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The degree of el.ecuificatfon of a body is represented by 1the el.ecr.rosir.uic potendal of the
clurged. body. The direction of Aow of charge betw"een cwo charged bodies placed in
coni:ac[ with each other is determined by decrrosrnric porenriaJ.
ELECTROSTATIC
POTENTIAL
AND
CAPACITANCE
ITOPJ,c 11
Electrostatic Potential, Electrostatic
Potential Difference and Electrostatic
Potential Energy
The: dc:ctmsratic potcncial at any point in the n:gion of dcctric fidd is c:qu.al to the:
amount of work done in bringing a unit positive test charge (with.om accdc:ration)
&om inlinity to that point.
Work done (W)
Electrostatic potcntia] (V) = - - - - - -
Charge (q 0
]t is a SC1.lar quantity. Its $] unit is volt (V} and ]V = l J/C and its dimensional
formula is (ML
2 T- A-I ).
Note
Electrostatic potential (V'l at a poinl is said to be OJ1e vo'lt, when one joule o wori<; is done in
moving one coulomb ol positive charge (withoot acaeleralion) lrom nfinity to that point
Work done ([W-] ..,.,) by an cxtema] furcc: in bringing (without accdcration) a
u.nit positive charge from infinity to a point is c:qual to the potcntia] (V) at that
point,
CHAPTER OHEOKLIST
- (^) Electrostatic Potential,
Electrostatic Potential
Difference and EJcctrostatic
PotcntiaJ E Crg)'
- (^) Didcctric and Capacitance
- e.
V = (W,,J.,rt=-[W..]. .1a:
q
o
o
where:, ( W - l e i. -. : is the work done: by the: dcctric fi d d
o n a chargc:d particle as that particle moves from infinity
to a point. A potc:mial (V) can be: positive, negative: o:r
zcm depending on the signs and magnitudes of q and W -
Note Electric potential is state dependent llunclion as electrosla1ic
forces a:re conSe1Va!iw forres. No work is crone in mo'iling a 1 .mit
positive test charge OVE:r a closed path in an eleclric field.
EXAMPLE I l I Potential at a point Pin space is given as
3 x 10 s V. Find the work done in bringing a charge of
2 x rn"""'c from infinity to the point P. Does the answer
depend on the path along which the chi'I.Ige is brought?
Sol Given,
Potential at the point P,
V = 3 X 11 f v, charge, lfo = 2 X 10--,;C
Work done in bringing l:he d1.arge from infinil:y to the
point Pis
W = q V = 2X10..,; X 3 X H r '
= 6 X 10 -
1
= Cl.6 J
No. the work done will be path independent.
ELECTROSTATIC POTE
TIAL DIFFERENCE
Electrostatic potential difference betwc:c:n two points P and
Q of a c h a w configmation consisting of charges
q , q q , q , and q is cqua] to the work done: by an c-xtemaJ
1 2 4
furc;c: in moving a unit positive test cha:rgc: again.st the:
dcct.rostatic force from point Q to P along any path
between these: two p o i n. Figure: shows that work done on
a test charge q
0
by the dcct:rostatic fl d d dl!le to any givrn
charge: c.onfiguratio n depends on1y on the po-sition of initial
point Q and pm,ition of final point P. Wo r k done ls
independent of the path ch0-srn. in going from Q to P.
6C/,
Electrostatic potential differenoe between two
poiril:S P and 0
If VQ ::md V p arc the dcctmstatic potcmfals at Q and P
rcspcctivdy, then dcct.rosta.tic potc:ntial differc:ncc: bc:tl',c:c:n
points Q 3Jnd P is
The: dimensional fo:rm ula fur electrostatic potc:ntial
differenGc: is given by
i\V = WQP = [ML2T-2]=(ML2T- 3 A- - I]
q 0
(AT]
The: SI unit of electrostatic potmtial diffcrmcc: is volt.
1V = ] JC - l= 1 - me-•
T h u s , dcct:rostatic potcntia] diffc:rc:nce bc:twccn any two
points in 3Jn dc:ctro-st3Jtic li d d is said to be: one volt, whc:n
one joule of work is done: by an c-xtcrnal force in moving a
positive charge: of one coulomb from one: point to the otl:-ic:r
aga:imt the dcct.rostatic force: of fidd without any
acccle.ration.
Note One elecl:ro:n•volt [1 el/) is the energy equal to the wmk
required to ma.ve a single elmrnmtary dlarge e such as an electron
or lhe proton through a pol€111ial dilterenoed exactly one volt {1V).
l e V- e (1\1)•(1.60 x 10".
1 gq (1J/C).. 1.60x ur
1 J
EXAMPLE 121 The potential difference between two
points is 20 V. How much work will be done in carrying a
charge of 400 μC: from one point to the another?
Sol Given & V = 20 V and q = 400 μC =
4CIOx10- C We know that
Electrostah
, c po
. ten
. t
" i
al di H
n:
\V
ork
d
one
Oharge
&V=W
q
w
2 0 = - - - -
400 X 10
W = 2 0 X 4 0 0 X 1 0 : - = 8 X 1 0 - J
EXAMPLE 13 I If 100 J of work mu.st be done to move
an electric charge of magnitude 4 C from a place A.,
where potential is -10 V to another place B where
potential is V volt. Find the Vctlue of V.
Sol Given, WAB=lOOJ, q= 4 C, VA=-rnV. V B = V =?
Since. W AB = q(V.ll - VA)
1 0 0 = 4 ( V H O ) V=15V
ELECTROSTATIC POTENTIAL
DUE TO A POINT CHARGE
Let P be the point at a distance r from the origin O 3Jt which
the dc:ctric potcnt.ia1 due to charge: + q is required.
