Chapter 30 : Gauss Law - HC Verma Solution, Physics Class 11 Notes | EduRev

HC Verma and Irodov Solutions

JEE : Chapter 30 : Gauss Law - HC Verma Solution, Physics Class 11 Notes | EduRev

 Page 1


30.1
CHAPTER – 30
GAUSS’S LAW
1. Given : E
?
= 3/5 E
0
i
ˆ
+ 4/5 E
0 j
ˆ
E
0 
= 2.0 × 10
3
N/C The plane is parallel to yz-plane.
Hence only 3/5 E
0
i
ˆ
passes perpendicular to the plane whereas 4/5 E
0 j
ˆ
goes 
parallel. Area = 0.2m
2
(given)
? Flux = A E
? ?
? = 3/5 × 2 × 10
3
× 0.2 = 2.4 × 10
2
Nm
2
/c = 240 Nm
2
/c
2. Given length of rod = edge of cube = l
Portion of rod inside the cube = l/2
Total charge = Q.
Linear charge density = ? = Q/l of rod.
We know: Flux ? charge enclosed. 
Charge enclosed in the rod inside the cube.
= l/2 ?
0
× Q/l = Q/2 ?
0
?
3. As the electric field is uniform.
Considering a perpendicular plane to it, we find that it is an equipotential surface. Hence 
there is no net current flow on that surface. Thus, net charge in that region is zero.
4. Given: E = i
ˆ
E
0
?
?
l= 2 cm, a = 1cm.
E
0
= 5 × 10
3
N/C. From fig. We see that flux passes mainly through surface 
areas. ABDC & EFGH. As the AEFB & CHGD are paralled to the Flux. Again in 
ABDC a = 0; hence the Flux only passes through the surface are EFGH. 
E = i
ˆ
x E
c
?
Flux = 
L
E
0
?
× Area = 
2
3
a
a 10 5
?
? ?
?
= 
?
3 3
a 10 5 ? ?
= 
2
3 3
10 2
) 01 . 0 ( 10 5
?
?
?
? ?
= 2.5 × 10
–1
Flux = 
0
q
?
so, q = ?
0
× Flux 
= 8.85 × 10
–12
× 2.5 × 10
–1
= 2.2125 × 10
–12
c
5. According to Gauss’s Law Flux = 
0
q
?
Since the charge is placed at the centre of the cube. Hence the flux passing through the 
six surfaces = 
0
6
Q
?
× 6 = 
0
Q
?
6. Given – A charge is placed o a plain surface with area = a
2
, about a/2 from its centre.
Assumption : let us assume that the given plain forms a surface of an imaginary cube. Then the charge 
is found to be at the centre of the cube.
Hence flux through the surface = 
6
1 Q
0
?
?
= 
0
6
Q
?
7. Given: Magnitude of the two charges placed = 10
–7
c.
We know: from Gauss’s law that the flux experienced by the sphere is only due to 
the internal charge and not by the external one.
Now 
?
?
?
0
Q
ds . E
?
= 
12
7
10 85 . 8
10
?
?
?
= 1.1 × 10
4
N-m
2
/C.
X
i
ˆ
×
Z
Y
× × × ×
× × × × ×
× × × × ×
× × × × ×
× × × ×
× × ×
j
ˆ
k
ˆ
l
l/2 l/2
E
?
B F
o,a,o
H
E
G
o
A
D
C
a,o,o
o,o,a
R
P Q
+
+
2R
Page 2


30.1
CHAPTER – 30
GAUSS’S LAW
1. Given : E
?
= 3/5 E
0
i
ˆ
+ 4/5 E
0 j
ˆ
E
0 
= 2.0 × 10
3
N/C The plane is parallel to yz-plane.
Hence only 3/5 E
0
i
ˆ
passes perpendicular to the plane whereas 4/5 E
0 j
ˆ
goes 
parallel. Area = 0.2m
2
(given)
? Flux = A E
? ?
? = 3/5 × 2 × 10
3
× 0.2 = 2.4 × 10
2
Nm
2
/c = 240 Nm
2
/c
2. Given length of rod = edge of cube = l
Portion of rod inside the cube = l/2
Total charge = Q.
Linear charge density = ? = Q/l of rod.
We know: Flux ? charge enclosed. 
Charge enclosed in the rod inside the cube.
= l/2 ?
0
× Q/l = Q/2 ?
0
?
3. As the electric field is uniform.
Considering a perpendicular plane to it, we find that it is an equipotential surface. Hence 
there is no net current flow on that surface. Thus, net charge in that region is zero.
4. Given: E = i
ˆ
E
0
?
?
l= 2 cm, a = 1cm.
E
0
= 5 × 10
3
N/C. From fig. We see that flux passes mainly through surface 
areas. ABDC & EFGH. As the AEFB & CHGD are paralled to the Flux. Again in 
ABDC a = 0; hence the Flux only passes through the surface are EFGH. 
E = i
ˆ
x E
c
?
Flux = 
L
E
0
?
× Area = 
2
3
a
a 10 5
?
? ?
?
= 
?
3 3
a 10 5 ? ?
= 
2
3 3
10 2
) 01 . 0 ( 10 5
?
?
?
? ?
= 2.5 × 10
–1
Flux = 
0
q
?
so, q = ?
0
× Flux 
= 8.85 × 10
–12
× 2.5 × 10
–1
= 2.2125 × 10
–12
c
5. According to Gauss’s Law Flux = 
0
q
?
Since the charge is placed at the centre of the cube. Hence the flux passing through the 
six surfaces = 
0
6
Q
?
× 6 = 
0
Q
?
6. Given – A charge is placed o a plain surface with area = a
2
, about a/2 from its centre.
