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CPP
ELECTROSTATICS –Gauss law)
LEVEL – I
1. A rectangular surface of 2 metre width and 4 metre length, is placed in an electric field of intensity 20
newton/C, there is an angle of 60º between the perpendicular to surface and electrical field intensity. Find the
total flux emitted from the surface. (in Volt- metre)

2. Mark the correct options:

(A) Gauss’s law is valid only for symmetrical charge distributions.
(B) Gauss’s law is valid only for charges placed in vacuum
(C) The electric field calculated by Gauss’s law is the field due to all the charges.
(D) The flux of the electric field through a closed surface due to all the charges is equal to the flux due to the
charges enclosed by the surface.

3. When no charge is confined with in the Gauss’s surface, it may implies that-
(A) E = 0
 
(B) E and ds are parallel
   
(C) E and ds are mutually perpendicular (D) E and ds are inclined at some angle

4. If the electric flux entering and leaving an enclosed surface respectively is 1 and 2, find the electric charge
inside the surface.

5. For which of the following fields, Gauss’s law is valid-

(A) fields following square inverse law (B) uniform field
(C) all types of field (D) this law has no concern with the field

6. If electric field flux coming out of a closed surface is zero, the electric field at the surface will be-
(A) zero (B) same at all places
(C) dependent upon the location of points (D) infinites

7. If three electric di-poles are placed in some closed surface, find the electric flux emitting from the surface.

8. A charge q is inside a closed surface and charge – q is outside. Find the outgoing electric flux is-

9. Three charges q1 = 1c, q2 =2c and q3 = –3c and four surfaces S1, S2, S3 and S4 are shown. Find the flux
emerging through surface S2 in N-m2 / C.

10. A square of side 20cm. is enclosed by a surface of sphere of 80 cm. radius. Square and sphere have the
same centre. four charges +2 × 10–6 c, –5 × 10–6 c, –3 × 10–6 c, +6 × 10–6c are located at the four corners of
a square, find the outgoing total flux from spherical surface in N-m2/C.
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11. A charged particle q is placed at the centre O of cube of length L (ABCDEFGH). Another same charge q is
placed at a distance L from O. Find the electric flux through BCFG.

12. In the given figure , charges q1 and -q1 are inside a Gaussian surface. Where as charge q2 is outside the
surface. Electric field on the Gaussian surface will be

(A) only due to q2

(B) zero on the Gaussian surface
(C) uniform on the Gaussian surface
(D) due to all

LEVEL – II 2Q S1
S4 +Q
1. Four closed surfaces S1, S2, S3 and S4 together with the charges 2Q, Q and Q are S3
shown in the figure. Find the electric flux through each surface. Q

S2

4
2. A charge of 1.2  10 C is placed inside a cylinder at the mid-point of the axis of cylinder. The flux through
one end of the cylinder is 4.5  10 N  m / C . What is the electric flux through the curved part of the
6 2

cylindrical surface?

3. Charge is uniformly distributed in a space. The net flux passing through the surface of an imaginary cube of
side “a” in the space is . What will be the net flux passing through the surface of an imaginary sphere of
radius “a” in the space.

4. A charge Q is uniformly distributed over a rod of length l. Consider a hypothetical cube of edge l with the
centre of the cube at one end of the rod. (a) Find the minimum possible flux of the electric field through the
entire surface of the cube. (b) Find the maximum possible flux of the electric field through the entire surface of
the cube.

5. Determine the electric flux through the closed surfaces A and B shown in
the figure below. Take, q1  1.0  10 10 C and q 2  1.0  10 10 C .

6. A point charge Q is located on the axis of a disc of radius R at a distance b from the plane of the disc (figure).
Show that if one-fourth of the electric flux from the charge passes through the disc, then R = 3b .
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7. A point charge q is kept on the vertex of the cone of base radius r and height r
r. Find the electric flux through the curved surface.

8. A disk of radius a/4 having a uniformly distributed charge 6C is placed in the x-y plane with its centre at (–a/2,
0, 0). A rod of length ‘a’ carrying a uniformly distributed charge 8C is placed on the x-axis from x = a/4 to x =
5a/4. Two point charges –7C and 3C are placed at (a/4, –a/4, 0) and (–3a/4, 3a/4, 0) respectively. Consider a
cubical surface formed by six surfaces x = ± a/2, y = ±a/2, z = ±a/2. Find the electric flux through this cubical
surface.
y

9. The electric field everywhere on the surface of a thin spherical shell of radius 0.750 m is measured to be
equal to 890 N/C and points radially towards the centre of the sphere. (a) What is the net charge within the
sphere’s surface? (b) Find the total electric flux passing through the shell.

