q:CV whereC=,

q{Qt-

*t"." i, : distance from axis of cylinder. =L 2reor I .'. E varies as :. Option (c) represents the correct answer. r'

(a), (c), (d):

.". A:

Qr

*:#l{e-ez)- Qt-JzlQ)f or L,v: g;1u; t

for

the shell. Let,B represent a point

47.

Q'

4n;oRT

q is constanl, C is decreasing. .'. U should increases. Option (d) is conect.

surface

46"

q * ' 4nEoR

The electric potential t',

therefore,

increases continuously as one goes from O to 3d.

oAllBC. Siope of(Ol)

> slope

oflB

V

lj

x-1d x:3d

Eleatrastatlcs

,219

Option (c) is correct.

. 52.

n" =

(d)

'

The option is incorrect, Refer to figure (lir).

(d)

: Letpotential at the origin=

:.

V

q +q ...v:L-lg-L*...f I [x6 2xo 3xo 4*o or v: q ft_1+1_1* Q

v= 53.

4nEnxo

l

4

centre is at the origin O.

zo:0

P (0,0,

P: F

z)

F

^Qq'g @;Zfo

""'(ii)

ut point,4 in the figure.

is continuous at r = R. Option (d) is inconect. Points

I

and B lie on the ellipsoidal cavity carved

inside the

Eo

| 4, n,= ' 4rso Rt

Era.r, forr (R, as(OA) infigure.

Option (a) is conect. lntensity outside sphere = E,

is *

Option (b) is correct.

+

*\

(c) Changes in $ and E are)

continuously present for

r > Ro.

Option (c) is incorrect. (d) For r 1 Ro, the potential { is constant and the electric intensity

E

is zero. Obviously, the electrostatic energy is zero for r < Ro. Option (d) is corect. Hence (a), (b) and (d) are conect.

E

lntensity inside sphere = E,

q

sphere.

field discontinued at r: Ro. Hence electric

+ Rt) = R,,

(a), (c) : Due to a charged nonconducting sphere,

atl

Option (a) is correct. (b) Electric field E = zero

it is clear that F is not froportional

Option (c) is conect. Hence options (a) and (c) are conect

(ii)

4, ,4trso R.'

(a), (b), (d) : The given g-r graph is for a spherical symmetric charge system, centred at origin. (a) The whole charge Q will be enclosed in a sphere of diameter 2Rn.

I Qqzn " r - 4nuo -R3ot Force (F) cc -zo or Force (F) a -displacement when:o << R. The motion is simple harmonic if zo K R..

:"

=

Hence option (d) is conect. Options (a) and (b) are'incorrect as the cavity is ellipsoidal. It may however, be noted that options (a) and (b) hold good in case ofa spherical cavity.

to displacement zo. Hence the motion is not simple harmonic. Option (b) is incorrect.

zl

Eo

s0

the above discussion.

From equation (ii),

(i)

:.

Eo

Total flux through cavity-surface =

The force Fattracts Ptowards centre O. At O, the force on P becomes zero. When P crosses O and moves to negative side of z-axis, the force is again towards O. Thus the motion of P is periodic for all values ofzo satis$ing 0 ( zo < oo. Option (a) is correct. Option (d) is inconect under

54.

suface =

Potential at I : Potential at B. Option (c) is correct. Accorciing to Gauss theorem,

-

4irlo

(

is incorrect in view of (a) and (c) above.

at

I

:. E=0 :. F=sE | or F: vr ' -

increases. Option (c) is correct.

points which lie on the conductor, is same, the potentials and are same.

v

Force on charge

r

within a perfect conductor. Since the potential of all

' 4^7P *49n ""'1tl

At centre of ring,

as (AB) in the figure.

(iv) Intensity

55. (c),(d):

= Charge on the ring. Elechic freld at p = F

-

(iii) Option (b)

.Oo

I

Let zu << R,

I

ln2'

... -lOz^ h.: "-

when R

r'

E2a--7,

.10

(a), (c): The poSitively charge metal ring lies in x-y plane. Its'

Let

4neo

.'. E, decreases as

4zes

4nesxsL 2 3

*]-9"

57.

(a) : $hen -g moves in an elliptical orbit about e, as in solar system, the angular momentum is a constant. But the linear velocities and hence linear momentum

will not

be a

constant.

'Ior:

Mrz

"! : r

Mvr

changes.

As r changes, v as well as o change but

(a)

ftrl:

constant.