Assumption : let us assume that the given plain forms a surface of an imaginary cube. Then the charge 
is found to be at the centre of the cube.
Hence flux through the surface = 
6
1 Q
0
?
?
= 
0
6
Q
?
7. Given: Magnitude of the two charges placed = 10
–7
c.
We know: from Gauss’s law that the flux experienced by the sphere is only due to 
the internal charge and not by the external one.
Now 
?
?
?
0
Q
ds . E
?
= 
12
7
10 85 . 8
10
?
?
?
= 1.1 × 10
4
N-m
2
/C.
X
i
ˆ
×
Z
Y
× × × ×
× × × × ×
× × × × ×
× × × × ×
× × × ×
× × ×
j
ˆ
k
ˆ
l
l/2 l/2
E
?
B F
o,a,o
H
E
G
o
A
D
C
a,o,o
o,o,a
R
P Q
+
+
2R
Gauss’s Law
30.2
8. We know: For a spherical surface 
Flux = 
?
?
?
0
q
ds . E
?
[by Gauss law]
Hence for a hemisphere = total surface area = 
2
1 q
0
?
?
= 
0
2
q
?
9. Given: Volume charge density = 2.0 × 10
–4
c/m
3
In order to find the electric field at a point 4cm = 4 × 10
–2
m from the centre let us assume a concentric 
spherical surface inside the sphere.
Now, 
?
?
?
0
q
ds . E
But ? = 
3
R 3 / 4
q
?
so, q = ? × 4/3 ? R
3
Hence = 
2 2
0
3 2
) 10 4 ( 7 / 22 4
1 ) 10 4 ( 7 / 22 3 / 4
?
?
? ? ?
?
?
? ? ? ? ?
= 2.0 × 10
–4
1/3 × 4 × 10
–2
× 
12
10 85 . 8
1
?
?
= 3.0 × 10
5
N/C ?
10. Charge present in a gold nucleus = 79 × 1.6 × 10
–19
C
Since the surface encloses all the charges we have:
(a) 
? ?
?
?
? ?
?
?
?
12
19
0
10 85 . 8
10 6 . 1 79 q
ds . E
?
E = 
ds
q
0
?
= 
2 15 12
19
) 10 7 ( 14 . 3 4
1
10 85 . 8
10 6 . 1 79
? ?
?
? ? ?
?
?
? ?
[ ?area = 4 ?r
2
]
= 2.3195131 × 10
21
N/C
(b) For the middle part of the radius. Now here r = 7/2 × 10
–15
m
Volume = 4/3 ? r
3
= 
45
10
8
343
7
22
3
48
?
? ? ?
Charge enclosed = ? × volume [ ? : volume charge density]
But ?= 
volume Net
e arg ch Net
= 
45
19
10 343
3
4
c 10 6 . 1 9 . 7
?
?
? ? ? ? ?
?
?
?
?
?
? ?
Net charged enclosed = 
45
45
19
10
8
343
3
4
10 343
3
4
10 6 . 1 9 . 7
?
?
?
? ? ? ?
? ? ? ? ?
?
?
?
?
?
? ?
= 
8
10 6 . 1 9 . 7
19 ?
? ?
?
ds E
?
= 
0
enclosed q
?
? E = 
S 8
10 6 . 1 9 . 7
0
19
? ? ?
? ?
?
= 
30 12
19
10
4
49
4 10 85 . 8 8
10 6 . 1 9 . 7
? ?
?
? ? ? ? ? ?
? ?
= 1.159 × 10
21
N/C
11. Now, Volume charge density = 
? ?
3
1
3
2
r r
3
4
Q
? ? ? ?
? ? = 
? ?
3
1
3
2
r r 4
Q 3
? ?
Again volume of sphere having radius x = 
3
x
3
4
?
Q
4 cm
r 1
O
? ? r 2
Page 3


30.1
CHAPTER – 30
GAUSS’S LAW
1. Given : E
?
= 3/5 E
0
i
ˆ
+ 4/5 E
0 j
ˆ
E
0 
= 2.0 × 10
3
N/C The plane is parallel to yz-plane.
Hence only 3/5 E
0
i
ˆ
passes perpendicular to the plane whereas 4/5 E
0 j
ˆ
goes 
parallel. Area = 0.2m
2
(given)
? Flux = A E
? ?
? = 3/5 × 2 × 10
3
× 0.2 = 2.4 × 10
2
Nm
2
/c = 240 Nm
2
/c
2. Given length of rod = edge of cube = l
Portion of rod inside the cube = l/2
Total charge = Q.
Linear charge density = ? = Q/l of rod.
We know: Flux ? charge enclosed. 
Charge enclosed in the rod inside the cube.
= l/2 ?
0
× Q/l = Q/2 ?
0
?
3. As the electric field is uniform.
Considering a perpendicular plane to it, we find that it is an equipotential surface. Hence 
there is no net current flow on that surface. Thus, net charge in that region is zero.
4. Given: E = i
ˆ
E
0
?
?
l= 2 cm, a = 1cm.
E
0
= 5 × 10
3
N/C. From fig. We see that flux passes mainly through surface 
areas. ABDC & EFGH. As the AEFB & CHGD are paralled to the Flux. Again in 
ABDC a = 0; hence the Flux only passes through the surface are EFGH. 
E = i
ˆ
x E
c
?
Flux = 
L
E
0
?
× Area = 
2
3
a
a 10 5
?
? ?
?
= 
?
3 3
a 10 5 ? ?
= 
2
3 3
10 2
) 01 . 0 ( 10 5
?
?
?
? ?
= 2.5 × 10
–1
Flux = 
0
q
?
so, q = ?
0
× Flux 
= 8.85 × 10
–12
× 2.5 × 10
–1
= 2.2125 × 10
–12
c
5. According to Gauss’s Law Flux = 
0
q
?