10. A charge q is placed just above the centre of a horizontal circle of radius ‘r’, and a hemisphere of this radius is
erected about the charge. Compute the electric flux through the closed surface that consists of hemisphere
and the plane circle, using Gauss’s law. Verify this result without using Gauss’s law.

11. Consider a cube of side ‘a = 1m’ located at the origin. An electric field given by E  bx 2 N / C ˆi , where b =
2.4  10 4 V / m 3 is a constant, is present in space. Calculate the net electric flux through the cube. What is
the net charge enclosed by the cubical surface.


12. The electric field in a particular space is E   x  2  N / C ˆi with x measured in meters. Consider a cylinder of
radius = 20 cm that is coaxial with the x-axis. One end of the cylinder is positioned at x = 0 and the other end
at x = 2.0 m. (a) what is the magnitude of the electric flux through the end of the cylinder at x = 2.0 m. (b)
What net charge is enclosed within the cylinder?

CPP
ELECTROSTATICS -SHEET: 7(Lecture – 7)
ANSWER KEY
Level-I
1. 80

1. Area of rectangular surface | A | 8m2

| E | 20N / C    60
 
  E  A
2 0
= 8m 20 N/C  cos 60
 = 80 v  m.
2. C, D 3. A, C, D
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4. (2 –1)/0(2  1) 0
qin
4. net  2  1 
0
qm 0  2  1 
5. A 6. C
7. Zero
7. Total charge enclosed by the surface = zero so, the electric flux net emiting through the surface will be zero.
8. q/ 0

9. –36  × 103

9.  (through S2)  q2  q3  2c  3c

0 0
1c
  4   4  10 6  k
4  0

  4  106  9  109  36   103 v  m .

10. Zero
11. Zero
q
11. Total flux coming out through the cube by the enclosed charge 
0
q
So, the total flux through BcFG  (due to the charge q at o) and due to the charge q at P outside the
6 0
q q q
cube through BcFG   so the net flux   0
6 0 6 0 6 0
12. D

Level-II

Q
1. 1  , 2  0, 3  0 and 4 = 0
0
qin 
1. 1  S1   
0 0
qin
2  S 2    zero
0
qin
3  S 3    zero
0
qin
4  S4    zero
0

6 2
2. 4.6 × 10 N – m / C
6 2
2. Flux through one end of cylinder = 4.5  10 nm /c through other end will be also equal due to symmetry
q 1.2  10 4
Total flux = in   4  9  109  1.2  10 4  4  4   10.8  105
0 4 0
5 6
Flux through the curved surface = net  total flux through the end  4  10.8  10  9  10
13.6  106  9  10 6  4.6  106 Nm2 / c
4
3. 
3
qin
3. 0 
0
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4 3
p R
0  3
0
4 3
p a
0  3
0
4  0
0  
3 0
4
0  
3
Flux through the cube
q
  in
0
Pa3

0
3
Pa = 0

4. (a) Q/20 (b) Q/ 20

4. Flux will be maximum when the rod will be placed along the diagonal of cube. Total length of the rod along the

diagonal keeping one end at centre inside the cube 
2


  1
(a) min (when the rod placed || to the edge  2   kt
0  2 0

min  .
2 0


(b) max 
qin
 2     1  
max 
0 0  2 0 2 0

5.  A  0 ; B  11.3V  m
qin q1  q2
5. A   0
0 0
qin q2 4q2
B   
0 0 4 0
 
 9  109  4  1010  11.3 V  m .

3b 0
6. Let assume R=3 b tan    3 d = 60 
b
Solid angle  = 2 (1  cos ) b
 = 2 (1  cos 600)
 1
 2  1    
 2  3b

Total flux due to Pt. charge   when the solid angle is 4 so, the flux for  solid
0

angle  proved.
4 0

7. zero
2C
8.
0
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qin qrod  qdisc  qPt c
8.  
0 0
2c  3c  ( 7c

0
2c  3c   7c  2
  c.
0 0
9. (a) 5.6 × 10-8C (b) 6.3 × 103 v – c

9. | E | 890N / C
r = 0.75 m

(a) angle b/w E and area vector is 
q
  ES  in
0
2
Qin =  E (4 0) r
9 1
 890   9
16 9  10
=  5.6  108 C
q 5.6  10 8  4
(b)   in  
0 4 0
  6.3  103 v  c

10. q/0
qin q
10.   q
0 0

11.   2.4  104 v  m qin = 0 

 
11. net  E  S
 2
S through the P/ane (x = 1) = 1 m

Ebi0
net = b v  m
4
= 2.4  10 v  m
qnet = 0
 
12. (a)   E  S y
2 2
2 Nm Nm
  2  2  ˆi    0.2  ˆi  0.16
C C
  qin
(b)   E  dS  x=0 x=2.0 m
0
qin
  x 2  x 0 
0
2 qin
 0.16  2 ()(0.2) =
0
 qin   0.080  C