DPP for NEET: Daily Practice Problems, Ch 32: Electrostatics- 1 (Solutions)

# Electrostatics- 1 Practice Questions - DPP for NEET

``` Page 1

1. (d)
12
Q QQ +=            .....(i)
and
12
2
QQ
Fk
r
=
..... (ii)
From (i) and (ii)
11
2
() -
=
kQ QQ
F
r
For F to be maximum
12
1
0
2
dFQ
QQ
dQ
= Þ ==
2. (a) The position of the balls in the satellite will become as
shown below
+ Q
L L
180°
+ Q
Thus angle
180 q=
o
and force
2
2
0
1
.
4
(2)
=
pÎ
Q
L
3. (b)
A
F = Force on
C
due to charge placed at
A
120°
A B
10 cm
C
+ 1µC
+2µC
1µC -
A
F
B
F
66
9
22
10 2 10
9 10 1.8
(10 10 )
N
--
-
´´
=´´=
´
B
F = force on
C
due to charge placed at
B
66
9
2
10 2 10
9 10 1.8
(0.1)
N
--
´´
=´´=
Net force on
C
22
( ) ( ) 2 cos120 1.8 = ++=
net A B AB
FFFFFN
o
4. (c)
22
23
.
ee
F k r kr
rr
=- =-
urr
\$
r
r
r
æö
\=
ç÷
ç÷
èø
r
\$
5. (c) After following the guidelines mentioned above
A
B
D C
+ Q
+ Q
+ Q
A
F
AC
F
D
F
C
F
22
net AC D A CD
F F F F FF = += ++
Since
2
2
AC
kq
FF
a
==
and
2
2
( 2)
D
kq
F
a
=
2222
2222
0
2 1 1 22
2
22
24
æö
+ æö
= + = +=
ç÷ ç÷
èø
èø pÎ
net
kq kq kq q
F
aaaa
6. (a)
6 6 12
22
00
1 ( 7 10 ( 5 10 ) 1 35 10
44
FN
rr
---
+´-´´
= =-
pe pe
6 6 12
22
00
1 ( 5 10 ( 7 10 ) 1 35 10
44
FN
rr
---
+´-´´
¢ = =-
p e pe
7. (c) Electric field outside of the sphere E
out
=
2
kQ
r
...(i)
Ele ctric field inside the dielec tric s hphere E
in
=
3
kQx
R
...(ii)
From (i) and (ii), E
in
=
2
out
rx
E
R
´
Þ At 3 cm, E =
()
2
3
3 20
100
10
´ = 120 V/m
Page 2

1. (d)
12
Q QQ +=            .....(i)
and
12
2
QQ
Fk
r
=
..... (ii)
From (i) and (ii)
11
2
() -
=
kQ QQ
F
r
For F to be maximum
12
1
0
2
dFQ
QQ
dQ
= Þ ==
2. (a) The position of the balls in the satellite will become as
shown below
+ Q
L L
180°
+ Q
Thus angle
180 q=
o
and force
2
2
0
1
.
4
(2)
=
pÎ
Q
L
3. (b)
A
F = Force on
C
due to charge placed at
A
120°
A B
10 cm
C
+ 1µC
+2µC
1µC -
A
F
B
F
66
9
22
10 2 10
9 10 1.8
(10 10 )
N
--
-
´´
=´´=
´
B
F = force on
C
due to charge placed at
B
66
9
2
10 2 10
9 10 1.8
(0.1)
N
--
´´
=´´=
Net force on
C
22
( ) ( ) 2 cos120 1.8 = ++=
net A B AB
FFFFFN
o
4. (c)
22
23
.
ee
F k r kr
rr
=- =-
urr
\$
r
r
r
æö
\=
ç÷
ç÷
èø
r
\$
5. (c) After following the guidelines mentioned above
A
B
D C
+ Q
+ Q
+ Q
A
F
AC
F
D
F
C
F
22
net AC D A CD
F F F F FF = += ++
Since
2
2
AC
kq
FF
a
==
and
2
2
( 2)
D
kq
F
a
=
2222
2222
0
2 1 1 22
2
22
24
æö
+ æö
= + = +=
ç÷ ç÷
èø
èø pÎ
net
kq kq kq q
F
aaaa
6. (a)
6 6 12
22
00
1 ( 7 10 ( 5 10 ) 1 35 10
44
FN
rr
---
+´-´´
= =-
pe pe
6 6 12
22
00
1 ( 5 10 ( 7 10 ) 1 35 10
44
FN
rr
---
+´-´´
¢ = =-
p e pe
7. (c) Electric field outside of the sphere E
out
=
2
kQ
r
...(i)
Ele ctric field inside the dielec tric s hphere E
in
=
3
kQx
R
...(ii)
From (i) and (ii), E
in
=
2
out
rx
E
R
´
Þ At 3 cm, E =
()
2
3
3 20
100
10
´ = 120 V/m
DPP/ P 32
89
8. (c) Electric lines force due to negative charge are radially
inward.
–
9. (a) In non-uniform electric field, intensity is more, where
the lines are more denser.
10. (b) According to the question,