Since the charge is placed at the centre of the cube. Hence the flux passing through the 
six surfaces = 
0
6
Q
?
× 6 = 
0
Q
?
6. Given – A charge is placed o a plain surface with area = a
2
, about a/2 from its centre.
Assumption : let us assume that the given plain forms a surface of an imaginary cube. Then the charge 
is found to be at the centre of the cube.
Hence flux through the surface = 
6
1 Q
0
?
?
= 
0
6
Q
?
7. Given: Magnitude of the two charges placed = 10
–7
c.
We know: from Gauss’s law that the flux experienced by the sphere is only due to 
the internal charge and not by the external one.
Now 
?
?
?
0
Q
ds . E
?
= 
12
7
10 85 . 8
10
?
?
?
= 1.1 × 10
4
N-m
2
/C.
X
i
ˆ
×
Z
Y
× × × ×
× × × × ×
× × × × ×
× × × × ×
× × × ×
× × ×
j
ˆ
k
ˆ
l
l/2 l/2
E
?
B F
o,a,o
H
E
G
o
A
D
C
a,o,o
o,o,a
R
P Q
+
+
2R
Gauss’s Law
30.2
8. We know: For a spherical surface 
Flux = 
?
?
?
0
q
ds . E
?
[by Gauss law]
Hence for a hemisphere = total surface area = 
2
1 q
0
?
?
= 
0
2
q
?
9. Given: Volume charge density = 2.0 × 10
–4
c/m
3
In order to find the electric field at a point 4cm = 4 × 10
–2
m from the centre let us assume a concentric 
spherical surface inside the sphere.
Now, 
?
?
?
0
q
ds . E
But ? = 
3
R 3 / 4
q
?
so, q = ? × 4/3 ? R
3
Hence = 
2 2
0
3 2
) 10 4 ( 7 / 22 4
1 ) 10 4 ( 7 / 22 3 / 4
?
?
? ? ?
?
?
? ? ? ? ?
= 2.0 × 10
–4
1/3 × 4 × 10
–2
× 
12
10 85 . 8
1
?
?
= 3.0 × 10
5
N/C ?
10. Charge present in a gold nucleus = 79 × 1.6 × 10
–19
C
Since the surface encloses all the charges we have:
(a) 
? ?
?
?
? ?
?
?
?
12
19
0
10 85 . 8
10 6 . 1 79 q
ds . E
?
E = 
ds
q
0
?
= 
2 15 12
19
) 10 7 ( 14 . 3 4
1
10 85 . 8
10 6 . 1 79
? ?
?
? ? ?
?
?
? ?
[ ?area = 4 ?r
2
]
= 2.3195131 × 10
21
N/C
(b) For the middle part of the radius. Now here r = 7/2 × 10
–15
m
Volume = 4/3 ? r
3
= 
45
10
8
343
7
22
3
48
?
? ? ?
Charge enclosed = ? × volume [ ? : volume charge density]
But ?= 
volume Net
e arg ch Net
= 
45
19
10 343
3
4
c 10 6 . 1 9 . 7
?
?
? ? ? ? ?
?
?
?
?
?
? ?
Net charged enclosed = 
45
45
19
10
8
343
3
4
10 343
3
4
10 6 . 1 9 . 7
?
?
?
? ? ? ?
? ? ? ? ?
?
?
?
?
?
? ?
= 
8
10 6 . 1 9 . 7
19 ?
? ?
?
ds E
?
= 
0
enclosed q
?
? E = 
S 8
10 6 . 1 9 . 7
0
19
? ? ?
? ?
?
= 
30 12
19
10
4
49
4 10 85 . 8 8
10 6 . 1 9 . 7
? ?
?
? ? ? ? ? ?
? ?
= 1.159 × 10
21
N/C
11. Now, Volume charge density = 
? ?
3
1
3
2
r r
3
4
Q
? ? ? ?
? ? = 
? ?
3
1
3
2
r r 4
Q 3
? ?
Again volume of sphere having radius x = 
3
x
3
4
?
Q
4 cm
r 1
O
? ? r 2
Gauss’s Law
30.3
Now charge enclosed by the sphere having radius
? = 
3
1
3
2
3
1
3
r
3
4
r
3
4
Q
r
3
4
3
4
? ? ?
? ?
?
?
?
?
?
? ? ?? = Q
?
?
?
?
?
?
?
?
?
? ?
3
1
3
2
3
1
3
r r
r
Applying Gauss’s law – E×4 ??
2
= 
0
enclosed q
?
? E = 
2 3
1
3
2
3
1
3
0
4
1
r r
r Q
??
?
?
?
?
?
?
?
?
?
?
? ?
?
= 
?
?
?
?
?
?
?
?
?
? ?
? ??
3
1
3
2
3
1
3
2
0
r r
r
4
Q
?
12. Given: The sphere is uncharged metallic sphere.
Due to induction the charge induced at the inner surface = –Q, and that outer surface = +Q.
(a) Hence the surface charge density at inner and outer surfaces = 
area surface total
e arg ch
= –
2
a 4
Q
?
and 
2
a 4
Q
?
respectively.
(b) Again if another charge ‘q’ is added to the surface. We have inner surface charge density = –
2
a 4
Q
?
, 
because the added charge does not affect it.
On the other hand the external surface charge density = 
2
a 4
q
Q
?
? as the ‘q’ gets added up.
(c) For electric field let us assume an imaginary surface area inside the sphere at a distance ‘x’ from 
centre. This is same in both the cases as the ‘q’ in ineffective.
Now, 
?
?
?
0
Q
ds . E So, E = 
2
0
x 4
1 Q
?
?
?
= 
2
0
x 4
Q
??
