ALLEN
Electrostatics
4.
ELECTROSTATICS
1.
Two point charges q1( 10 mC) and q2(–25 mC)
are placed on the x-axis at x = l m and x = 4
m respectively. The electric field (in V/m) at
a point y = 3 m on y-axis is,
1
é
9
2 -2 ù
ê take 4 pe = 9 ´ 10 Nm C ú
0
ë
û
R
5
R
(3)
2
ˆ ´ 10 2
(2) (81iˆ - 81j)
(3) (63iˆ - 27ˆj) ´ 10 2
(2) R
(4) R 2
Charges –q and +q located at A and B,
respectively, constitute an electric dipole.
Distance AB = 2a, O is the mid point of the
dipole and OP is perpendicular to AB.
A charge Q is placed at P where OP = y and
y >> 2a. The charge Q experiences and
electrostatic force F. If Q is now moved along
LL
EN
ˆ ´ 10 2
(4) ( -81iˆ + 81j)
2.
For a uniformly charged ring of radius R, the
electric field on its axis has the largest
magnitude at a distance h from its centre. Then
value of h is :
(1)
5.
(1) ( -63iˆ + 27ˆj) ´ 10 2
Charge is distributed within a sphere of radius
A
r
R with a volume charge density r(r) = 2 e
-2r / a
y
the equatorial line to P' such that OP'= æç ö÷ ,
,
è3ø
æy
ö
the force on Q will be close to : ç >> 2a ÷
è3
ø
where A and a are constants. If Q is the total
charge of this charge distribution, the radius
R is :
P
a
Q ö
æ
(1) log ç 1 ÷
2
è 2 paA ø
æ
è
(2) a log ç 1 æ
Q P'
Q ö
÷
2 paA ø
A
ö
Q ÷
ç1÷
2 paA ø
è
E
(4)
3.
O
–q
B
+q
1
A
node06\B0B0-BA\Kota\JEE Main\Jee Main-2019_Subject Topic PDF With Solution\Physics\English\5-Electrostatics
(3) a log ç
1
a
1
æ
ö
log ç
Q ÷
2
ç1÷
è
2paA ø
(1)
6.
F
3
(2) 3F
(3) 9F
(4) 27F
Four equal point charges Q each are placed in
the xy plane at (0, 2), (4, 2), (4, –2) and
(0, –2). The work required to put a fifth charge
Q at the origin of the coordinate system will
Three charges +Q, q, + Q are placed
respectively, at distance, 0, d/2 and d from the
origin, on the x-axis. If the net force
experienced by + Q, placed at x = 0, Ls zero,
then value of q is :
be :
Q2
(1) 2 2pe
0
Q2 æ
1 ö
(2) 4 pe ç 1 +
÷
5ø
0 è
(1) +Q/2
(2) –Q/2
(3) –Q/4
(4) +Q/4
Q2 æ
1 ö
(3) 4 pe ç 1 +
÷
3ø
0 è
Q2
(4) 4pe
0
ALLEN
2
Electrostatics
7.
A charge Q is distributed over three concentric
spherical shells of radii a, b, c (a < b < c ) such
that their surface charge densities are equal to
one another. The total potential at a point at
distance r from their common centre, where
r < a, would be :
10.
The charges Q + q and +q are placed at the
vertices of a right-angle isosceles triangle as
shown below. The net electrostatic energy of
the configuration is zero, it the value of Q is:
Q
Q
(1) 4 pe (a + b + c)
0
+q
Q(a + b + c)
(2) 4 pe (a 2 + b 2 + c 2 )
0
(1)
Q ab + bc + ca
(3) 12pe
abc
0
- 2q
(2) –2q
2 +1
(3)
-q
(4) +q
1+ 2
The given graph shows variation (with distance
r from centre) of :
LL
EN
11.
+q
Q (a 2 + b2 + c 2 )
(4) 4 pe (a 3 + b3 + c 3 )
0
r0
Two electric dipoles, A, B with respective
r
r
dipole moments d A = -4qaiˆ and d B = -2qaiˆ
placed on the x-axis with a separation R, as
shown in the figure
R
A
B
(2) Potential of a uniformly charged spherical
shell
X
(4)
R
2 -1
A
(2)
R
2 +1
(1)
(3)
9.