Consider first the equilibrirrm of bead at.p, having charge

220

Ifthe 4,. Different forces acting on it are shown in the figure. force F Q, Md Qzare similar in nature, then electric "harges repulsive and is given bY will be c

'--4fl, t'

or v =

At .orv:4trtor(r" 4 *^5-** 4*rne;n'from(i)and(ii)

R

Of vt

Resolving the forces

n parallel

andg-

mg

+R2)

4neoQ2

61. Initially, potential difference across

two equations for

.'. Total energy in capacitors

(Z- F) cosc

: R, = zgcos30o + (7n-.F) sino ' Similarly, for the equilibrium of bead at

mgsin60"

: (I-

.'.Initial

.....(iii)

I " :. uA: rQC)v':

ar

mgcos60o

T: F+ mg: (qgJP)+ mg

From equation (ii) R, = mgcos3A" + Rr: nzgcos30o + rng sin60o

(?'- F) sin60'

.tjmg

:. n,=

Potential of.B

u,:tsr' :l

. "u2

equation (vi)

......(ix)

62.

Let qr= Charge on inner sPhere or = Surface density on inner sphere r = Radius of inner sphere Let q,o,4nd R be these quantities for out€r sphere.

Qz n, 4t :

ffi- o;V

Qz

,2

-.

or ez: or

"

:.

Qt

4rsrR

.....(D

U,

rr:.,[t*(c)tq)l .. uD-"1 4"%(AD

)

_ -2x9x +

q):,Q

Potential at common centre due to Q, and, q, V'r,Vr= Pote'ntials at common centre due to q, and qr. Qt tr... ,t*

.'. y- =

K,

J

potential energy at D =

loe x(5 x 10-5)2

AD

.'.

....(ii)

Total energy

at

: -_E AD

Kinetic energy at D = zero

Let V :

qtrsor

=_9J

"' .'. Total energy at f, = (z$ - !) = -5 J

q,)

ffi*h"r"@, orz Similarly, h: f;Vt

L

Kr:4

side

-(Z_-at;

4z:

potential and kine$c energies while

5

R"

R2(q, +

5

lOe x (5 x 10-5)2

Kinetic energy at C = Add I to each

has

(ii)

q " (-ilf o ,^. : r*lt"4neo(AC) )

_ -2x 9x

4rtb ---_12 +R2 Of '

4't

-q

......

!!:1. Uz

o.

5cv2 5

The moving charge

;.

oz

h'-

(JB

at C and at D. The sum-total of the two eneigies at C and D remains constant. Potential energy at C = U,

This shows that q, and qrmust have unlike charges for beads to remain stationary. No. It moves along the line of force only if it is'straight.

o,:

:+

,CY V 3C3

3cv2 * c'.' ur: rr^ = ' -7-,, "2 lcv' -n"z3"'

'..'..(viD

mgP

New capacity

.'. FinaltotalenergY= (JA+

From equation (iv) R, = mgicos60o + mgcos60o: mC-{viii) (b) If the string is cut, then Z beiomes:zero. Hence from

Qflr:'-

i

Initial charge stored

:

....(i)

3CV2

:. un:f, (f)'

'.....(vi)

is I/.

=(I(Jn')'z - cvz energy(Ur):ClP

Let the switoh between A and B be opened. Potential ofl is still V..Capacity ofl is (3Q.

p

F; sina

each capacitor

",.....(i) ......(iD

.'.'(w) + (I- F) cosa string the in Tension where 7: Dividing equation (iii) by (i),'we get : tan60o = tand' ......(v) i.e. a:60" From equation (iii) T - F : ng Rr=

60.

QQ+ R\

| = -'---------7-

and perpendicular to AB, we get following equilibrium mgcos60" =

59.

from (iii) and (iv)

#4,

#.

as the charge stops

there'

D: + * 0 = -rf AD AD

......(ii)

Equate energies atC md D /.q

-s = -A;

....(iiD

or

AD=9m

LAD},(OD': UDz -UOY or (oD)'=Q)2'Uo)z or (OD)z=Q)2-QF=72 or (OD)=8.48m' From

....(iv)

V= V,* V,

63.

The potential energy

will

be minimum when the two bigger

Electrostatief

221

charges are located at maximum distance in the system.

cm = x

x l0-2 m,

be the distance between

r*t}ofzq"q

= +neo 1 x ,,_g'r100[2, 8 . 16l 4nen Lx 9-x 9J

Potential energv

du

&

=o

2q

Let.r

and q.