13. (a) Let the three orbits be considered as three concentric spheres A, B & C.
Now, Charge of ‘A’ = 4 × 1.6 × 10
–16
c
Charge of ‘B’ = 2 ×1.6 × 10
–16
c
Charge of ‘C’ = 2 × 1.6 × 10
–16
c
As the point ‘P’ is just inside 1s, so its distance from centre = 1.3 × 10
–11
m
Electric field = 
2
0
x 4
Q
??
= 
2 11 12
19
) 10 3 . 1 ( 10 85 . 8 14 . 3 4
10 6 . 1 4
? ?
?
? ? ? ? ?
? ?
= 3.4 × 10
13 
N/C
(b) For a point just inside the 2 s cloud
Total charge enclosed = 4 × 1.6 × 10
–19
– 2 × 1.6 × 10
–19
= 2 × 1.6 × 10
–19
Hence, Electric filed, 
E
?
= 
2 11 12
19
) 10 2 . 5 ( 10 85 . 8 14 . 3 4
10 6 . 1 2
? ?
?
? ? ? ? ?
? ?
= 1.065 × 10
12
N/C ˜ 1.1 × 10
12
N/C
14. Drawing an electric field around the line charge we find a cylinder of radius 4 × 10
–2
m.
Given: ? = linear charge density
Let the length be l = 2 × 10
–6
c/m
We know 
0 0
Q
dl . E
?
?
?
?
?
?
?
? E × 2 ? r l = 
0
?
? ?
? E = 
r 2
0
? ? ?
?
For, r = 2 × 10
–2
m & ? = 2 × 10
–6
c/m
? E = 
2 12
6
10 2 14 . 3 2 10 85 . 8
10 2
? ?
?
? ? ? ? ?
?
= 8.99 × 10
5
N/C ? 9 ×10
5
N/C
?
–q
a
Q 
+q
10
–15
m
5.2×10
–11 
m
C
A
N
B
2S
P
1S
1.3×10
–11 
m
2×10
-6
c/m
l
4 cm
Page 4


30.1
CHAPTER – 30
GAUSS’S LAW
1. Given : E
?
= 3/5 E
0
i
ˆ
+ 4/5 E
0 j
ˆ
E
0 
= 2.0 × 10
3
N/C The plane is parallel to yz-plane.
Hence only 3/5 E
0
i
ˆ
passes perpendicular to the plane whereas 4/5 E
0 j
ˆ
goes 
parallel. Area = 0.2m
2
(given)
? Flux = A E
? ?
? = 3/5 × 2 × 10
3
× 0.2 = 2.4 × 10
2
Nm
2
/c = 240 Nm
2
/c
2. Given length of rod = edge of cube = l
Portion of rod inside the cube = l/2
Total charge = Q.
Linear charge density = ? = Q/l of rod.
We know: Flux ? charge enclosed. 
Charge enclosed in the rod inside the cube.
= l/2 ?
0
× Q/l = Q/2 ?
0
?
3. As the electric field is uniform.
Considering a perpendicular plane to it, we find that it is an equipotential surface. Hence 
there is no net current flow on that surface. Thus, net charge in that region is zero.
4. Given: E = i
ˆ
E
0
?
?
l= 2 cm, a = 1cm.
E
0
= 5 × 10
3
N/C. From fig. We see that flux passes mainly through surface 
areas. ABDC & EFGH. As the AEFB & CHGD are paralled to the Flux. Again in 
ABDC a = 0; hence the Flux only passes through the surface are EFGH. 
E = i
ˆ
x E
c
?
Flux = 
L
E
0
?
× Area = 
2
3
a
a 10 5
?
? ?
?
= 
?
3 3
a 10 5 ? ?
= 
2
3 3
10 2
) 01 . 0 ( 10 5
?
?
?
? ?
= 2.5 × 10
–1
Flux = 
0
q
?
so, q = ?
0
× Flux 
= 8.85 × 10
–12
× 2.5 × 10
–1
= 2.2125 × 10
–12
c
5. According to Gauss’s Law Flux = 
0
q
?
Since the charge is placed at the centre of the cube. Hence the flux passing through the 
six surfaces = 
0
6
Q
?
× 6 = 
0
Q
?
6. Given – A charge is placed o a plain surface with area = a
2
, about a/2 from its centre.
Assumption : let us assume that the given plain forms a surface of an imaginary cube. Then the charge 
is found to be at the centre of the cube.
Hence flux through the surface = 
6
1 Q
0
?
?
= 
0
6
Q
?
7. Given: Magnitude of the two charges placed = 10
–7
c.
We know: from Gauss’s law that the flux experienced by the sphere is only due to 
the internal charge and not by the external one.
Now 
?
?
?
0
Q
ds . E
?
= 
12
7
10 85 . 8
10
?
?
?
= 1.1 × 10
4
N-m
2
/C.
X
i
ˆ
×
Z
Y
× × × ×
× × × × ×
× × × × ×
× × × × ×
× × × ×
× × ×
j
ˆ
k
ˆ
l
l/2 l/2
E
?
B F
o,a,o
H
E
G
o
A
D
C
a,o,o
o,o,a
R
P Q
+
+
2R
Gauss’s Law
30.2
8. We know: For a spherical surface 
Flux = 
?
?
?
0
q
ds . E
?
[by Gauss law]
Hence for a hemisphere = total surface area = 
2
1 q
0
?
?
= 
0
2
q
?
9. Given: Volume charge density = 2.0 × 10
–4
c/m
3
In order to find the electric field at a point 4cm = 4 × 10
–2
m from the centre let us assume a concentric 
spherical surface inside the sphere.
Now, 
?
?
?