2 +1
2R
2 -1
(3) Electric field of uniformly charged spherical
shell
(4) Electric field of uniformly charged sphere
12.
Determine the electric dipole moment of the
system of three charges, placed on the vertices
of an equilateral triangle, as shown in the
figure:
An electric field of 1000 V/m is applied to an
electric dipole at angle of 45°. The value of
electric dipole moment is 10–29 C.m. What is
the potential energy of the electric dipole ?
y
+q
(1) – 9 × 10–20 J
(2) – 7 × 10–27 J
(3) – 10 × 10–29 J
(4) – 20 × 10–18 J
r
(1) Potential of a uniformly charged sphere
The distance from A at which both of them
produce the same potential is :
2R
r0
(1) (ql)
ˆi + ˆj
2
(3) - 3q l ˆj
–2q
l
l
l
+q
x
(2)
3ql
(4) 2ql ˆj
ˆj - ˆi
2
node06\B0B0-BA\Kota\JEE Main\Jee Main-2019_Subject Topic PDF With Solution\Physics\English\5-Electrostatics
8.
E
ALLEN
13.
Electrostatics
There is a uniform spherically symmetric
surface charge density at a distance R0 from
the origin. The charge distribution is initially
at rest and starts expanding because of mutual
repulsion. The figure that represents best the
speed V(R(t)) of the distribution as a function
of its instantaneous radius R (t) is :
15.
3
A positive point charge is released from rest at
a distance r 0 from a positive line charge with
uniform density. The speed (v) of the point
charge, as a function of instantaneous distance
r from line charge, is proportional to :-
V(R(t))
r0
V0
(1)
(1) v µ e + r / r
0
R0
R(t)
è 0ø
LL
EN
V(R(t))
ærö
(2) v µ ln ç r ÷
ær ö
(3) v µ ç r ÷
è 0ø
(2)
R(t)
R0
ær ö
(4) v µ ln ç r ÷
è 0ø
V(R(t))
16.
(3)
R0
A
node06\B0B0-BA\Kota\JEE Main\Jee Main-2019_Subject Topic PDF With Solution\Physics\English\5-Electrostatics
E
(1) –48 V
(2) –520 V
(3) 180 V
(4)
R0
14.
R(t)
17.
An electric dipole is formed by two equal and
opposite charges q with separation d. The
charges have same mass m. It is kept in a
uniform electric field E. If it is slightly rotated
from its equilibrium orientation, then its angular
frequency w is :(1)
r
E = ( Ax + B ) ˆi , where E is in NC–1 and x is in
metres. The values of constants are
A = 20 SI unit and B = 10 SI unit. If the
potential at x = 1 is V1 and that at x = –5 is V2,
then V1 – V2 is :-
R(t)
V(R(t))
The electric field in a region is given by
qE
qE
(2) 2
(3)
2md
md
2qE
(4)
md
qE
md
(4) 320 V
The bob of a simple pendulum has mass 2g
and a charge of 5.0 µC. It is at rest in a uniform
horizontal electric field of intensity 2000 V/m.
At equilibrium, the angle that the pendulum
makes with the vertical is : (take g = 10 m/s2)
(1) tan–1(5.0)
(2) tan–1(2.0)
(3) tan–1(0.5)
(4) tan–1(0.2)
ALLEN
Electrostatics
18.
A solid conducting sphere, having a charge
Q, is surrounded by an uncharged conducting
hollow spherical shell. Let the potential
difference between the surface of the solid
sphere and that of the outer surface of the
hollow shell be V. If the shell is now given a
charge of –4 Q, the new potential difference
between the same two surfaces is :
(1) V
(2) 2V
(3) –2V
(4) 4V
1
(1) E µ
D
20.