8q*q *(zq)(sd] --l

* (9 x) -

8l .-.-...'_l (9 - x)') zqq

or .F =

zsfrca +'loal - JI or 4neol9 "' 36 -t ---

F:

E=O

(0.03)z

q'

.....(t)

Let 0 denote the angle which the string makes with the horizontal in position of equilibrium.

or cos0=

l+no21oy

=9[o

_ b+cl

=0.0173

100J3

or g=89o

.....(ir)

, 4nb2(-o)

b

=

c

)

b

lqnaz(o) *4nb2(-o)

il+

rr _ t

+rc2(o)l

lqn aQn.*acl ' ';J

4neof b

b

*qnc2d)]

c

)

-u*"]

l +na21o1,

qnb2(-o)

.

+nc2 .-\1

' 4reol c c c ) ol o' _ _+ct=_t b2 1 of -o'-b'+c'1 " '- enlc c I eol " J |

(ii)

.

Given: Vt = Vc

9(o €o

-

uac)=

g(4 €o\c

6'

c

*J )

-o2h2 a_b+c:__ +c cc - (q+b\(a - b\ a - t): or(a+[)=c. c Charge (-2Q)is located at(-3a,0) inx-yplane. Charge (Q) is located at (+ 3a,0) in x-y plane,

(a)

Locus ofzero potential: Let P be a general point, denoted by (*, y), where potential due to the two charges is zero.

v

(2 cos30")/ J -----------z-

residing

_ I

_ t

(9 x loe) x.q2 x J:

/: cose= I

be due to

4"%lb

)

lor3

will

v^: I

,l ,' .1- Jj2

9r F = J3 *

F

or F=mgcot0 o. JT * t1tt q, = (l x l0-3X10) cot 89o or q=3.16 x l0-eC. (i) Potential of any

4zeof a

cos3Oo

l4neoa"

----+Icos0

* tz * t"f v": n -J-lqn 4neola b ,)

Eaeh particle of mass m and charge q is suspended by a 100 cm long string, m: I gram. They are located at comers ofan equilateral triangle ofsidelength 3 cm. Three forces act on each particle. (i) mg vertically dovmwards (ii) Tension Z (iii) Electrostatic force F due to charges at other corners ofthe Fo triangle. F will be perpendicular to zg.

2{

F mS

on all the three shells.

The charges be placed in the sequence of2q, q and 8q in a straight line suchthat 2 q Nd q are separated by 3 cm while g and 8q are separated by 6 cm. '' Elechic field at position of q = zero.

or

Icos0:f'

( *o)x13xl0-r)'z

^ or L:

il

T sin} -- mg

shell

Electric field at q:

F=

:.

charges

or ?s,=9 -x or x=3 cm E=

TfinA

stated above.

or tarr}=

when U is minimurn-

dU az x l00f-2 ' dx 4*o L*' 8 ^ -2 or o:-:+ x' (9 - ,)" or 8*=2(9- xY

Consider the equilibrium of a particle under the three forces

(-20A (aa,0)

JEE

t IT-

222

C

h aote rwi

se S ol uti ons

V=Vn+Vu

'^=olffil "=^lw5=1 vA+vB=0

2l

ar |

Y-4rt%lx-3a-

3r+-)

Potential onx-axis is zero at two places atx =

- *)' *y'l: (3a+x)2 +y2 or 4{9a2 + x2 - 6m + yzl: ga2 + }+ 6uc + Y2 or 36az + 4x2 -24ax + 4y2 = ga2 + x2 +6m + y2 or 27qv + 3x2 * 3om + 3y? : o o, *'*y2-lom+9a2=o or (x - 5a\2 + 1y - 012 = (4a)2

or

4l(3a

This equation represents a circle with centre at (54, 0) radius (4a). Radius of circle = 4a Location ofcentre ofcircle = (54, 0) (b) Expression V(x) at a general point on x-axis: On.x-axis, y

:

0

a andx:9a.The

figure represents the Z-x graPhs.