0
q
ds . E
But ? = 
3
R 3 / 4
q
?
so, q = ? × 4/3 ? R
3
Hence = 
2 2
0
3 2
) 10 4 ( 7 / 22 4
1 ) 10 4 ( 7 / 22 3 / 4
?
?
? ? ?
?
?
? ? ? ? ?
= 2.0 × 10
–4
1/3 × 4 × 10
–2
× 
12
10 85 . 8
1
?
?
= 3.0 × 10
5
N/C ?
10. Charge present in a gold nucleus = 79 × 1.6 × 10
–19
C
Since the surface encloses all the charges we have:
(a) 
? ?
?
?
? ?
?
?
?
12
19
0
10 85 . 8
10 6 . 1 79 q
ds . E
?
E = 
ds
q
0
?
= 
2 15 12
19
) 10 7 ( 14 . 3 4
1
10 85 . 8
10 6 . 1 79
? ?
?
? ? ?
?
?
? ?
[ ?area = 4 ?r
2
]
= 2.3195131 × 10
21
N/C
(b) For the middle part of the radius. Now here r = 7/2 × 10
–15
m
Volume = 4/3 ? r
3
= 
45
10
8
343
7
22
3
48
?
? ? ?
Charge enclosed = ? × volume [ ? : volume charge density]
But ?= 
volume Net
e arg ch Net
= 
45
19
10 343
3
4
c 10 6 . 1 9 . 7
?
?
? ? ? ? ?
?
?
?
?
?
? ?
Net charged enclosed = 
45
45
19
10
8
343
3
4
10 343
3
4
10 6 . 1 9 . 7
?
?
?
? ? ? ?
? ? ? ? ?
?
?
?
?
?
? ?
= 
8
10 6 . 1 9 . 7
19 ?
? ?
?
ds E
?
= 
0
enclosed q
?
? E = 
S 8
10 6 . 1 9 . 7
0
19
? ? ?
? ?
?
= 
30 12
19
10
4
49
4 10 85 . 8 8
10 6 . 1 9 . 7
? ?
?
? ? ? ? ? ?
? ?
= 1.159 × 10
21
N/C
11. Now, Volume charge density = 
? ?
3
1
3
2
r r
3
4
Q
? ? ? ?
? ? = 
? ?
3
1
3
2
r r 4
Q 3
? ?
Again volume of sphere having radius x = 
3
x
3
4
?
Q
4 cm
r 1
O
? ? r 2
Gauss’s Law
30.3
Now charge enclosed by the sphere having radius
? = 
3
1
3
2
3
1
3
r
3
4
r
3
4
Q
r
3
4
3
4
? ? ?
? ?
?
?
?
?
?
? ? ?? = Q
?
?
?
?
?
?
?
?
?
? ?
3
1
3
2
3
1
3
r r
r
Applying Gauss’s law – E×4 ??
2
= 
0
enclosed q
?
? E = 
2 3
1
3
2
3
1
3
0
4
1
r r
r Q
??
?
?
?
?
?
?
?
?
?
?
? ?
?
= 
?
?
?
?
?
?
?
?
?
? ?
? ??
3
1
3
2
3
1
3
2
0
r r
r
4
Q
?
12. Given: The sphere is uncharged metallic sphere.
Due to induction the charge induced at the inner surface = –Q, and that outer surface = +Q.
(a) Hence the surface charge density at inner and outer surfaces = 
area surface total
e arg ch
= –
2
a 4
Q
?
and 
2
a 4
Q
?
respectively.
(b) Again if another charge ‘q’ is added to the surface. We have inner surface charge density = –
2
a 4
Q
?
, 
because the added charge does not affect it.
On the other hand the external surface charge density = 
2
a 4
q
Q
?
? as the ‘q’ gets added up.
(c) For electric field let us assume an imaginary surface area inside the sphere at a distance ‘x’ from 
centre. This is same in both the cases as the ‘q’ in ineffective.
Now, 
?
?
?
0
Q
ds . E So, E = 
2
0
x 4
1 Q
?
?
?
= 
2
0
x 4
Q
??
13. (a) Let the three orbits be considered as three concentric spheres A, B & C.
Now, Charge of ‘A’ = 4 × 1.6 × 10
–16
c
Charge of ‘B’ = 2 ×1.6 × 10
–16
c
Charge of ‘C’ = 2 × 1.6 × 10
–16
c
As the point ‘P’ is just inside 1s, so its distance from centre = 1.3 × 10
–11
m
Electric field = 
2
0
x 4
Q
??
= 
2 11 12
19
) 10 3 . 1 ( 10 85 . 8 14 . 3 4
10 6 . 1 4
? ?
?
? ? ? ? ?
? ?
= 3.4 × 10
13 
N/C
(b) For a point just inside the 2 s cloud
Total charge enclosed = 4 × 1.6 × 10
–19
– 2 × 1.6 × 10
–19
= 2 × 1.6 × 10
–19
Hence, Electric filed, 
E
?
= 
2 11 12
19
) 10 2 . 5 ( 10 85 . 8 14 . 3 4
10 6 . 1 2
? ?
?
? ? ? ? ?
? ?
= 1.065 × 10
12
N/C ˜ 1.1 × 10
12
N/C
14. Drawing an electric field around the line charge we find a cylinder of radius 4 × 10
–2
m.
Given: ? = linear charge density
Let the length be l = 2 × 10
–6
c/m
We know 
0 0
Q
dl . E
?
?
?
?
?
?
?
? E × 2 ? r l = 
0
?
? ?
? E = 
r 2
0
? ? ?
?
For, r = 2 × 10
–2
m & ? = 2 × 10
–6
c/m
? E = 
2 12
6
10 2 14 . 3 2 10 85 . 8
10 2
? ?
?
? ? ? ? ?
?