+
+
+
+
+
+
+
+
+
+
+
+
Four point charges –q, +q, +q and –q are placed
on y-axis at y = –2d, y = –d, y = +d and
y = +2d, respectively. The magnitude of the
electric field E at a point on the x-axis at
x = D, with D >> d, will behave as :-
(3) E µ
q
m
L
(1) 2p
æ qE ö
g2 + ç
÷
èmø
(2) 2p
2
L
qE ö
çg ÷
mø
è
(3) 2p æ
1
(2) E µ 3
D
1
D2
(4) E µ
1
D4
22.
d
–q
Q
A
If D >> d, the potential energy of the system
is best given by :
1 é q 2 qQd ù
- –
4pe 0 êë d 2D 2 úû
1 é q 2 qQd ù
(2) 4pe ê + d + D 2 ú
0 ë
û
1 é q 2 2qQd ù
(3) 4pe ê – d + D 2 ú
û
0 ë
(4) 2p
L
g2 -
q2 E 2
m2
In free space, a particle A of charge 1 mC is
held fixed at a point P. Another particle B of
the same charge and mass 4 mg is kept at a
distance of 1 mm from P. if B is released, then
its velocity at a distance of 9 mm from P is :
23.
(1) 2.0 × 103 m/s
(2) 3.0 × 104 m/s
(3) 1.5 × 102 m/s
(4) 1.0 m/s
A uniformly charged ring of radius 3a and total
charge q is placed in xy-plane centred at origin.
A point charge q is moving towards the ring
along the z-axis and has speed u at z = 4a. The
minimum value of u such that it crosses the
origin is :
1/2
(1)
2 æ 1 q2 ö
ç
÷
m è 15 4pe0 a ø
(2)
2 æ 2 q2 ö
ç
÷
m è 15 4pe0 a ø
(3)
2 æ 4 q2 ö
ç
÷
m è 15 4pe0 a ø
(4)
2 æ 1 q2 ö
ç
÷
m è 5 4pe0 a ø
1/2
1/2
1 é q 2 qQd ù
(4) 4pe ê – d – D 2 ú
û
0 ë
L
qE ö
æ
çg + m ÷
è
ø
é
ù
1
= 9 ´ 109 Nm 2 C -2 ú
ê Take
4pe 0
ë
û
D
(1)
L
E
A system of three charges are placed as shown
in the figure :
+q
A simple pendulum of length L is placed
between the plates of a parallel plate capacitor
having electric field E, as shown in figure. Its
bob has mass m and charge q. The time period
of the pendulum is given by :
LL
EN
19.
21.
1/2
node06\B0B0-BA\Kota\JEE Main\Jee Main-2019_Subject Topic PDF With Solution\Physics\English\5-Electrostatics
4
E
ALLEN
24.
Let a total charge 2Q be distributed in a sphere
of radius R, with the charge density given by
r(r) = kr, where r is the distance from the centre.
Two charges A and B, of –Q each, are placed
on diametrically opposite points, at equal
distance, a, from the centre. If A and B do not
experience any force, then :
(1) a =
25.
Electrostatics
3R
2¼
26.
r
A point dipole p = -p 0 xˆ is kept at the origin.
The potential and electric field due to this
dipole on the y-axis at a distance d are,
respectively: (Take V = 0 at infinity) :
r
r
|p|
-p
(1) 4pe d 2 , 4pe d3
0
0
(2) a = R / 3
(3) a = 8–1/4R
(4) a = 2–1/4 R
Shown in the figure is a shell made of a
conductor. It has inner radius a and outer radius
b, and carries charge Q. At its centre is a dipole
r
p as shown. In this case :
(2) 0,
r
p
4pe0 d3
r
r
|p|
p
(3) 4pe d 2 , 4pe d3
0
0
r
-p
4pe0 d3
LL
EN
(4) 0,
p
(1) Electric field outside the shell is the same
as that of a point charge at the centre of the
shell.
(2) Surface charge density on the inner surface
of the shell is zero everywhere.
(3) Surface charge density on the inner surface
is uniform and equal to
(Q / 2)
.
4pa 2
(4) Surface charge density on the outer surface
r
E
A
node06\B0B0-BA\Kota\JEE Main\Jee Main-2019_Subject Topic PDF With Solution\Physics\English\5-Electrostatics
depends on p
5
ALLEN
Electrostatics
2.