+q: +1C
- ----·········•·····---·····-·--
0 p B A
le
Thus,
I , - • • · · l · - - · · · I
j
V
ELECTROSTATIC POTENTIAL
DUE TO A SYSTEM OF CHARGES
Let there be a number of point charges 1,, 1
2
,..., 1. at
distances r p r 2
t r 3 1
... 1 rn respectively from the point P,
where electric potential is to be calculated.
Potential at P due to charge 1,,
l .:!J....
,-
41tEo r 1
p
3x
q,
Similarly,
Q.,
A system
ofcharges
V 2 = - l!
f.2.,
41tE 0
r2P
V = - 1
:f.!!_
n 41t£0 rnP
Sol
Important Results
2 3 1 1 2 3
tf lj , r , r ,... • , areposition vectorsof thechargesQi ,q , q ,... ,
q.,respectivety,^ then^ electrostatic^ potential^ at^ point^ P
whose position vector isr 0
, would be
V = - 1 - f q
4 r t £ o ;. 1 lro-r;I
tf we have to calculate electricpotential due to a
continuous charge distribution characterised by volume
charge density
p(r). we divkle theentire volume intoa large number of
small
volume elementseach of volume i:>.V.
Chargeon eachelement:: pi:>.V. for a uniformly charged conducting spherical shell, the
electric field outside the shell is as, if the entire charge is
concentrated at the centre. Thus, the potential outside the
shell isgiven by
V = -
1
- - (-.-, RI
4m;, r
where, q is the total charge on the shell and R is itsradius.
The electric field inside the shell is zero. This implies that
potential is constant inside the shell {asnowork isdone in
movinga charge inside the shell) and therefore equal to its
value at the surface, which is
V = - 1 !!_
4n"o R
Using superposition principle, we obtain resultant potentiaJ
at point P due to total charge configuration as the algebraic
sum of the potentials due to individual charges.
V =V, +V
2
+V
3
+···+ V.
V = - 1 - .2.J...+_ 1! f. 2. + _ 1 _.
21...+41tE··· 0 r 1 p^ 41t£ 0 r2P^ 41tE^ O r
3
p
I
41tEo r,p
The net electrostatic potential at a point due to multiple
charges is equal to the algebraic sum of the potentials due to
individual charges at that particular point.
Mathematically, it is expressed as
v n e r = i. v i
i = I
EXAMPLE I 6 I Two point charges of 4 μC and- 2 μC
are separated bya distance of 1 rn in air. Find the location
of a point on the line joining the two charges, where
the
electric Sol Let potential the electrostatic is zero. potential be zero at point P
between
the two charges separated bya distance x metre.
p 4μC - 2μC
At point P,
Vp = V, + V 2 = O
_ t 11. + - '-· q , = O
4tt.o 'i 4 tto
' I 4 X t o - • I (- 2 X l 0 -
6 ) => - - · - - - + - - - ' - - - - - 0
41Uo 0
x 41t£ 0
( 1 - x )
4 x 1 0 - •
X
2 x 1 0 - •
(I - X)
x (1- x)
2(1-X)= X
2 =3x
or x = -
:. Electrostatic potential is zero at a
distance 2/3 m from
charge 4μC between the two charges.
EXAMPLE 17 I A charge Q is distributed over two
concentric hollow spheres of radii r and R (>r) such that
the surface densities are equal. Find the potential at the
common centre.
Sol. Let q, andq
2
be the charges on them.
0 1
= 0 2
q, -
q , 4nr2-
4rrR
2 qi
r
q, = R'
i.e. charge on them is distributed
in above ratio
? R'
or q , = - - Q a n d q 2
= - - Q
,2+R 2
,2+R
Potential at V = cent Potential re due to q, + Potential due to
V q, - I _q,+
I __ q , __
Q(R+r) 411 .£ 0
r 411£
0
R
4tt 0
(r
2 +R
2 )
EXAMPLE I 8 I Twospherical metal shells with different
radii r and R are far apart and connected by a thin
conducting wire. A charge Q is placed on one of them.
The charge redistributes so that same is on each sphere.
How much charge is on the sphere with radius r?
Wire
Sol. The electrical potential of a spherical shell with charge q
and radius r is kq/r, where k =1/(411£
0
Since, the shells are joinedby a conductor the charge will
distribute between them so that they attain the same
electrical potential.
Let the charge on the sphere with radius r be q, and that
on the another sphere q •· Then, equating the potentials
gives q, Ir=q, /R.
q, = q, (r/R) ...(i)
·: The total charge equals the original charge.
Q=q, +q, = q , = Q - q ,
By F.q. (i), q, =(Q - q, Xrl R)
Solving for q, gives q,(I + r/R) = Q(r!R)
=> q,
=Qr/(R+r)
which is the required charge.
Note The value of k is nol needed, staling proportionally is
sulficie<il.
ELECTROSTATIC POTENTIAL
DUE TO AN ELECTRIC
DIPOLE
Let us consider an electric dipole consisting of charges+ q
and - q separated by a distance 2a.
The dipole moment Ip I = q X 2a.
-
B
-
(
·
+
·
q
1
)
·· - ··
'
2
..•.
.
- - ; -
:·
,.,p
p ...
.
a r ,.. /
8 I
,,,,,'
2a
O ,
,,,
,/
,,,,..,,
, '
.
A ( - q )
Electric potential at point P due to electricdipole
Let O be the centre of the dipole, P be any point near the
electric dipole inclined at an angle8 as shown in the figure.
Let P be the point at which e l = r i c potential is required.