= 8.99 × 10
5
N/C ? 9 ×10
5
N/C
?
–q
a
Q 
+q
10
–15
m
5.2×10
–11 
m
C
A
N
B
2S
P
1S
1.3×10
–11 
m
2×10
-6
c/m
l
4 cm
Gauss’s Law
30.4
15. Given :
? = 2 × 10
–6
c/m
For the previous problem.
E = 
r 2
0
? ?
?
for a cylindrical electricfield.
Now, For experienced by the electron due to the electric filed in wire = centripetal 
force.
Eq = mv
2
?
?
?
?
?
?
?
?
? ?
? ?
?
radius assumed r ?, v
, kg 10 1 . 9 m , know we
e
31
e
?
2
1
Eq = 
r
mv
2
1
2
? KE = 1/2 × E × q × r = 
2
1
× 
r 2
0
? ?
?
× 1.6 × 10
–19
= 2.88 × 10
–17
J. ?
16. Given: Volume charge density = ?
Let the height of cylinder be h.
? Charge Q at P = ? × 4 ??
2
× h
For electric field 
?
?
?
0
Q
ds . E
E = 
ds
Q
0
? ?
= 
h 2
h 4
0
2
? ? ? ? ? ? ?
? ?? ? ?
= 
0
2
?
??
?
17.
?
?
?
0
Q
dA . E
Let the area be A. 
Uniform change distribution density is ?
Q = ?A
E = dA
Q
0
?
?
= 
A
a
0
? ?
? ? ? ?
= 
0
?
??
18. Q = –2.0 × 10
–6
C Surface charge density = 4 × 10
–6
C/m
2
We know E
?
due to a charge conducting sheet = 
0
2 ?
?
Again Force of attraction between particle & plate
= Eq =
0
2 ?
?
× q = 
12
6 6
10 8 2
10 2 10 4
?
? ?
? ?
? ? ?
= 0.452N
19. Ball mass = 10g
Charge = 4 × 10
–6
c
Thread length = 10 cm
Now from the fig, T cos ? = mg
T sin ? = electric force
Electric force = 
0
2
q
?
?
( ? surface charge density)
T sin ? = 
0
2
q
?
?
, T cos ??= mg
Tan ? = 
0
mg 2
q
?
?
? = 
q
tan mg 2
0
? ?
= 
6
3 12
10 4
732 . 1 8 . 9 10 10 10 85 . 8 2
?
? ?
?
? ? ? ? ? ?
= 7.5 × 10
–7
C/m
2
l
P x
?
0<x<d
?
d
x
×
×
60°
T Cos ? ?
mg
10 cm
T Sin ??
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
Page 5


30.1
CHAPTER – 30
GAUSS’S LAW
1. Given : E
?
= 3/5 E
0
i
ˆ
+ 4/5 E
0 j
ˆ
E
0 
= 2.0 × 10
3
N/C The plane is parallel to yz-plane.
Hence only 3/5 E
0
i
ˆ
passes perpendicular to the plane whereas 4/5 E
0 j
ˆ
goes 
parallel. Area = 0.2m
2
(given)
? Flux = A E
? ?
? = 3/5 × 2 × 10
3
× 0.2 = 2.4 × 10
2
Nm
2
/c = 240 Nm
2
/c
2. Given length of rod = edge of cube = l
Portion of rod inside the cube = l/2
Total charge = Q.
Linear charge density = ? = Q/l of rod.
We know: Flux ? charge enclosed. 
Charge enclosed in the rod inside the cube.
= l/2 ?
0
× Q/l = Q/2 ?
0
?
3. As the electric field is uniform.
Considering a perpendicular plane to it, we find that it is an equipotential surface. Hence 
there is no net current flow on that surface. Thus, net charge in that region is zero.
4. Given: E = i
ˆ
E
0
?
?
l= 2 cm, a = 1cm.
E
0
= 5 × 10
3
N/C. From fig. We see that flux passes mainly through surface 
areas. ABDC & EFGH. As the AEFB & CHGD are paralled to the Flux. Again in 
ABDC a = 0; hence the Flux only passes through the surface are EFGH. 
E = i
ˆ
x E
c
?
Flux = 
L
E
0
?
× Area = 
2
3
a
a 10 5
?
? ?
?
= 
?
3 3
a 10 5 ? ?
= 
2
3 3
10 2
) 01 . 0 ( 10 5
?
?
?
? ?
= 2.5 × 10
–1
Flux = 
0
q
?
so, q = ?
0
× Flux 
= 8.85 × 10
–12
× 2.5 × 10
–1
= 2.2125 × 10
–12
c
5. According to Gauss’s Law Flux = 
0
q
?
Since the charge is placed at the centre of the cube. Hence the flux passing through the 
six surfaces = 
0
6
Q
?
× 6 = 
0
Q
?
6. Given – A charge is placed o a plain surface with area = a
2
, about a/2 from its centre.
Assumption : let us assume that the given plain forms a surface of an imaginary cube. Then the charge 
is found to be at the centre of the cube.
Hence flux through the surface = 
6
1 Q
0
?
?
= 
0
6
Q
?
7. Given: Magnitude of the two charges placed = 10
–7
c.
We know: from Gauss’s law that the flux experienced by the sphere is only due to 
the internal charge and not by the external one.
Now 
?
?
?
0
Q
ds . E
?
= 
12
7
10 85 . 8
10
?
?
?
= 1.1 × 10
4
N-m
2
/C.
X
i
ˆ
×
Z
Y
× × × ×
× × × × ×
× × × × ×
× × × × ×
× × × ×
× × ×
j
ˆ
k
ˆ
l
l/2 l/2
E
?