(c)

Speed

ofparticle: Putx = 5a,

o n=2-lt orv:l6reoa 8al neolZa-21 Potential on the circle will be zero. Potential on circumference is less than potential at centreHence the particle will cross the circle. ZiqLV

..-.."...: m

Let speed ofparticle = v

or

Speed

=ffi

or

Speed

=ffi

67. (a) A charge is uniformly distributed

over a non-conducting

ol t x .'.v= 4nenl3a -

spherical volume of radius R metre. Energy of the system ccinsists of two parts:

(i)

Energy stored within sphere: L/,

V:0

(ii)

Energy stored outside the sphere

a|x = a

V ,+ - q as x tends to -3a

V-++cnasxtendsto+3a Forx > 3a, consider Z

t,: Y-

Total energy ofsystem 0

2l I 4ttEoLx-3a-- .-3"J ol

V=0atx:9a

(i)

To calculate U,

:.

:

(Jt+

tr/=0whenxtendsto-m.

ElectricfieldE=:-.9., 4re0 R'

This is the field'in uniformly

chargednon-conductingsphere',',, .'. Energy stored per unit volume

.'. Volume of element

The results required for tracing the curve are as below:

.'.

(iii) Forx 1-3a,

or

(i) Atx:3a,V:+a (ii) Atx:1a,V=-q

1 - 21 v=9-[ 3a+xi 4*oL3a - x

(iv)

For

(v)

For.r > 3a,

-Jq1x13a,

| --l-l n=Ql '-4n;oba-x- t"-r)

lJ2.

lnside sphere:

Forx<-3a,considerl/

21 Ql 1 u: ' 4neoLZa -, - 3a+x)

= U,

Energy stored

=

:

: !r^n' 2"

4n4dr

(lr,t')14n2dr)

\2"

drJ,=:,,(#) ,4nr2dr I .4.roa, or d(.t.' 8r8o

R"

or ldu,= rr8eteoR""o ^l- .4 or q =

02 -

.-R

o'futr'lo

.i,'r,

Etectrostatics

223 -

'

(ii)

68.

rQ2

or u,:

40neo

.....(i)

R

To calculate U, : Outside sphere:

ElectricfieldE

Volume

When the slab is placed inside l, only 0.02 m2 is

| Q -. 4*o'7 =

Energy per unit volume

filled up with it while

remaining 0.02 m2 is with

Irot'

air. Capacitorl is

a combination of two

= 4t2dr ,

capacitors C* and Cn in

r2

/

du2= ttn'zarlf,'ol@

(i) Capacitorl: Capacity and energy: The plate area of A = 0.04 m? The plate area of B = 0.02 m2 Dielectric slab can fill only 0.02 m2

parallel.

3)

KeoA,

dd

(b)

or

!+ rt

d(lr: '

or Ior, =

Snen

1@ O' edr " t_ -

8rreo

j

d'

.......(iD

( r)

[.**J . u= ,,_-3 M2x4Gor U= 4: GM2 """ 20R "' -.:

..,,'.(iii)

"u.tf. -i,* .r u=

of

(c) Energy

l:

l.s *

charged

Part (a) deals with a

8aeolt

J.

where (J'1= Energy of capacitor without slab (Jo=Energy of capacitor with slab Once the capacitor is charged, its the charge remains stored in it even when the battery is removed.

Q= (2.0

of Awhen

' lOi) x (110) :....(iii)

slab is removed

eo(A, + Az)

C'u

.....,(i9

1 (2.2 " Or (l't = -x 2 0.4

10-7 )2

"

roe

"

Q?

Q2

tO-e;" lttO;2

lO-s Jor W-4.84x,10rJ Workdone in removing slab = 4.84 lQ- 5 J. + Spherical conductor

or U:

(2.0 x

or Ui :6.05 x l0-5 J ;. W=6.05 x l0r J-l.2lx

with a conducting

4neoR

Workdone in removing slab: W:

ix

,r=+*

non--

sphere.

2

......(i)

(8.85 x 10-12) (0.02 + 0.02) 8.85 x l0 ot C'A = 0.4 x 10-e F

conducting sphere.

or [J=!*

(ii)

or

: : +where C = 4neoRforsphere. 2C

Part (c) deals

I

IJn: l.2l x l0r

Un=

or C,.Ad-

1032 J.

spherical conductor Energy

.

or or

C'a = Capacitance

* l03r) * l0 , lu

F

Q= CeV or or Q:2.2 x 10-7 C

?

3r1z.s

Co= 2.0 x 10-e

Charge

." lul =;ttrRc 1u 1=

I

Il/ = U|-U,t

Uis the enerry required to disassembf"

,=+(#)+

" to-r2) I l0-")

(8.85 x

u.n2n= lc.v2

analogous to Q

Here G is analogous.