- q
Potential at P due to - q charge, V 1
4n£ r 0 1
Potential at P due to+ q charge, V 2
= q
(^4) 1t£or
As, potential is related to work done by the field,
electrostatic potential also follows the superposition
principle. Therefore, potential at P due to the dipole,
Vp = V
1
+ V
2
4 1 t £ 0 '
Now, by geometry, ,,
,, = r + a + 2ar cosa
Similarly, rl = r
2
2
- 2 a r c o s ( 1 8 0 ° - 8)
or rl = r
_2
2
- 2 a rc o s 8 [ · : c o s ( 1 8 0 ° - 8 ) = - c os 8 ]
a n d
EQUIPOTE TIAL SURFACES
A.Dy surface: which has same dc:ctrrnstatic potential at c:vc:.ry point,
on it is called. an cquipotc:ntial surface:. For a s1nglc: charge: q, the:
potc:nt:J
a
I
s gi
vc:n
b y
V
]
q
Th ,
i, s
tn
d' tcatcs
th at
V
ts a
4,re 0
r
constant, if r is constant. Thus , the: c:quipotc:.ntia1 surface: for
single point charge arc: spherical surfac.cs cc:ntrc:d at the:
d1.a.rgc:.Thc: c:quipotc:ntial rnrfuccs can be drawn th.rough any
rc:gion in which thc:rc: is electric field. lf all the: points at same:
potential in the elc:ctric fi d d arc joinc:d, then an c:qu:ipotcntia]
surface is obtained..
The: shape of equipotential surface dl!lc: to a
(i) line charge is cylindrica] (ii) point charge is spherical.
D ilfcrcnt properties of c:quipotc:ntial surfaces arc: given as below:
(i) Equipotc:n.tial sl!lrfuccs do not intcrsc:ct each o thc:r as it
gives two dirrctiom of dcct:r1c fi d d at intc:rsccting point
which is not possi blc:.
Oi) Equipotc:ntial surfac.cs arc d o s d y spac.c:d in the: rc:gion of
strong c:lc:ctric field and widdy spaGc:d in the: region of
weak dc:ctric field.
(iii) For any charge: configuration, c:quipotential surface:
thmugh a point is normal to the: dc:ctric fi d d at that point
and d:irc:ctc:d from one: equipotential surfucc: at highc:r
potential to the: othc:r equipotential surface at lowc:r
potential.
(iv) o work is rc:quirc:d to move: a test charge
on an c:quipmc:n tial surfaGc:.
(v) For a uniform dc:ctric fic:ld E, lc:t along X-axis, the
c:quipotc:ntial surfaces arc: normal to the X-axis, i.e. pl=c:s
para.lid to the JZ-planc.
Equipotential Surface in Different Ca e
Case I The equipotential swfaces produced by a point
charge or a spherically sym men-real charge di!
mibm:ion is a famjly of concentric spheres as
shown below in die figure.
' j 1
I I '
Equrpolerntial
srnaces
Cast1 JI The eq uiporemial surlaces for a uniform
electric field. are as shown below in figure
by cl.otmd Ii nes.
Equipotential surtaces
Equipotential surfaces for a unifo,m elect 11 ic
tiield
Case III The eq uiporemial surlaces due to cwo
idemrcal positive c h a r s a r e a. shown
below
Equipotential surfaces due to two positive charges
Cast1 N The eq uiporemial surfaces for an el.ectric
rupole are as shown below in die figure
by cl.oned lines.
E(J.llpoU!fllial surfaces
...... ,/
Equipotential surfaces due to an,elec,tric dipole
Electric fidd is always pc:rpc:nd.icular to an
c:quipotcntia1 surfac.c: and as a rcsu.1t, work done in
moving a charge: bc:twcc:n two points on an
c:quipotcntia1 surfac.c: is zero.
Note This topic has been trequentty asked in pre1,11ous
yews 2014 , 2013 , 201 1, 2010.
EXAMPLE I 101 Turo charges 2μC and - 2 lll'C are
placed at points A and B, 5 cm apart. Depict an
eqnri.potential snrface of the system. Del!hi 2013
Sol. Eqruipotenti.al surface means the surface where
potential remains same at each point
Here, this is the system of two equal and oppo.site
charges.
2 c
A
B
C
Equipore11,1Jial surfaces fen a point
dl'ilarg,e
The potential at C (m.id-point of AB),
V - 1 ( < 1 1 + 11._)
4111::o r, 1'"
l [ .2x 10 - • { - 2 x 1 0 -
= 411£ 2 5 X 1 0 - i +. 2 5 x t o -
2
I=
O
Thllls, potenti.aJ is zero at each point on the line which
piasses through the 1nid-poi11t of AB and perpendicular
to it So. a plane piassing through the mid-point C of AB
is an equipotential surface.
'
- 5 cm 1 2.5 mc
A C B
Relation between Electric Field and
Electric Potential
Let l!lS comidc:r a positive test d:i.argc: (q 0
) moves a di.stance:
(ds) from one cqlllipotc:ntial surface to anothc:r.Thc:
displacement (ds) makes an angle:(0) with the di.rc:ction of
the dectric fic:ld (E).
upposc: a positive tc:.st charge (q O
) moves through a
ditfc.rentia1 displacement ds from one cqllipotc:ntial surface:
to the: adjacc:nt smfacc:.
\Ve know that the work done: by the dccrric fidd on the tc:st
charge: during its movement is - q 0
dV. We sc:c: rhat the:
work done: by rhc dc:ctri.c field may also be w r i m n as the:
sc.alar product ( % E. d s ) o r qr,E COIS'. eds.
'
. '
'^ ' ' '
I
' '
'
I
L
I
'I
\ ds \
0
I I
/! , y.
., s
'
.
.
I
'I
Two equipotential SI.Jrlac.!!S
Displacement of charge between
two equipotential s1.Jrfaoes
dV
Equating thc:sc: two c:xpres.si.o ns or the wo:rk yields
E G,o .f. l, ds
E c o s 8 = - -
ds
Since, E co..• 0 i.s the component of E in the direction of d.s,
therefo:re
a
v
as
E =--
s
whc:rc: Ex, E., and E • arc the: x, y and z..c;omponcnts of E
at any point, thc:n
"" .... ....