mg
eE mgE
e
= Þ=
11. (b) Because E points along the tangent to the lines of force.
If initial velocity is zero, then due to the force, it always
moves in the direction of E. Hence will always move
on some lines of force.
12. (b) The field produced by charge - 3Q at
A
, this is
E
as
mentioned in the example.
2
3Q
E
x
\= (along AB directed towards negative
charge)
A
x
B
Q
–3Q
Now field at location of 3Q - i.e. field at
B
due to
charge Q will be
2
'
3
QE
E
x
==
(along
AB
directed
away from positive charge)
13. (c) The electric field is due to all charges present whether
inside or outside the given surface.
14. (b)
q q
q
q
q
q
E
E
E
E
E
E
0
net
E Þ=
–q q
q
–q
q
q
E
E
E
E
E
E

Þ
2E
2E
2E
120°
2
net
E E =
2q 2q
q
2q 2q
q
E
2E
2E
E
2E
2E
0
net
E Þ=
q 2q
2q
q 2q
q
E
2E
2E
E
2E
E

Þ
E
2E
2E
0
net
E =
E
15. (b)
A
B
D C
O
A
E
B
E
C
E
D
E
Þ
A
B
D
C
2E 2E
net
E
,,
2, 34
A B CD
E EE EE EEE = = ==
16. (b)
\$ \$ \$ \$
000 0
22
2 22
E k k kk
s s ss
=- - - =-
eee e
ur
17. (b)
18. (c) According to Gauss law . E ds
ò Ñ
=
1
0
q
e
ds
ò Ñ
= 2 prl; (E is constant)
\ E. 2prl =
1
0
q
e
Þ E =
0
2
q
r pe
i.e. E µ
1
r
19. (c) Let sphere has uniform chare density r
3
3
4
Q
R
æö
ç÷
p èø
and E
is the electric field at distance x from the centre of the
sphere.
Page 3

1. (d)
12
Q QQ +=            .....(i)
and
12
2
QQ
Fk
r
=
..... (ii)
From (i) and (ii)
11
2
() -
=
kQ QQ
F
r
For F to be maximum
12
1
0
2
dFQ
QQ
dQ
= Þ ==
2. (a) The position of the balls in the satellite will become as
shown below
+ Q
L L
180°
+ Q
Thus angle
180 q=
o
and force
2
2
0
1
.
4
(2)
=
pÎ
Q
L
3. (b)
A
F = Force on
C
due to charge placed at
A
120°
A B
10 cm
C
+ 1µC
+2µC
1µC -
A
F
B
F
66
9
22
10 2 10
9 10 1.8
(10 10 )
N
--
-
´´
=´´=
´
B
F = force on
C
due to charge placed at
B
66
9
2
10 2 10
9 10 1.8
(0.1)
N
--
´´
=´´=
Net force on
C
22
( ) ( ) 2 cos120 1.8 = ++=
net A B AB
FFFFFN
o
4. (c)
22
23
.
ee
F k r kr
rr
=- =-
urr
\$
r
r
r
æö
\=
ç÷
ç÷
èø
r
\$
5. (c) After following the guidelines mentioned above
A
B
D C
+ Q
+ Q
+ Q
A
F
AC
F
D
F
C
F
22
net AC D A CD
F F F F FF = += ++
Since
2
2
AC
kq
FF
a
==
and
2
2
( 2)
D
kq
F
a
=
2222
2222
0
2 1 1 22
2
22
24
æö
+ æö
= + = +=
ç÷ ç÷
èø
èø pÎ
net
kq kq kq q
F
aaaa
6. (a)
6 6 12
22
00
1 ( 7 10 ( 5 10 ) 1 35 10
44
FN
rr
---
+´-´´
= =-
pe pe
6 6 12
22
00
1 ( 5 10 ( 7 10 ) 1 35 10
44
FN
rr
---
+´-´´
¢ = =-
p e pe
7. (c) Electric field outside of the sphere E
out
=
2
kQ
r
...(i)
Ele ctric field inside the dielec tric s hphere E
in
=
3
kQx
R
...(ii)
From (i) and (ii), E
in
=
2
out
rx
E
R
´
Þ At 3 cm, E =
()
2
3
3 20
100
10
´ = 120 V/m
DPP/ P 32
89
8. (c) Electric lines force due to negative charge are radially
inward.
–
9. (a) In non-uniform electric field, intensity is more, where
the lines are more denser.
10. (b) According to the question,
mg
eE mgE
e
= Þ=
11. (b) Because E points along the tangent to the lines of force.
If initial velocity is zero, then due to the force, it always
moves in the direction of E. Hence will always move