(8.8s

Energy in capacitor = U, Qz neoR

This potential energy is stored in electric field. (b) Compare electrostatic forces and gravitational forces for analogous quantities.

Here Mis

or L, ^ or

enerry: Ur + Uz

or tJ= Q' * Q' or (J:3 40zreoR 8zceoR 20

eoA"

or Co=$trxo.o2+o.o21 n

"2

02 I or Ur: . - -=8zreo -R Total

,

(c)

(iii)

Energy stored whenl and B are connected: U Capacitor B is filled with dielechic slab.

. n _ KeoA,

224 (9) x (8.85

*

10-12)

or

LB

or

C" :1.8 * lg-e F two capacitors, A and B, are in parallel.

The

on increase. This gradual increase in q results in currenl drawn from the' battery.

......(v)

y length of plate is dipped in oil, the system be equivalent to two capacitors in parallel, one with air and the other with oil as dielectric so that If

L = LAi. *"o"

.'. u:-l02 2C x -,-ral 1g-7'z or U:1.1 x l0rJ or (J:: I * e.2 '-,'-, 2 2.2x10

distance x from its aentre on its axis is given by

P ut't. \lR. + x.",

vo:*;

zfiN'

.

It

I

''-

-_---

Jn'liFnf

,, =tl+y(rK-l)l . dq ,ov -tK_lt- g_ Andhence,, =A = So, 4=

71.

/- 8.85x10-t2x500 0.01

x 111-11x0.001 =4.425 l0-eA

Consider a strip PQR at distance

r

from end E. Let its width

4eo

O.'

I

Kinetic energa of charged particle = i ^f For minimum speed'of particle so that it does not return

to P,

" or v'=

2

Let the resultant capacitance = C"

.

?"

The kinetic energy ofthe charged particle is converted- into : ' its potential en ogy x .'. Potential energy ofcharge (q)= qV

2aV

2q"?u

m

mx4ao

l_l_l co ct

,%*

attached to the circuit tr= constant. Now a;

to lowering of plates in the oil, capacity will increase gradually and as

......(i)

c2

Capacity of a parallel plate capacitor

=

ftffi

By similarhiangles EQRand EGF,

QR GFI=!

or

QR

=

whereGF=d

4I'

= Plat.-*eparation for C,

Pe=pn-eR:d-+=r(+) separation for

qlv

As here battery remains

due

so,

dt

l,

or v= - t or v= 4eo ,L 2ro 4"0

or v=

0.01

Strip Pp forms a capacitance C,. Ship QR forms a capacitance Cr.

v: vo- vp

Lûz = ov

0.01

be dx.

Potential difference between points O and P = V

"'

xKxIxY

eo

z

2nR?,,

4neo

.

, alO.

* no:+ fr

e,, x I x (l-y) =----6If---

=ft u*tr-rlrl

The ring is located in the y-z plane. Particle of mass zr and charge +q is projected from point. P(R\6,0,0) is projected on the positive.r-axis directly towards the origin O. Total charge on ring = (2nR))u Potential due to a ring at a

at any instant,

will

Energy stored in system = U

v(x): 4rEEo .t

will

capacitor

... C = C1+ or C= (0.4 x'l0a)+(l-8 x 10") ......(vi) or .C = 2.2 x t1-e F

69.

: C{ the charge

q

* 0.02

q.

n - Kreo(b dx)t - KreoA(&) -tdQ-n aQ-i Kzto(b-dx)l

c,"xdxd ... 1 =l*

co cr

I c2

-

KrerA(dx)

(;.bt : A\

Electrostatics

of-

225

1 tiu --,x\

ai*$=|Ai o, 1a =2onoRz or o!: lnonz z-', 2onoRz =

Co KpoA(&) KreoA (d*)

t

or

xd

d lt-*

Co -:-l

New charge density

e6A(&)

taAKrKz or c^ = " -{,Kzl + (& - Kr)xl'.14*1

13,

the

d(Kt- K)

:.06 x

+ Krt, K1t) - inKrt)

lnKt

or

K2

.....(i)

After connection: The charges Q, and Q are reilisfiibuted

till

attain the same potential Z.

and

Qi

V,

"'# #:+eo'oi=+ + 8i

.....(v)

qr:90 x l0{C qz=210 x l0{C qr= 150 x l0{ C (ii) Initidl snd final eleQtrostatic Initlai enerry = UA+

ot LI. =

or Q=20noR2

g

Lqr*\qt+3ez=0

or ur :

orQ=l6nR2+4noR2

Qi

......(iii)

qo

By solving (iii), (iv) and (v), we get

.