E =E. ,. i+ E , j+Ezk
a
v
av
av
E" = - ax > E y = - d)'> E z = -
a;
av- av - av
E = -
[
i h i + c ) y j+az"k
Fo r the: simple situation in which the: electric fi d d E 1s
uniform. [ £ = -
l
Nc:gativc sign shows that the direction of c:lectric field E in
the dirc:ction of decreasing potc:ntial.
Since:, AV i.s 1 1 ri v e , th.en t'l V = - Ill VI
We can rewrite this c:quation as given bdow
Further, the: magnitude of an d c c t r k fidd is givc:n by
change: in magnirndc: of potcnti.a] per l!lnit displacement
normal to th.e c:qu:ipotcntial surface: at the: point. This is
callc:d potential gradient, i.e.
IEI=-
ld
ds
VI
=- (Potential gradient)
We thus arrive at two important condusions con.cc:rning
the rdation bc:tlvc:c:n dc:ctric field and potc:ntial which. a.re as
given below
(i) El.ectric fidd is in the direction in which th.c:
potentia1 dc:cn ascs stccpcst.
(ii) Its magnirudc is given by the change in the
magnitude of potc:ntial pc:r un.it displ.acc:rnc:nt normal
to the cquipotcntia1 surface at the point.
EXAMPLE I 11 I A small p.articile carrying a negative
charge of 1 .6 rn-
1 g
1
Cissl!lspendedin equrilibrium between
the horiwntal metal plates 5 cm apart, having a potential
difference of 3000 V across tltem. Find the mass of the
particle.
S'al Here q = - l. 6 x 10 -
1 C,
dr=S cm= 5 X1O-
2 rn
dV = 3000V
E = , - Q V = - 3 0 0 0 = - 6 X 1 0 4 V m - 1
di'" 5 X 10 -
2
al:ld
EXAMPLE 1141 In a hydrogen atom, the electron and
proton are bol!lnd at a distance of about 0.53 A.
(i) Estimate the potential energy of the syste.m in eV,
taking the z.em of the potential energy at infinite
separation of the electron from proton..
(ii.) What is the minimum work requited to free the
electron, given that its kinetic: energy in the orbit is
half the magnitude of potential energy obtained in (i)?
(ill) What are the answers to (i) and (ii) above, if the zero of
potential energy is taken at l.06 A sepa1ation?
CHRT
Hi11h: The potential energy of any objcd at any pmnt i5 equal
to tht differenct in its polrnl:ial enc g y at infinily and at that
point. Wor.k done is equal to the total energy of the :s:y:skm.
Sol. Charge on electron, q = - 1 J'i x 1 o-
19 C and cbarge on
proton, q P = 1.6 x Hr
19 C
19 c 1.6 )
(10-
19 c p
(i) Potential energy of the system
= Polential energy at jnfmity
a distance of0.53 A
= 0 - - l q,,q,,
41t£9X10 0 r
9 X (-1. 6) X 10 ,-
19 X L6X 10 -
1
0.53 X 1 0 - "l
=--43. 47 X 10 -
19 J
43 .•i.7X 10 -
19
[·.· 1 eV = 1.6 X 10 -
19
J]
1.6 X 10 -
1
{ii) Th.e ki.net:ic energy=- !.x Potent:ial energy
2
1
= - - x (-27.16) = 13.58 eV
2
TotaJ energy= KE+ PE= 1 3 5 8 - .27. 16 = - 13.58 eV
Thus, the minimum work: done required to f:ree the
e1edron is 13.58 eV.
(iu) Potential energy at separation of 1. 06 A
1 ,q,,q,,
4ffico LOl'i X 10 - io
1 X 1.6 xrn -
1 9
1.06 xto-•
0
= - 21.73 X W -
1
J
21.73 X: 10 -
1
1.6 Xl0-
1 9'
Electrostatic Potential Energy of a
System ofThree Point Charges
Let us now consider a systc.m of duce point charges qt, q 2
and q
3
having position vectors r 1
, r 2
and r 3
, rcspcctivdy as
from origin.
,Q,
Three point charges system
To bri.ng q
1
first from infinity to pos:itionr 1
, n o w o r k is
rc:quirc.d because we bring cha.rgc: qt from in6.n.ity to a
particular location. where potcmia] is zero.
Thercfurc, lV
1
= 0
2 2
Thc work clone in b:ringin.g q from infinity to position r is
given by W 2
=q 2
Vt h)
l qtq
4«Eo
r
Charges q 1 and q 2 produce a potential which at any point P
is givrn by
Yi 2 = -
2
)
' 4n:e O 1"13 r
3
Work done in bringi.ngq from infinity to posiion 1'3 isq 3
times V 1 , 2 at 1'3,
W3 =q3 V 1 , 2 (r3)
l [q 1 q 3
= 4rceo q2'13 ) r13 + r
The total work donc in assembling thc charges at the
given location.s (eql!lal to the poten.t:ial energy of the
sy:stem) is obtained. by adding the work done in different
steps.
U = W
1
+lV
2
+W
3
U = O + - · - · q l q 2 + - · - ( q l q 3 + q2 q3 )
4rce 0 "Li
4:itE 0
r13, r
Thus, the potential energy of the system at 1.06 A
= PE at distance 1.06 A - PE at distance o.53 A
= - 13_58 - (- 27.16) = 13.SS eV
Thus, on shifiing the zero of potential energy, work
required to free electron remains same and it is equai.
to 13.58 eV.
Th i s :resu1t can also bc expressed in :sum.m.ati.on furm. as
[
(^3) 't'
3 ·q ,
l
q - ]
]
- - -
U L , ,. ,. -
4 11:eo i = t j = l r -
·
i"'-; 'l
Due to the: conservative: nature: of electrostatic furcc, the
value of U is indepc:ndc:m of the manner in which the
rnnfi.guration is assemblc:d.