SOLUTION
Ans. (3)
dr
E1
r
y=3
q2
Q = ò rdv
E2
y=2
R
q1
(0,0)
q1
x=3 x=4m
x=1 x=2
R
A
-2 r / a
( 4pr dr )
R
= 4pA ò e-2r / a dr
LL
EN
R
(4 + 3 )
2
2
æ
ö
ç e -2 r / a ÷
= 4pA ç
2 ÷
çç - ÷÷
è a ø0
V/m
r
\ E 2 = 9 ´ 10 3 cos q2 ˆi - sin q2 ˆj
(
æ aö
è
ø
= 4pA ç - ÷ ( e
2
)
Magnitude of E1 =
1
4p Î0
= 9 ´ 10 ´ 10 ´ 10
)
10 ´ 10 -6
12 + 32
(
)
( )
\ tanq1 = 3
3.
Ans.
(3)
d
-7
= 9 10 ´ 10 2
r
\ E1 = 9 10 ´ 10 2 é cos q1 -ˆi + sin q1ˆj ù
ë
û
10
3
q1
1
3 ˆù
é 1
E1 = 9 ´ 10 ´ 10 2 ê
- ˆi +
jú
10 û
ë 10
( )
E1 = 9 ´ 10 2 éë -ˆi + 3ˆjùû = éë-9iˆ + 27ˆjùû 10 2
r r
r
2
\ E = E1 + E 2 = 63iˆ - 27ˆj ´ 10 V / m
(
\ correct answer is (3)
)
- 1)
æ
ö
ç
÷
a
1
R = log ç
Q ÷
2
çç 1 ÷÷
è 2paA ø
r
3 ö
æ4
\ E 2 = 9 ´ 10 3 ç ˆi - ˆj ÷ = 72iˆ - 54 ˆj ´ 10 2
5 ø
è5
(
-2R / a
Q = 2paA(1 – e–2R/a)
3
Q tan q2 =
4
)
2
0
9 ´ 10 9 ´ ( 25 ) ´ 10 -6
9
2
0
1 q2
4pÎ0 r 2
E2 = 9 × 103 V/m
(
( 4pr dr )
= ò r2 e
x
to q 1 & q 2 respectively magnitude of
E2 =
-2 r / a
0
q2
q2
r
r
Let E1 & E 2 are the vaues of electric field due
E2 =
A
= ò r2 e
y=3
A
1.
Ans. (4)
Fa
Fb
+Q
d/2
For equilibrium,
r r
Fa + FB = 0
r
r
Fa = - FB
kQQ
kQq
=2
2
d
(d / 2)
Þq =-
Q
4
q
d/2
+Q
node06\B0B0-BA\Kota\JEE Main\Jee Main-2019_Subject Topic PDF With Solution\Physics\English\5-Electrostatics
6
E
ALLEN
4.
Electrostatics
Ans. (3)
Electric field on axis of ring
E=
7.
7
Ans. (2)
kQh
(h + R )
2 3/2
2
a
b
for maximum electric field
r
c
P
dE
=0
dh
Þh=
R
2
Potential at point P, V =
kQ a kQ b kQ c
+
+
a
b
c
Q Qa : Qb : Qc : : a2 : b2 : c2
{since sa = sb = sc}
Ans. (4)
Electric field of equitorial plane of dipole
r
KP
=- 3
r
é
ù
a2
\Qa = ê 2
Q
2
2 ú
ëa + b + c û
LL
EN
5.
é
ù
b2
Qb = ê 2
Q
2
2 ú
ëa + b + c û
r
KP
\ At P, F = - 3 Q.
r
At P1 , F1 = 6.
Ans. (2)
(0,2) Q
r
KPQ
( r / 3)
3
é
ù
c2
Qc = ê 2
Q
2
2 ú
ëa + b + c û
= 27 F .
V=
Q(4, +2)
8.