.'.Totalcharge= Q+Qz

Letthenewcharges be

lo{ * qlr*

ql _ 4, ,* 4z ,=o 3xl0*= 2xl0- 2xl0*

"' Charge on smaller sphere = o[4mnl

Totalcharge=

....(iD

C

For plates 4 and 5, ......(iv) 3.6 x 104 = gr+ 4, Kirchof;Ps second law, applied to the clobed loop yields

Total charge on the system oftwo spheres remaihs oonstent before and after connection, Before connection: Charge on bigger sphere : cl4n(2R)'11

Vr=

l0{

Let final

For platca 2 and3,

(&- K)

eoAKtKz

qB* (2 x l0{X180) = 3.6 x

(5,6j

'r

:'

....(i)

C

comp6fi6ntr, Final charge = Initial chsrg€ Letthe plates ofcapacitors be design*ed as (1,2), (3, 4) and

ioAK#:2 or C- d(Kt-K)' nnK,/-lnr^n t

4noR2

l0{

differcnt

d(&- K)-

Q, =

=3x

charge is conscwed when shared between

or c = #5*g(Krt

:.

:, qr= Cnv^ 6r qr= (3 x l0{X100)

xds;.

d

l6nRz

6

= q^

The

+ (t
Qt:

5o

chargBs on capacitors be qp e,

a* ra -esAKyK2| ... Jr-oa IrXrt+1yr_ yry_

72.

for bigger sphere

:. qr=C,V,

capacitor.

=

Initial charge onA

or (i)

Th6 iiiiEgiattbn of Co between ihe limits.x :0 to x = I gives the capacliriiice Of ihE wedges-filled

C

(ii)

Initial charge on B = q,

resulant capabitatrcb 6f the strip oonsidbrEd;

or

=

# 4}naRz . " v 3x4";4F

(82 + x,t4 - $zx}

Co denotes

......

3

Kz)

\

aoA(dx\L

ld Co

+-l'l

...'. (vi)

.....(vii) ..... (viii) energies:

(JB

*

I"^rrur' f;cugrr)' l,10{; x Gool2 +

I,ft

f;"

12

or

U1= 1.5 x l0-2 + 1.0 x 3.24 x l0-2

ay

U; + 4,,74 x

the two sirheres

Finatrxwq/*|

t}a

I

* 1s{; * ltso;2

....(ix)

i

G 4 2cn*!?9c cA*!'d

lo-6)2*_____---_______ 1x(2to* 1o{)2 _ or f./^:'ltxlox ; ' z, (3 x lo*) 2 x (2 x 1o-")

+ -A ''

2x(lxl0-)

226

8l x r0-4 - .11L:q *225x104 ,; _ --T7! or uz: zx2. 24

rr _ q3" -(QR\' '.^ (/@2c-\;I

rcz* .ûf1-=.-* U.:10 urLtzl or

" or U^: or LJ^: v@ _ Yo2R2 r2rz*4neoR "@

ttzz

*erl

to-4 x 2160 or [Jr= 1-8 x 10-2 J or (J.r: t# 74.

......(x)

When a conducting sphere S, is brought in with another conducting sphere ,Sr, a transfer of charge occurs from higher potential to lower potential till the two spheres attain equal

potential. Let S, contain charge p before with S;. On , let a charge qr shift frorn S, to Sr.

\a - il

=

k=*; $ Qr:t(^*) \ *n","

kl? - Q2- q)l _ tqz'

rR

RQ

-

Rq, + Rq.t

' v*: *l'tF;l '"0

At

x:

rH =

:

,.:nl*-,-[#)'*..

......(i)

6A

....'(ii)

.rE

P (m, q)

Gain of electrostatic

+(*.J']

or

k----!-+4

VH)

msH:ry-*1,-17*F.n)

or H=2a+2H -2^[7+II'z ZJa'+ H2 =2a+ H

whererr, is common ratio.

R_)'l

l

or

4(a2

+ H')

or

4H2

:

: 4o' a 112 + 4aH

H2 + 4aH

o' n =!

after n such s = U,

22

,, _q; _ s; ". un-n-a$n%N

.......(iii)

(b)

I . [c{{, - f-i-]'}l' * \R+ri 4neoR 2

or u, = 1

- Hl *l'p;77 '"0

or msH:n^1, - l77E.nf

in G. P.,

of$

H,

or mgH : qLV or mgH:q(Yo-

On nd',

rln.'l

-,1

potential energy.