If we write the: distance I r 1
I as r l j , theabove: equation
may be: expressed as for system of n point charges system.
U = -
[ (^) n n
I -
q, q
1 ]
4ttE (^) O ; =I j: rlj
El.cctmstatic potc:ntial c:nc:rgy of a .systc:m of N point charges
is equal to the: total amount of work done in assembling all
the charges at the: givc:n pmitiom frnm infinity.
U = [1: I
q,v.]
0
4ttE i = I 1 _'= 1 '
J
j ;t I
N I
qj
wh.crc, V , = I : - · -
1 j = I 4 , t j "'- / , o '"if
= Potential at r 1
due to all othc:r chargc:s
The: SI unit of dcctrmtatic potc:ntial cncrgy is joule: (J).
Another convenient unit of energy is electron volt (eV).
[ I eV = l.6 X rn,-I? X] V = 1.6 X 10-
1 9
§ ]
EXAMPLE 1151 Three charges (allq = lOC) are placed
at the ,edge of an equilateral triangle of side 2 m. Find the
net potential energy of the system.
o.1- Given, d1arge, q "'10 C (q 1 " ' q l "'q)
Each sid.e of equilateral triangle, r = 2 ,n
Potential energy (PE) "'?
Potential energy between two charges is given by
PE"' kq,ql [·: r"' distance between q 1
and.
ql]
r .•. J' of system wiU be three times the potential energy
between the h\•o charges as the equal charge is placed. al
the vertices of equilateral Mangle.
S o
..
P E
... - - - 3 X 9 ' X 1 0 xtoxto
r r
2
"'135 X 10 ,1 J
EXAMPLE I 16 I Three point charges q, 2q aJld ,8q are
to be placed on a 9 cm long straight line. Find the
positions where the charges should be placed such
that the potential energy of this system is minimum.
In this sit1!lation, what is the electric field at the
position of the charge q due to the other two charges?
o[ Consider l:he gi.ven situation as: shown in 6gme_
2 q 8q
or potential energy to be lllinimum the bigger charges
shou!l.d be farthest Let x be the distance of q from 2q_
Tihen potential energy of the system shown in figure
would be
u ; ; ; ; il(2q)(q)+ (8q)(q)+ (2q)(8q)Jl
X (9 - x) 9
Here, K
" ' _ l _
4 1 t
8 or Uto be lllinimum .! + - - shou!l.d be minilllum_
X 9 - : C
_!! [ +- 8 ]-
o
,dx x ·9-r
x - 2 ( 9 - xi
or
X 1
;;;
9 - X 2
r = 3 c m
Le. distance of charge q from 2q shouM be 3 cm_
:. E!ec,trk fieid. at q,
_ E =.
K ...
( ._.
2 ...
q ..
) ...
K """
( "
8 '"
q ""
) "'-'--
=O
(3 XHl-
2
EXAMPLE I 17 I If one f of(6 Xthe 10- l two)2 electrons of
H
2
molecule is removed, w,e get a hydrogen molecular ion
H;. In the ,gro1!lnd state of an H;, the two protons
are
separated by roughly 1.5 A.and the ,e.ectron is rol!lghly 1
A
from each proton. Determine the potential energy of the
system. Specify your ch□ioe of the zero of potential
energy. Cfflt.T
at let there are two protons p 1
and p 2
with an eiedron ,e.
,!il
,•,
A (^) ........._
f 3
, (^) ...... ,
,,/_
,
P 1
"'
L5A
P
Distance beh,•een two protons is given by
1
r .. 1- 5 A= l.S xto-• m
1
Distance between proton p and electron ,t is given by
,; ""I - txrn-• m
Distance between proton P; (lil]d eied:ron e is given by
1"::f"' 1
A
lXJO-iD m
Tlhe total potential energy of the sys,tem,
U = - i - _ - ,[q,,,q,,, + q, , q. + q,,_q•] (i)
-t:11' I I 'i Fz
r
Given. q,, "" q,, "'·u; x to-
19 C 1
and q,. o , - H i X
lO-a C
POTENTIAL ENERGY IN AN
EXTERNAL FIELD
A single: charge: or a system of charges possess dcctrosratic
potential energy in the presence of an external electric 6cld,
these arc discwscd as follows.
Potential Energy of a Single
Char< 6
e in E.xterna1 Field
Porcnrial energy of a single chargcq ar a point wirh position
vector r in an exrernal field = q • V (r), where V(r) is the
potential ar the point due to external decrric field E.
Potential Energy of a Syste1n of Two
Ch,uges in an Exte111al Field
For a system of two charges q
1
and q
1
, rhc potential
energy
1s g L v c n a s ' - - - - - - - - - - - - - -
where:, qp 'h = rwo point charges at position
vc:cmrs r
1
and r
2
- rcspectivdy
V (r 1
) = potential ar r 1
due to the external field
and V(r 2 ) = potential at r 2 due ro the
external field.
Potential Energy of a
Dipole in an E.-.."ternal Field
Consider a dipole with charges + q and - q placed in a
uniform external decnic fidd as shown in che figure. In a
uniform dcccric fidd, the dipole experiences no force, bur
cxpericnccs a torque T given by 't = p XE.This corquc will
tend to rotate the dipole:. Suppose: an C:X'[c:rnal torque: ' t a , i s
,;;;; pc. t-( ; O U J 9 ;
= p£(cos 1
)
Thework done W is stored as rhc potential energy of the
system. Therefore, the porcntial energy of the dipole placed
in external field E is given by
[ U(8) = p£{cos 1
) ]
J,. Particular
Coses
(i) When the dipole is initially aligned along the eleclric field,
i.e.9, =()>and we have to set it at angJe6 wilh E, i.e.
o,
=8. W = - p f( co s 0 - cos()>)
= - pE(cosa - I)
This work doneisstored inthe dipole inthe form
or potential energy.