on some lines of force.
12. (b) The field produced by charge - 3Q at
A
, this is
E
as
mentioned in the example.
2
3Q
E
x
\= (along AB directed towards negative
charge)
A
x
B
Q
–3Q
Now field at location of 3Q - i.e. field at
B
due to
charge Q will be
2
'
3
QE
E
x
==
(along
AB
directed
away from positive charge)
13. (c) The electric field is due to all charges present whether
inside or outside the given surface.
14. (b)
q q
q
q
q
q
E
E
E
E
E
E
0
net
E Þ=
–q q
q
–q
q
q
E
E
E
E
E
E

Þ
2E
2E
2E
120°
2
net
E E =
2q 2q
q
2q 2q
q
E
2E
2E
E
2E
2E
0
net
E Þ=
q 2q
2q
q 2q
q
E
2E
2E
E
2E
E

Þ
E
2E
2E
0
net
E =
E
15. (b)
A
B
D C
O
A
E
B
E
C
E
D
E
Þ
A
B
D
C
2E 2E
net
E
,,
2, 34
A B CD
E EE EE EEE = = ==
16. (b)
\$ \$ \$ \$
000 0
22
2 22
E k k kk
s s ss
=- - - =-
eee e
ur
17. (b)
18. (c) According to Gauss law . E ds
ò Ñ
=
1
0
q
e
ds
ò Ñ
= 2 prl; (E is constant)
\ E. 2prl =
1
0
q
e
Þ E =
0
2
q
r pe
i.e. E µ
1
r
19. (c) Let sphere has uniform chare density r
3
3
4
Q
R
æö
ç÷
p èø
and E
is the electric field at distance x from the centre of the
sphere.
90
DPP/ P 32
Applying Gauss law,
E.4p x
2
=
0
q
e
=
0
V r¢
e
=
3
0
4
3
x
r
´p
e
(V = V olume of dotted sphere)
x
R
\ E =
0
3
x
r
e
Þ E = µ x
20. (b) T sin q = qE
+
+
+
+
+
+
+
q
T
mg
T cos  q
qE
T sin  q
and T cos q = mg
Þ tan q =
qE
mg
=
q
mg
0
2
æö s
ç÷
e
èø
Þs µ tan q.
21. (d) Next flux through the cube
0
net
Q
f=
e
; so flux through
one face
face
0
6
f=
e
q
22. (d). For A : Power consumed P = I
2
R
But
0
q
=f
e
,
so
2
00
dq
q t I 2t
dt
= ae Þ = = ae
2 22
0
P 4 Rt Þ = ae
For B : Assuming initial charge in reservoir be q
0
then
electric flux through a closed
Spherical surface around S
2
will be
2
00
S
2
0
qt - ae
f=
e
For C :
S
2
d
2t
dt
f
= -a
23. (c). The time period will change only when the additional
electrostatic force has a component along the direction
of the displacement, which is always perpendicular
to the string.
24. ()
25. (b) Net charge inside the sphere
sphere
dV =r
ò
Due to spherical symmetry , we get
2
0
4 ()
R
Q r r dr = pr
ò

2
0
4 ()
R
A r R r dr =p-
ò
44
4
34
RR
A
æö
=p- ç÷
ç÷
èø
\
4
3Q
A
R
=
p
26. (a) According to Gauss law
2
. 4 ()
s
EdS r Er =p
ò
uuruur
Ñ
2
0
0
4 ()
r
r r dr pr
=
Î
ò
Þ
2
2 0
0
4 ()
4 ()
r
A r R r dr
r Er
p-
p=
Î
ò

34
0
4
34
A rRr
æö
p
=- ç÷
ç÷
Î
èø
Hence,
2
0
()
34
A rRr
Er
æö
=- ç÷
ç÷
Î
èø
, for
0 rR <<
But
4
3Q
A
R
=
p
\ We get,
2
2
0
3 11
()
34
Q rr
Er
RR
R
éù
æ ö æö
=- êú
ç ÷ ç÷
è ø èø Îêú
ëû
27. (b) The electric field outside the sphere is given by :
2
(),
kQ
Er
r
= for rR ³
28. (c) If electric lines of forces cross each other, then the
electric field at the point of intersection will have two
direction simultaneously which is not possible
physically.
29. (c) Electric field at the nearby point will be resultant of
existing field and field due to the charge brought. It
may increase or decrease if the charge is positive or
negative depending on the position of the point with
respect to the charge brought.
30. ()
```