B F
o,a,o
H
E
G
o
A
D
C
a,o,o
o,o,a
R
P Q
+
+
2R
Gauss’s Law
30.2
8. We know: For a spherical surface 
Flux = 
?
?
?
0
q
ds . E
?
[by Gauss law]
Hence for a hemisphere = total surface area = 
2
1 q
0
?
?
= 
0
2
q
?
9. Given: Volume charge density = 2.0 × 10
–4
c/m
3
In order to find the electric field at a point 4cm = 4 × 10
–2
m from the centre let us assume a concentric 
spherical surface inside the sphere.
Now, 
?
?
?
0
q
ds . E
But ? = 
3
R 3 / 4
q
?
so, q = ? × 4/3 ? R
3
Hence = 
2 2
0
3 2
) 10 4 ( 7 / 22 4
1 ) 10 4 ( 7 / 22 3 / 4
?
?
? ? ?
?
?
? ? ? ? ?
= 2.0 × 10
–4
1/3 × 4 × 10
–2
× 
12
10 85 . 8
1
?
?
= 3.0 × 10
5
N/C ?
10. Charge present in a gold nucleus = 79 × 1.6 × 10
–19
C
Since the surface encloses all the charges we have:
(a) 
? ?
?
?
? ?
?
?
?
12
19
0
10 85 . 8
10 6 . 1 79 q
ds . E
?
E = 
ds
q
0
?
= 
2 15 12
19
) 10 7 ( 14 . 3 4
1
10 85 . 8
10 6 . 1 79
? ?
?
? ? ?
?
?
? ?
[ ?area = 4 ?r
2
]
= 2.3195131 × 10
21
N/C
(b) For the middle part of the radius. Now here r = 7/2 × 10
–15
m
Volume = 4/3 ? r
3
= 
45
10
8
343
7
22
3
48
?
? ? ?
Charge enclosed = ? × volume [ ? : volume charge density]
But ?= 
volume Net
e arg ch Net
= 
45
19
10 343
3
4
c 10 6 . 1 9 . 7
?
?
? ? ? ? ?
?
?
?
?
?
? ?
Net charged enclosed = 
45
45
19
10
8
343
3
4
10 343
3
4
10 6 . 1 9 . 7
?
?
?
? ? ? ?
? ? ? ? ?
?
?
?
?
?
? ?
= 
8
10 6 . 1 9 . 7
19 ?
? ?
?
ds E
?
= 
0
enclosed q
?
? E = 
S 8
10 6 . 1 9 . 7
0
19
? ? ?
? ?
?
= 
30 12
19
10
4
49
4 10 85 . 8 8
10 6 . 1 9 . 7
? ?
?
? ? ? ? ? ?
? ?
= 1.159 × 10
21
N/C
11. Now, Volume charge density = 
? ?
3
1
3
2
r r
3
4
Q
? ? ? ?
? ? = 
? ?
3
1
3
2
r r 4
Q 3
? ?
Again volume of sphere having radius x = 
3
x
3
4
?
Q
4 cm
r 1
O
? ? r 2
Gauss’s Law
30.3
Now charge enclosed by the sphere having radius
? = 
3
1
3
2
3
1
3
r
3
4
r
3
4
Q
r
3
4
3
4
? ? ?
? ?
?
?
?
?
?
? ? ?? = Q
?
?
?
?
?
?
?
?
?
? ?
3
1
3
2
3
1
3
r r
r
Applying Gauss’s law – E×4 ??
2
= 
0
enclosed q
?
? E = 
2 3
1
3
2
3
1
3
0
4
1
r r
r Q
??
?
?
?
?
?
?
?
?
?
?
? ?
?
= 
?
?
?
?
?
?
?
?
?
? ?
? ??
3
1
3
2
3
1
3
2
0
r r
r
4
Q
?
12. Given: The sphere is uncharged metallic sphere.
Due to induction the charge induced at the inner surface = –Q, and that outer surface = +Q.
(a) Hence the surface charge density at inner and outer surfaces = 
area surface total
e arg ch
= –
2
a 4
Q
?
and 
2
a 4
Q
?
respectively.
(b) Again if another charge ‘q’ is added to the surface. We have inner surface charge density = –
2
a 4
Q
?
, 
because the added charge does not affect it.
On the other hand the external surface charge density = 
2
a 4
q
Q
?
? as the ‘q’ gets added up.
(c) For electric field let us assume an imaginary surface area inside the sphere at a distance ‘x’ from 
centre. This is same in both the cases as the ‘q’ in ineffective.
Now, 
?
?
?
0
Q
ds . E So, E = 
2
0
x 4
1 Q
?
?
?
= 
2
0
x 4
Q
??
13. (a) Let the three orbits be considered as three concentric spheres A, B & C.
Now, Charge of ‘A’ = 4 × 1.6 × 10
–16
c
Charge of ‘B’ = 2 ×1.6 × 10
–16
c
Charge of ‘C’ = 2 × 1.6 × 10
–16
c
As the point ‘P’ is just inside 1s, so its distance from centre = 1.3 × 10
–11
m
Electric field = 
2
0
x 4
Q
??
= 
2 11 12
19
) 10 3 . 1 ( 10 85 . 8 14 . 3 4
10 6 . 1 4
? ?
?
? ? ? ? ?
? ?
= 3.4 × 10
13 
N/C
(b) For a point just inside the 2 s cloud
Total charge enclosed = 4 × 1.6 × 10
–19
– 2 × 1.6 × 10
–19
= 2 × 1.6 × 10
–19
Hence, Electric filed, 
E
?
= 
2 11 12
19
) 10 2 . 5 ( 10 85 . 8 14 . 3 4
10 6 . 1 2
? ?
?
? ? ? ? ?
? ?