(n) When the dipole is inttially at rightangle lo E, i.e.9,
= goo
and we have to set it at angle8 withE, i.e. 8 2
= 8.
W = - p £ ( c o s 8 - cos9QO)
= - p£cos
Potential energy of dipole, U = W = - pE cosO
U= - p • E
Obviously, potential energy of an electric dipole is a
scalar
quantity. ft ismeasured in jouJe.
Important Results
Some importantresults related to electric dipole are as
given
below
(i) Electric potential at any point on the bisector of
dipole is
zero.
(ii) A dipole experiences a net force in a non•uniform
electric
field.
(iii) A dipole experiences maximum torque at
theposit.ion where potential energy is zero.
EXAMPLE I 18 I An electric dipole of length 4 cm, when
placed with its axis making an angle of 60° witha uniform
electric field, experiences a torque of 4 3 N-m. Calculate
the potential energy of the dipole, if it hascharge± 8 nC.
Delhi 2014
Sol Given, length, 2a = 4 cm = 4x 10-
2 m
Angle, 0 = 60°
torque t = 4,.f,N-m
Charge, Q = 8 x l 0 - •c
We know that, t = Q (2a) E sin 0
Electric field, E = t
Q(2a) sin
N/C
4 3
8 x 10 - •x 4 x 10 -
2 x sin60°
:. Potential
energy, U = -
pE cos 0
=- Q (2a) E
cos 0
= - 8 X 10
X 4 X 10 -
2
X
r 4 3Xcos60°
7
lsx 10 -
9 X 4X 10 -
2 X sin 60°J
= T =- 4 J
EXAMPLE 1191 A point charge q is fixed at origin. A
dipole with a dipole moment p is placed along the X-axis
far away from the origin with p pointing along positive
X-axis. Find
(i) the kinetic energy of the dipole when it
reaches a distance d from the origin.
(ii) the force experienced by the charge q at this
moment.
Delh
i
Sol (i) Applying energy conservation principle, increase in
kinetic energy of the dipole = decrease in electrostatic
potential energy of the dipole.
Kinetic energyof dipole at distancedfrom origin
= U , - U
1
= 0 - (-p · E ) = p·E
(I q•)
qp
= (pi)· - - - i = - -
4Jt€o J2 4Jt£od
(ii) Electric field at origin due to the dipole,
E
= _ 1 _2p-,.
d'
[
·;
E
u;,
ff p·^1
Force on charge q.
F=qE = _. E i _ i
21u:()d
1
ITOPIC PRACTICE 1 I
OBJECTIVE Type Questions
I. Which of the following is not a unit of
electrostatic potential?
(a) Volt
(b) Joule/coulomb
(c) Newton/Coulomb
{d) Newton - metre/ Coulomb
2. Work d o n e b y a n external force in bringing a
unit positive charge from infinity to a
point is
{a) equal to the electrostatic potential ( V) at that
point
(b) equal to the negative of work done by
electrostatic forces
(c) Both (a) and (b)
{d) Neither (a) nor (b)
- To find the value of potential at a point, the
external force at every point of the path is to be
equal an d opposite to the
(a) work done
(b) electrostatic force on the test charge at that point
(c) Both (a) and (b)
{d) Neither (a) nor (b)
4. If electrostatic potential at the surface of a
sph e re of 5 c m radius is 50 V, then the potential
at the centre of sphere will be
{a) 10 V
(c) 250 V
(b) 50V
(d) zero
5. The electrostatic potential of a uniformly
charged thin spherical shell of charge Q and
radius R at a distance r from the centre is
{a) g for points outside and _ g
forpoints
411t 0 r 4!!t 0 R
inside the shell
(b) g for both points inside and outside
the shell
41t£ 0
(c) zero r for points outside and g for points inside
the
41t£ 0 r
shell
{d) zero for both points inside and outside the shell
22. Two uniformly large parallel thin plates having
charge densities+ CJ and - CJ are kept in the
XZ·plane at a distanced apart.Sketch an
equipotential surface due to electric field
between the plates. If a particle of mass m and
charge - q remains stationary between the
plates. What is the magnitude and direction of
this field? Delhi 2011
23. Find out the expression for the
potential energy of a system of three charges q'"
q
and q
located atr 1 , r 2 andr 3 with respect
to the common origin 0. Delhi 2010
24. Two point charges q, and q
2
are located at r
1
and
r
2
, respectively in an external electric
field E.
Obtain the expression for the total work done
in assembling this configuration. Delhi 2014 C
25. Adipole with its charges,- q and +q, located at
the points (0, - b, 0) and (0, +b, 0) is present in a
uniform electric field E. The equipotential
surfaces of this field are planes parallel to the
YZ-planes.
(i) What is the direction of the electric field
E?
(ii) How much torque would the
dipole experience in this field?
Delhi 2010
26. If a point charge +q is taken from A to
C and then from C to B, points A and B lying
on a circle drawn with another charge +q at its
centre, then along which path more work will
be done?
27. Do free electrons travel to region of
higher potential or lower potential?
NCERTExemplar
28. Prove that a closed equipotential
surface with no charge within itself, must
enclose an equipotential value.
LONG ANSWER Type 1 Questions
29. A cube of side b has a charge q
at each of its vertices. Determine the potential
and electric field due to this charge array at the
centre of the cube. NCERT
30. (i) Derive the expression for the electric
potential due to an electric dipole at a point
on its axial line.
(ii) Depict the equipotential surfaces due to an
electric dipole. Delhi 2017
31. Give the simplified expression for the
following and draw the graph for variation of
potential with distance.
(i) Electrostatic potential due to a point charge
q at a distance r from it.
(ii) General expression for electric
potential due to a dipole.