Q é (a + b + c) ù
ê
ú
4 p Î0 ë a 2 + b 2 + c 2 û
\ correct answer is (2)
Correct Ans. (3)
According to JEE-Mains Ans. key (1 or 3)
4qa
A
node06\B0B0-BA\Kota\JEE Main\Jee Main-2019_Subject Topic PDF With Solution\Physics\English\5-Electrostatics
(0,–2) Q
E
Potential at origin =
(Potential at ¥ = 0)
Q(4, –2)
2x = R + x
KQ KQ KQ KQ
+
+
+
2
2
20
20
R
x=
æ
1 ö
= KQ ç 1 +
÷
5ø
è
\ Work required to put a fifth charge Q at
Q2 æ
1 ö
origin is equal to
1+
ç
÷
4 pe 0 è
5ø
2qa
V = ( R + x ) = ( x2 )
9.
2 -1
R
4qa
2qa
dist =
R
+R =
2 -1
2R
2 -1
Ans. (2)
rr
U = - P.E
= –PE cos q
= –(10–29) (103) cos 45º
= – 0.707 × 10–26 J
= –7 × 10–27 J.
8
Electrostatics
10.
Ans. (1)
ALLEN
14.
Ans. (3)
E
é q 2 Qq Qq ù
+
+
ú=0
a a 2û
ëa
U = Kê
a 2
2
a
+q
F = qE –q,m
md
ædö
moment of inertia (I) = m ç ÷ ´ 2 =
2
è2ø
+q
md 2
.a
(qE) (d sin q) =
2
æ 2qE ö
÷ sin q
è md ø
a= ç
LL
EN
–2q
for small q
P2
Þ Angular frequency w =
15.
æ 3ö
2 qd çç 2 ÷÷ =
è
ø
3 qd
Sol.
Ans. (1)
At any instant 't'
Total energy of charge distribution is constant
A
2 KQ2 æ 1 1 ö
.ç
- ÷
m 2 è R0 R ø
\ V=
KQ 2 æ 1 1 ö
1 1
- ÷ =C
ç
m è R0 R ø
R0 R
1
mV 2 = -q ( Vf - Vi )
2
l
E = 2 pe r
0
l
ær ö
ln ç 0 ÷
2 pe 0 è r ø
1
-ql æ r0 ö
mv 2 =
ln ç ÷
2
2pe 0 è r ø
1
KQ2 KQ2
2
mV
=
\ 2
2R 0
2R
\ V=
Ans. (4)
DV =
1
KQ 2
KQ 2
2
mV
+
=
0
+
i.e. 2
2R
2R 0
ær ö
v µ ln ç ÷
è r0 ø
16.
Ans. (3)
r
Sol. E = ( 20x + 10 ) ˆi
1
Also the slope of v-s curve will go on
decreasing
\ Graph is correctly shown by option(1)
2qE
md
V1 – V2 - ò ( 20x + 10 ) dx
-5
V1 – V2 = - (10x 2 + 10x )-5
1
V1 – V2 = 10(25 – 5 – 1 – 1)
V1 – V2 = 180 V
node06\B0B0-BA\Kota\JEE Main\Jee Main-2019_Subject Topic PDF With Solution\Physics\English\5-Electrostatics
+q
æ 2qE ö
÷q
è md ø
Þa= ç
x
|P1| = q(d)
|P2| = qd
|Resultant| = 2 P cos30º
13.
2
Now by t = Ia
y
+q
d/2
Sol.
Ans. (2)
Ans. (3)
P1
F = qE
q
a
–q 2
Þ Q=
2 +1
11.
12.
d/2
Q
1 ù
é
Þ q = - Q ê1+ ú
2û
ë
+q
m
E
ALLEN
17.
Electrostatics
Ans. (3)
9
æ kQ 4kQ ö æ kQ 4kQ ö
Vin - Vout = ç
÷ -ç
÷
r2 ø è r2
r2 ø
è r1
E
q
Sol.
=
qE
kQ kQ
r1
r2
æ1 1ö
= kQ ç - ÷ = V
è r1 r2 ø
mg
qE 5 ´ 10-6 ´ 2000
=
mg
2 ´ 10-3 ´ 10
tan q =
1
Þ q = tan–1 (0.5)
2
19.
Sol.
Hence, we also obtain that potential difference
does not depend on charge of outer sphere.