= 1.065 × 10
12
N/C ˜ 1.1 × 10
12
N/C
14. Drawing an electric field around the line charge we find a cylinder of radius 4 × 10
–2
m.
Given: ? = linear charge density
Let the length be l = 2 × 10
–6
c/m
We know 
0 0
Q
dl . E
?
?
?
?
?
?
?
? E × 2 ? r l = 
0
?
? ?
? E = 
r 2
0
? ? ?
?
For, r = 2 × 10
–2
m & ? = 2 × 10
–6
c/m
? E = 
2 12
6
10 2 14 . 3 2 10 85 . 8
10 2
? ?
?
? ? ? ? ?
?
= 8.99 × 10
5
N/C ? 9 ×10
5
N/C
?
–q
a
Q 
+q
10
–15
m
5.2×10
–11 
m
C
A
N
B
2S
P
1S
1.3×10
–11 
m
2×10
-6
c/m
l
4 cm
Gauss’s Law
30.4
15. Given :
? = 2 × 10
–6
c/m
For the previous problem.
E = 
r 2
0
? ?
?
for a cylindrical electricfield.
Now, For experienced by the electron due to the electric filed in wire = centripetal 
force.
Eq = mv
2
?
?
?
?
?
?
?
?
? ?
? ?
?
radius assumed r ?, v
, kg 10 1 . 9 m , know we
e
31
e
?
2
1
Eq = 
r
mv
2
1
2
? KE = 1/2 × E × q × r = 
2
1
× 
r 2
0
? ?
?
× 1.6 × 10
–19
= 2.88 × 10
–17
J. ?
16. Given: Volume charge density = ?
Let the height of cylinder be h.
? Charge Q at P = ? × 4 ??
2
× h
For electric field 
?
?
?
0
Q
ds . E
E = 
ds
Q
0
? ?
= 
h 2
h 4
0
2
? ? ? ? ? ? ?
? ?? ? ?
= 
0
2
?
??
?
17.
?
?
?
0
Q
dA . E
Let the area be A. 
Uniform change distribution density is ?
Q = ?A
E = dA
Q
0
?
?
= 
A
a
0
? ?
? ? ? ?
= 
0
?
??
18. Q = –2.0 × 10
–6
C Surface charge density = 4 × 10
–6
C/m
2
We know E
?
due to a charge conducting sheet = 
0
2 ?
?
Again Force of attraction between particle & plate
= Eq =
0
2 ?
?
× q = 
12
6 6
10 8 2
10 2 10 4
?
? ?
? ?
? ? ?
= 0.452N
19. Ball mass = 10g
Charge = 4 × 10
–6
c
Thread length = 10 cm
Now from the fig, T cos ? = mg
T sin ? = electric force
Electric force = 
0
2
q
?
?
( ? surface charge density)
T sin ? = 
0
2
q
?
?
, T cos ??= mg
Tan ? = 
0
mg 2
q
?
?
? = 
q
tan mg 2
0
? ?
= 
6
3 12
10 4
732 . 1 8 . 9 10 10 10 85 . 8 2
?
? ?
?
? ? ? ? ? ?
= 7.5 × 10
–7
C/m
2
l
P x
?
0<x<d
?
d
x
×
×
60°
T Cos ? ?
mg
10 cm
T Sin ??
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
Gauss’s Law
30.5
l
2 cm
20. (a) Tension in the string in Equilibrium 
T cos 60° = mg
? T = 
? 60 cos
mg
= 
2 / 1
10 10 10
3
? ?
?
= 10
–1
× 2 = 0.20 N
(b) Straingtening the same figure.
Now the resultant for ‘R’
Induces the acceleration in the pendulum.
T = 2 × ?
g
?
? = 2 ?
2 / 1
2
0
2
m 2
q
g
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= 2 ?
2 / 1
2
2
10 2
3
2 . 0 100
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
= 2 ?
2 / 1
) 300 100 ( ?
?
= 2 ?
20
?
= 2 × 3.1416 × 
20
10 10
2 ?
?
= 0.45 sec. ?
21. s = 2cm = 2 × 10
–2
m, u = 0, a = ? t = 2 ?s = 2 × 10
–6
s
Acceleration of the electron, s= (1/2) at
2
2 × 10
–2
= (1/2) × a × (2 × 10
–6
)
2
? a = 
12
2
10 4
10 2 2
?
?
?
? ?
? a = 10
10
m/s
2
The electric field due to charge plate = 
0
?
?
Now, electric force = 
0
?
?
× q = acceleration = 
e 0
m
q
?
?
?
Now
e 0
m
q
?
?
?
= 10
10
? ? = 
q
m 10
e 0
10
? ? ?
= 
19
31 12 10
10 6 . 1
10 1 . 9 10 85 . 8 10
?
? ?
?
? ? ? ?
= 50.334 × 10
–14
= 0.50334 × 10
–12
c/m
2
22. Given: Surface density = ?
(a) & (c) For any point to the left & right of the dual plater, the electric field is zero.
As there are no electric flux outside the system.
(b) For a test charge put in the middle.
It experiences a fore 
0
2
q
?
?
towards the (-ve) plate. 
Hence net electric field 
0 0 0
2
q
2
q
q
1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
23. (a) For the surface charge density of a single plate.
Let the surface charge density at both sides be ?
1
& ?
2
= Now, electric field at both ends.
= 
0
2
0
1
2
&
2 ?
?
?
?
Due to a net balanced electric field on the plate 
0
2
0
1
2
&
2 ?
?
?
?
? ?
1
= ?
2
So, q
1
= q
2
= Q/2
? Net surface charge density = Q/2A
+
+
+
+
+
+
+
–
–
–
–
–
–
–
60
m
10 
Eq ?
m
Eq
60
R
60
A
Q
Y X
? 1
Q
? 2
Read More
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