32. Given figure shows a charge array known as an
electric quadrupole. For a point on the axis of
the quadrupole, obtain the dependence of
potential on r for r/a » I and contrast your
results with that due to an electric dipole and
an electric monopole (i.e. a single charge).
A a B a
C
- p
q - q - q q
NCER
T
33. Define an equipotential surface.
Draw equipotential surfaces
(i) in case of a single point charge
(ii) in a constant electric field in 2-direction.
Why the equipotential surfaces about a
single charge are not equidistant?
(iii) Can electric field exist
tangential to an equipotential surface?
Give reason.
All
lndia 2
34. Three charges - q, + Q and- q are placed
at
equal distance on straight line. If the potential
energy of the system of the three charges is
zero, then what is the ratio of Q: q?
35. Four point charges Q,q, Q and q are placed at
the corners of a square of side a as shown in
figure.
Find the
(a) resultant electric force on a charge Q and
(b) potential energy of this system.
CBSE 2018
Or
(a) Three point charges q, - 4q and 2q are
placed at the vertices of an equilateral
triangle ABC of side I as shown in the figure.
Obtain the expression for the magnitude of
the resultant ele:crric force acting on the
charge q.
(b) Find out. the amount of the work done to
separate the charges at infinite distance.
CB E20.l8:
LO G N WER Type H Qm ticmr
6. hree concentric ·metal shells A, B and of
radius a, b nd c (a< b < c have surface charge
den ities +'Cl', - c and +er, respectively.
(i) Find the potential o.fthree shells at A, B
and _
,(ii) I the shells A and C a.re t the same
potential, obtain the relation between the
radii t.1, b and c.
37. Two metal spheres, on.e ofradius .R a:nd the
other ofradius2R, both have a.me surface charge
densityO'. They are brought in contact and
separated. What will be new surface charge
densities on them? NCERTEx:empfar
38. (a) U e Causs' law to derive the expre sion for
the electric field (E) due to a
straight uni.form[ y charged infinite line of
charge density i,. C/m.
(b) Draw a graph to how the valiation
of E with perpendi.cul r distance r om
the line af charge.
(c) Find the work done in bringing a charge q
r o m p rpendiculardislancer
1
tor
2
(r
2
> r
1
).
CBSE2.fU
NUMERICAL PROBLEMS
19. What is the work done in moving a2 μC point
,charge from comer A to ,corner Bof a square
ABCD, when a l.O μC charge exists at the centre
of the square?
2 μ C. A G J • B
1 0μC
D C
40. Theelectric potenli.. J at OJ ·rn from a point
charge is SO V. What is the magnitude and sign
of the charge? AJUmlia 2011
4l. Two charges 6 x 10-s. C a n d - 3 x r n -
1 1
Care
located 11. 6 cm apart. At what poi.nt s) on the line
jaining the U \ l o charges is the ele,otric potential
zem? Toke the potential at infinity to be zero.
CERT
42. A reguhu hexagon of side 10 cm has a charge
5μC at ,each o its vertice. Calculate the
potential at the centre ofthe hexagon.
3. A charge of 8 mC is located at the origin.
CaJculat,e me work done in taking a smaU
charge of - 2 x11.0-
9 c from a point P 0, 0, 3) (in
cm) to a point Q ,(0, 4, 0) (in cm), via a point
R (0, 6 , 9) (in cm). CEH.T
44. Thecircular arc is shown in the figure given
below, has a uniform charge pe unit length of
1 xl0-'
8
C/m.Find the potential at the cencre Oo
the arc.
Ll
o
2 m
4 ·S
.
A small partic]e carrying a negative charge of
LG x rn,-
19
C is :suspended in equilibrium
between the bori.mntaJ metal pl. tes 10 cm
apart, having a potential difference of 4000
across them, find the mass ofthe particle.
46. An infinite plane sbeet of charge density
io -
8
C/ m
2
is held i.n air. In this situation, how
far apart are two equ.ipotential surfaces whose
potential difference is 5 V?
7. A lest charge q is moved without acceleration.
from A to C along the path from A to B and then
from B to C i.n electric field E as shown in the
figure.
',
I
'•'•c E
{i) Calculate the potential difference between A
andC.
(ii) At which point (of the two) is the electric
potential more a.nd why? Alli India
- Eled:ric lield is always normal h:i the equipotential
imrface at every point, btl'cause no v,rnrk is
done, as
W' ;;;;;; q (V,A - VB)
v. - v.1 ;;o 1
hence W ;;U
If the field. we:r.e not t1orrna1 to tlhe equipotential s n r f a ,
it would have a non-zero co.mponent along the surface.
So, to move a test du.rge agairu:t this component a
work would have to be done.
- The electric potential energy at any point lying at a
distance ,. from the oonrce charge q js equal to
tihe amou1lt ofwor\k done in moving unil posihve
test charge from .infinity to 01.at point without any
accele.ration against electrostatic force.
J.6. Potential is :maximum al A. as potential decreases in tile
direction of field or we an say lhat V,. > V
11
"' Ve.
1 7. Rt"fer to graph on page 64.
J. 8. Refer ta tel([ on page 68 (case [).
19. Equipotential surfaces d.o no! intersed each olher as it
give h.vo directions of electric ·1eld.at. jn tersect.ing point
which is not possible.
20. As, the collection of charges 11t a great distance, so
it has sphe,kal equipotential surfaC€.
- (i) Equipoteu• al snrfaet"s of the system Cdipole.),
Equrpahmtial
sufacas
- q charge balances, when
qE " ' m g
E "'!;!!!
q
11t<" direction of eJectric field is along vertically
do1,1vnward direction.
Neta The XZ-plane is so chosen lhat Ille dirnction o! elE:c:tric lield
due to t' ND plallas is alcmg wrlically downward direction,
o1hervns.e Wllighl (mg) of ch.M!J'fil partide could not be
balanc1:1d.