\ P.d. remains same
Ans. (4)
LL
EN
tan q =
18. Ans. (1)
Sol. As given in the first condition :
–q
d
+q
d
Solid
Hollow
(uncharged)
r1
r2
d
q1
q2
q1
E1
–q
D
+Q
Electric field at p = 2E1cosq1 –2E1cosq2
2Kq
D
2Kq
D
= (d 2 + D 2 ) ´ (d 2 + D 2 )1/ 2 – [(2d)2 + D2 ] ´ [(2d)2 + D2 ]1/ 2
æ kQ ö æ kQ ö
Vin - Vout = ç
÷-ç
÷
è r1 ø è r2 ø
= 2KqD éë(d 2 + D2 ) –3/2 - (4d 2 + D2 )–3/2 ùû
A
node06\B0B0-BA\Kota\JEE Main\Jee Main-2019_Subject Topic PDF With Solution\Physics\English\5-Electrostatics
E2
E1
q2 p
+q
Both conducting spheres are shown.
E
X
d
Q
–Q
E2
æ1 1ö
= kQ ç - ÷ = V
è r1 r2 ø
2KqD éæ
d2 ö
ê
1
=
+
ç
÷
D 3 êè
D2 ø
ë
In the second condition :
Applying binomial approximation Q d << D
=
Q
–4Q
r1
r2
æ 4d 2 ö
- ç1 + 2 ÷
D ø
è
–3/2
2KqD é 3 d 2 æ 3 ´ 4d 2 ö ù
- ç1 ê1 ÷ú
D3 ë 2 D2 è
2D2 ø û
2KqD é12 d 2 3 d 2 ù
= D3 ê 2 D2 - 2 D2 ú
ë
û
=
Shell is now given charge –4Q.
–3/2
9kqd 2
D4
ù
ú
úû
10
Electrostatics
20.
Ans. (4)
ALLEN
24.
Ans. (3)
a
D
–q
D >> d
Q
e0
k 4 pa 4
E = 4 ´ 4 pe
0
Utotal = Uself of dipole + Uinteraction
kq 2 æ kQ ö
–ç
÷ qd
d è D2 ø
k=
+
+ + +
-12
-
v = 4 × 10
v = 2 × 103 m/s
23. Ans. (2)
Sol. Ui + Ki = Uf + Kf
2
6
kq 2
1
kq 2
+ mv 2 =
3a
16a 2 + 9a 2 2
1
kq 2 æ 1 1 ö 2kq 2
mv 2 =
ç - ÷=
2
a è 3 5 ø 15a
v=
4kq 2
15ma
=
1
´ (4 ´ 10 -6 )v 2
2
E=
26.
KQ
r2
Ans. (4)
E
Sol.
d
p0
V=0
r
KP
E=- 3
r
r
p
=4pe 0 d3
node06\B0B0-BA\Kota\JEE Main\Jee Main-2019_Subject Topic PDF With Solution\Physics\English\5-Electrostatics
(9 ´ 10 ) ´ 10
9 ´ 10 -3
9
1 2
mv
2
A
-12
+
(9 ´ 10 ) ´ 10
10 -3
9
+ + +
kq1q 2
r
– – ––
–
–
–
–
+ p
+
+
++ ++
Total charge of dipole = 0, so charge induced
on outside surface = 0.
But due to non uniform electric field of dipole,
the charge induced on inner surface is non zero
and non uniform.
So, for any abserver outside the shell, the resultant
electric field is due to Q uniformly distributed on
outer surface only and it is equal to.
2
Sol. WE = –[DU] = Ui – UF =
U=
+ +
+ +
Sol.
æ qE ö
g2 + ç ÷
èmø
Ans. (1)
+ + +
Q
+ + +
2
l
= 2p
R = a8
Ans. (1)
+
qE ö
÷
èmø
l
0
QQ
(2a)2
1/4
25.
æ
Sol. geff = g 2 + ç
22.
QE = 4pe
Option (4)
Ans. (1)
T = 2p g
eff
2Q
pR 4
LL
EN
21.
2
1
é q 2 qQd ù
+ 2 ú
D û
ëd
= – kê
+ +
= –
R
2Q = ò0 kr 4pr dr
+
d
2
+ +
+q
Sol.
ò kr 4 pr dr
Sol. E 4pa2 = 0
E