1

aRl,ot Qn:;L'

*: #

Potential due to charged disc = Y,

conserved. Lo ss of gravitational endrgy

^' p+rJ

R_,( o_)'*r_i_)'l ulV + , * [a *.J ' [^ *../ ar:_^[

Let the energy

surface charge density o where

ofdisc, energy is

On third ,

a( ---rnnJ -'' " (l_{)

z(4ne)r2

dropped along the axis

,^* or a^--r[r-l't ttt42 t/l(R+"/ . r-i-)'-l (R.r/.1

,s. =

Q'a -

""o When the particle is

-l

q"(r + R): gl

2x(4neo)x(R)

(a) Value of H. The non-conducting disc of radius a has positive

Ll: 'o

or rqz:RQ - anr*^g(-!-\

or

'

At .r : 0, at centre of disc,

On second . similarlY,

or rqz:

.

Lt l

Jl

(b) I-irniting value of energy as ,? -) Let us calculate q, when r tends to co"

@.

U-H $aph and equilibrium position. At height 11,

Ur= QVr+ mgH

or u, : +:;1,t7;7F

- ul" *gu

At equilibrium posifion,'494 = g ciH

For G.P.,

-

S* =

;-= l-rt

where.r, = common ratio

t--l-l

eR eR .' 4o= f;;l | o' Q*: Lr "'L ' - _i_ R+rl

enw+ Hr -t)**s or o = z,sl --:L --- - t] * *g Lla' + H" J

':;;[(!

.......(iv)

Electrostatics

or or

.

:

---2H

227

- 2+l=0

,la' + H" 4H2

+=l a" + H"

o, n=

2k x (t x

_

4 ./3

l0{)

x (0.1 x

t0{)cos

B

'J x-+-

.....(v)

where ft =

-l4r1eo

2

Accordinglo equation (iv),

x uH =

#;lJilF_Hl+mgn

,.:)Q,.:)'"

'r

or (/, = 2*sU7 + H, - nf

+ mgn

The equation represents a parabola.

AtH:0,

Uo=2mga

U

z -15

2

2

\6

or r= t.l-F lz

:

U-in =

fr, r-

4al3

"tequilibrium,

Let the particle be, at some instant, at a point P distant x from

the origin. Of the four charges placed alonglaxis, the like charges repel each other while the unlike charges attract each other. At some value

ofr,

the attracfive and repulsive forces

balance each other. Therefore the attractive forceswill move the charge towards origin.

Net force towards origin = 2 ^F' cos p Net force away from origin = 2 Fcos a For net force to become zero,

v

-f 8 or v =2xex,o'*,o*li_ or

V:2.7 x 104 volt

(4 displaced)

*

oosa:

r0{)

(0.1 x

or

ufr=g

-r+I-,

t., n

l"*t

l**;

2k x 19 x 10-6) x (0.1 x l0-6)cos a

T

['-*;J

,

I

il .....(i0

1

x (2.7 x to4y =

}rc

x l0-4) v; :

....(iii)

(9 x loe)x(g x 10-6)

2x19xl0e)x(1x10-6)

if -9] :,r,.,r['nrnt'] ', ;, or

oosp-l-

:2x

"ln

+0.lpC

x

r

I

v: i*3

or

. ,,

2Fcoscr=2F iosF

)

Workdone = kinetic enerry gained

-tpcJ+

tM

.

!

Kinetic energr at the origin: Let potential at origin be Vn

+81rc,-,lt

net

r $a" n_ 2x(9xtOe)x(gxt0{) Yx(9"10e)x(lxl0{) 527 -t

vo=3m/s

-\ra,-J+

, tle

t:

.8$c,lTI o

to r =

(i)

force on it becomes zero. Let electric potential ut

!3 mga.

The figure representsthe graph.

76.

.......

When the particle travels from infinity

al\3 AtH

m

vo=2xexro'" ro*fs

4F*

V07r2,4 x..101.-vo!t; Difference of potential = 2.7 x 104 - 2.4 x 104 or Potential differenee = 0.3 x lOa .'. Potentialenerry=(0.1 x 10{) x (0.3 x lOa) or Potential energy :3 x l0{ J. .'. K.E. = 3 x .'. Minimum velocit5r vo = 3 69t Kinetic eners/ at origin = 3 x 10r J. or,,; .

l0r

J

228

77.