The s\kelcb of equipotential imrface due !o electric; field
between Uie pfate!: is shown in figure below.
'f
E (1/efiically
dOWTJWard)
z
- Refer lo text an page 71.
24. Refer to ta::t an page 73.
- (i) 'l11e direction of e]ectrk 1eld is perpencli.C!ular to their
equipotential.surface.. So, the direction of e]ecl:ric fldd
is aJoug X-axis as its length should be perpeudicu!ar
ta equiPQl:t"ntial snrface lying in YZ-plane.
(ii) le ng th of lhe dipol<" .. 2b
As. dipole's axis is along lihe Y-axis.
:. E.lec!.ric dipole moment, p - q( 2b f
and electric field, E "" E t
t"' p >< E "' q( 2b)j x E
"' + 2qbE (j x j)
"'2qbE - k )
Torque, 1 1:I-" 2qbE
26 ,. Consider 1'1e situation as shown in ligore.
(ii Eqnipoteiitial surfaces get closer Io €a.ch other near
lhe point charges as strong e]ect.ric fi.e]d is prodm:ed
there.
W o r k d o n e f o r l h e p a t h A C
(for a given equipotential surface]
w h e n , !:nu.I! .ii r represents strong e]eclrk field and
vice-vel"!;a.
22. Here, - q charge experiences f o r in a direction
opposite to the direction of electric field.
- - - - - - - - + a
qE
w= ...
+q(
c - .4)
Similarly, Wc 8
=+ q{V
8
Vi::l
VA'"' V
IW..ic I "'IWCIJI
dov
Y .n
ln w
ti a
c r
a d
l El -
q
} mg
--------
a
27. The free electrons experience electrostatic force in a
direction opposiite to tJ1e directfon of electric field, being
ofnega.Hve charge. The electric field is always directed
from higher potential fo lower potential.
Therefore, electrostatic force and hence, direction of
travelling of electrons is from lower potential ta the
region of higher potential.
HiHts: In this problem, we need to know tha:t the
e.lectl'ic field intensity E and electric potenlfal Va.re·
related as E .. --
dV
and the fi.eld lines a.re
always
dr
perpendicular from one equipotential surface
mamtained al high electrosta:Hc potential to ot!her
eg_nipotenitial snrface maintained at a low
electl'ost:a:tic potential.
3:0. (i) Refer ta text on page 67_
(ii) Refer to text on page 68 (Case N)
- Refer text on page 64 for the graph.
(i) Refer to text on pages 63 and 64_
(ii) Refer ta text on pages 66 and 6 7_
32. Gh•en, AC "' 2n, BP "' r
AP= r + ,a and PC "' r- .:i
A !i B a C
q q
p
Let's assume oontradicting statement that lhe potential
is not same inside lhe dosed equipolential surface. Lei
the potential. jnsl insiide the surface be different lo t:hal
on the surface having a potential gradient ( : : ) -
Consequently, electric field oomes inito existence, which
is given by, E "' - dV_
dr
Consequently, fleld lines point inwards or outwards
fron1 the surface_ These lines cannot be formed on the
surface, as the surface is equipotential.. H is possible only
when the other end of the field lines are originated from
the cha:rges inside. This contradicts the original
assumption_ Hence, the entire vohm1e inside nmst be
e,g_nipotenitiaL
Cons,ider a rube of side b and its centre be Q_ The cha:rge
q is placed at each of the corners_
Side of the cube "' b
Q - b - - ·
- - - , - - - - -
Length of the 111ain diagonal of the cube
;;;; ll + b
2
Distance of centl'e O from each of the vertices is
b./
r ; ; ; ; -
2
Potential at point O due lo one cha:rge, V "' -
1
4nE 0 r
Potential at point O due lo all.charges placed al the
vertices of the·rube,
V ' " ' 8V ..
8 X l X q _- 8 q _ x _
2 = - [from Eq_
(i)]
4ffE 0
r 41tE 0
4q
= .fi1t£ b 0
The electric field due to one vertex is balanced by the
electric fie.Id dne to the opposite \fertex because aH
charges are pos,ithre in nature_ Thu.s, the resnlta:nt
electl'ic fie.Id al the cerntre O of the cube is zero_
The potential at P is
V_
V .. Potenlfa[ at P due to A + Potential at P dne to B
= l [1 - !1 + _:! ]
411:1:: 0
AP BP CP
I [ 1 2 1
]
= 41r:£,i• q (r+ u - ) ; + (r -
u)
q lr(r - a ) - 2(r+ a ) ( r - a ) + r ( r + a)l
"'4SJCE 11 il r(r + a ) ( r - a) J
q rr
2
2
2
2
"'4m:..il r ( r 2 - .: i 2 ) J
q-2a2 q-
2112
a. - - - - , - - a. - - - - - - - - , - - 2 2
411:!;i r ( r - a )
41tf: 0 - r- r
2 ( 1 - - ;
, ;
:i
2 )
According to the g_nestion,
r (^) q- 2
2
a
ff - > > I., .:i< < r_ Therefore, V "' ......-- -. 1
4ffEa-r
v...-
1
rj
As, we know that e.lectric potential at a point on axial
line clue to an electric dipole is
V.,.2._
r
]n case of electric monopole, V .., !.._
r
Then, we condude that for larger r, the electric: potential
due to a quadrupole is inversely proportional to the cube
of the distance r, while due to an electric dipole, it is
inversely proportional ta the sg_na.re of rand i1werse.ly
proportional to he distance r for a monopole_
33. (i) Refer ta text on page 68_
(ii) Equipotential surfaces when the eiedric field is in
Z-d.irec,tion_
X
y
Epuipols11lial
suiaces
,J-----f- + - - + - - + 1 - ---- ,..z
E