## Physics Class 12

105 videos|425 docs|114 tests

## FAQs on Electrostatics- 1 Practice Questions - DPP for NEET

 1. What is Electrostatics?
Ans. Electrostatics is the branch of physics that deals with the study of electric charges at rest and the forces and fields associated with them.
 2. What are the key concepts in Electrostatics?
Ans. The key concepts in Electrostatics include electric charge, electric field, electric potential, Coulomb's law, and Gauss's law.
 3. How does Coulomb's law explain the interaction between electric charges?
Ans. Coulomb's law states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. It mathematically explains the interaction between electric charges.
 4. What is the difference between conductors and insulators in the context of Electrostatics?
Ans. Conductors are materials that allow the easy flow of electric charges, while insulators are materials that do not allow the flow of electric charges. In Electrostatics, conductors can redistribute charges on their surface, while insulators keep the charges localized.
 5. How does Gauss's law help in determining the electric field due to a charge distribution?
Ans. Gauss's law is a fundamental law in Electrostatics that relates the electric flux through a closed surface to the charge enclosed by that surface. It provides a convenient way to calculate the electric field due to a charge distribution by considering a Gaussian surface surrounding the charge distribution.

## Physics Class 12

105 videos|425 docs|114 tests

### Up next

 Explore Courses for NEET exam

### Top Courses for NEET

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;