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
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
DPP/ P 32
. 6 (c) Electric lines force due to negative charge are radially
inward.
–
7. (b) According to the question,
mg
eE mgE
e
= Þ=
. 8 (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)
 
 
 
 
9. (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 = = ==
0 1 . (b)
$ $ $ $
000 0
22
2 22
E k k kk
s s ss
=- - - =-
eee e
ur
1 1 . (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
2 1 . (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
DPP/ P 32
. 6 (c) Electric lines force due to negative charge are radially
inward.
–
7. (b) According to the question,
mg
eE mgE
e
= Þ=
. 8 (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)
 
 
 
 
9. (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 = = ==
0 1 . (b)
$ $ $ $
000 0
22
2 22
E k k kk
s s ss
=- - - =-
eee e
ur
1 1 . (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
2 1 . (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.
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
3 1 . (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.
4 1 . (d) Next flux through the cube 
0
net
Q
f=
e
; so flux through
one face 
face
0
6
f=
e
q
. 5 1 (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
. 6 1 (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
7 1 . (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
8 1 . (b) The electric field outside the sphere is given by :
2
(),
kQ
Er
r
= for rR ³
9 1 . (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.
0 2 . (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.