Potential enerry between two charges

=

4_2

ffi

tq.

ln the present case, the charge-pairs arc

E lF.+

8^ - no L2--8x7 1V2 - zo Of these 28 pair3, we haver/

4{o

(i) 12 pairs of dissirpiiar charges placed at a mutual ofa. iiii t-r i p.i* of similar charges placed at a munral separation of .J2a. (iii) 4 pairs of dissimilar charges placed at a mutual separdion of separation ''

y-d or... p-

80

Workdone by electric field.

^!-u.

For (i) above, P.E. = For (ii) above.

t2(il{.-q) 4nssa

-

For (iii) above, P.E. =

(r

or ll =

x 0.707

sYstem =

"12

U

.'. Volume of liquid in bubble = 4ra2t Volume ofthe droplet =

izl 5.824{4

\+nsoa) -

|

.'. Workdonetoseparatethecharges =

5.8241i

2\

.'.

__r-_l

*i

\4rena. oa)

q

where q denotes charge on the bubble.

Charge is conserved in this transformation

Potential of &oPlet

"ni.ry distance dfrom origin,

= o-(-F.a.) or AU =t,il(ffii

orpotentiar

=

ffi

4rc,âV 1 =Wrdrp

.'. Potentialofdropla

OD

ffi

tl.

(b)

Force on charge (+ P) : Intensity of field due to diPole

E= |

4ttso

I

is

= ,l+)

tue , you all

convention. Statement . 2 is tnte. You theoreni.

f

24i

o\"' nl.;J,ffi'1 -.(

fiiii::ii:**] \[

are familiar with this

alllknowthis, recali the shell

But statement - 2 is not th6 correct exPlanation of

d'

Force on charge (+

('.

Due to,Sr,

=

(b) : Statement -

=

, tt'

fu.n 4neod'

.'. Kinetic ener$/ atthe point=

Dueto,S2,

3

;. V'= : 4fiEoA

The free dipole, far au'ay from the origin, tavGls to$'ards origin, where acharge (+ Q)is fixed. Therewill beadecreOse inthe elecfiostatic polenlial energ/.-o-fth9 dipgte. Thiswill be' oittt" aip6ii. wiien *C aipolb is'at a equal to kinetic

7t" (a)

.:.

1tP = 4tw2t

& = 13az tyll3 ............,r..,.,-..r........'. (i) Let the potential of liquid-bubble = I/

\4neoa)

:

I*t 3

or,

;. w = r.rro(-t)r.

or AU

cz)a

eo

and liquid in the outside surface. When the bubble collapses into a droplet, the volume of liquid remains same in the transformation. La thickness ofiiquid in surfaee = f'

\ - - ,.rro(-o' \Uwoa)

'. Bindingenergy ofsystem =

û

qo(or

cos45o

A liquid-bubble contains air inside

W-=- M' x0.577 G;ffi=-@, ^ '.''

Toal Potential eners/ of

.'.

.'. Work done = (qoE) * a

=-12q2 4nera

p.E.: l=z(q\!\ = !zq' arsi{za 4*eoa

f

62

E=--: 'eo o^ E-. -s0

Q\ = QE

_l2N; - 4*oT,

statement -1.

(a) : Electric field at a distance R from the ceitre

ofthe nuoleus is,

:. E= Er- E,

-lZe

E=--.

4neo

Pz'

which is independent of a.

Eec&ll

t,. 'i

,.

Z2g/

.

:

l. :

81,

::-''

'

Q='lpr4wzar Now

*

'

throughout the rrcltmre of nucleus.

.

*=U5,*d'"fi$t*

lhis will 'bc'possiblc only when: ,s=8.:..

,

'

at

w9 uv2.& =!l eo6 r[r"*l] - rc fro*t1-. uF=q;F + "= *ZyffiJ

E4lz

iR'

Siace

3?2 sN o=#. A=ze=-T,

't

s4. (c) !,Elcciric field inside a

t_ _ Eaz=iEa

Giv€n)

uniformly dileFdsph€r€ is given aY'

E=* J30

For f,

*

ri

p should be

eonsteirt

,

t5. By Gausstheorcrn, fE'as-=fty

9,,,=,,!jcnrrl

. o=l!r*-n4*zyr=\dRt 3 .j

:

(b) : Ifa = 0, theltoat ctrarge on the'nuclcus'is given by

'

;'t#+1"i;r1 * ai2

-2o+,=s

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