Download mtg jee previous year question and more Exams Physics in PDF only on Docsity!
1.2 Electric Charges 1.3 Conductors and Insulators 1.4 Charging by Induction 1.5 Basic Properties of Electric Charge 1.6 Coulomb’s Law 1.7 Forces between Multiple Charges 1.8 Electric Field
1.9 Electric Field Lines 1.10 Electric Flux 1.11 Electric Dipole 1.12 Dipole in a Uniform External Field 1.13 Continuous Charge Distribution 1.14 Gauss’s Law 1.15 Applications of Gauss’s Law
Electric Charges
and Fields
Topicwise Analysis of Last 10 Years’ CBSE Board Questions (2016-2007)
8 Maximum weightage is of Applications of Gauss’s Law. 8 Maximum VSA type questions were asked from Gauss’s Law.
8 Maximum SA II and LA type questions were asked from Applications of Gauss’s Law. 8 No VBQ type questions were asked till now.
7 6 5 4 3 2 1 0
VSA SA I SA II VBQ^ LA
Topic
Number of questions
1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.
2 CBSE Chapterwise-Topicwise Physics
8 Charge :^ Electric charge is an intrinsic property of elementary particles of matter which gives rise to electric force between various objects. X Quantization : Charge is always in the form of an integral multiple of electronic charge and never its fraction. q = ±ne where n is an integer and e = 1.6 × 10 –19^ C. X Millikan’s oil drop experiment showed the discrete nature of charge. Charge cannot be fractional multiple of e. X Conservation of charge : Net charge of an isolated physical system always remains constant. Charge can neither be created nor destroyed. It can be transferred from one body to another. X Electric charge is additive, i.e., total charge is the algebraic sum of the individual charges. X Electric charge is invariant as it does not depend upon the motion of the charged body or the observer. 8 Coulomb’s^ inverse^ square^ law^ :^ It^ states that the electrostatic force of attraction or repulsion acting between two stationary point charges is given by
F
q q
r
1 2
pe^2
where F denotes the force between two charges q 1 and q 2 separated by a distance r in free space. e 0 is a constant known as permittivity of free space. Free space is vacuum and may be taken to be air practically. 1 4 pe 0
= 9 × 10^9 N m C
2 2
X If free space is replaced by a medium, thene 0 is replaced by (e 0 K) or (e 0 er) where K is known as dielectric constant or relative permittivity. F
q q r K
q q r
q q r r
1 2 2 0
1 2 2 0
1 2 pe pe pe e^2
K = e
e 0
or er = e
e 0
K = 1 for vacuum (or air) and K = ∞ for conductor/metal.
e 0 = 8.85 × 10–12^ C^2 N–1^ m–2^ and its dimensional formula is [M–1L–3T^4 A^2 ]. Vector form of the law (q 1 and q 2 are like charges) (i)
F
q q
r r
12 r^ r
0
1 2 1 2
3 1 2
pe | |
1 2 21
pe^3
q q
r
r
(ii)
F
q q
r r
r r
q q
r
21 r
0
1 2 2 1
3 2 1
1 2 (^0 )
3 12
pe | | pe
F 12 F 21
- q 2 8 Electrostatic force due to continuous charge distribution The region in which charges are closely spaced is said to have continuous distribution of charge. It is of three types : X Linear charge distribution
dq = l dl where, l = linear charge density
dF q^ dq r
r dF
q dl r
r
0 (^2 )
0 pe pe^2
( ) l | |
Net force on charge q 0 , (^) F q^ dl r l r
= (^0) ∫ 4 pe 0 2
l | | X Surface charge distribution
r q 0
dS dF
dq = s dS where, s = surface charge density
Net force on charge q 0 , (^) F q^ dS r S r
= (^0) ∫ 4 pe 0 2
s | |
QUICK RECAP
4 CBSE Chapterwise-Topicwise Physics
X Two forces [qE and (– qE)] equal and opposite, separated by a distance constitute a couple (torque).
X Torque on a dipole = pE sin q numerically. Vectorially,
Torque ( )
t =
p ×E
X The direction oft is perpendicular to the plane
containing
p and
E.
X The torque tends to align the dipole in the direction of field. X Torque is maximum whenq = 90° i.e., dipole is perpendicular to E. \ Maximum torque = pE. When q = 0° or 180° then tmin = 0. X When dipole is parallel to electric field, it is in stable equilibrium. When it is antiparallel to electric field, it is in unstable equilibrium. 8 Gauss’s law :^ For a closed surface enclosing a net charge q, the net electric flux f emerging out is
given by f
e
= (^) ∫ ⋅ =
E dS^
q
S 0 X If a dipole isenclosed by a closed surface, flux f is equal to zero. Here the algebraic sum of charges (+ q – q = 0) is zero. X The electric field lines due to positive and negative charges and their combinations are :
(i) (ii)
(iii)
(iv) X Flux from a cube (i) If q is at the centre of cube, total flux
(f) = q
e 0
(ii) From each face of cube, flux =
q
X Electric field due to a thin, infinitely long straight wire of uniform linear charge density l,
E
r
0
l
2 pe
, where r is the perpendicular
distance of the observation point from the wire. X Electric field due to uniformly charged thin spherical shell of uniform surface charge density s and radius R at a point distant r from the centre of the shell is given as follows : At a point outside the shell i.e., r > R,
E
q
r
0
4 pe^2
At a point on the shell i.e., r = R,
E
q
R
0
4 pe^2
At a point inside the shell i.e., r < R,
E = 0
Here, q = 4pR^2 s X Electric field due to a thin non conducting infinite sheet of charge with uniform surface charge density s is
E =
s
X Electric field between two infinite thin plane parallel sheets of uniform surface charge density s and – s is
E = s/e 0. X Gaussian surface
- For a sphere or spherical shell : A concentric sphere.
- For cylinder or an infinite rod : A coaxial cylinder.
- For a plate : A cube or a cuboid.
Electric Charges and Fields 5
1.6 Coulomb’s Law
VSA (1 mark)
1. Two equal balls having equal positive charge ‘q’ coulombs are suspended by two insulating strings of equal length. What would be the effect on the force when a plastic sheet is inserted between the two? (AI 2014)
SA I (2 marks)
2. Plot a graph showing the variation of coulomb
force (F) versus
r^2
^
^
, where r is the distance
between the two charges of each pair of charges : (1mC, 2mC) and (2mC, – 3mC), interpret the graphs obtained. (AI 2011)
3. An infinite number of charges, each of q coulomb, are placed along x-axis at x = 1m, 3m, 9m and so on. Calculate the electric field at the point x = 0, due to these charges if all the charges are of the same sign. (Delhi 2009)
1.8 Electric Field
SA I (2 marks)
4. A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at large distances from the ring, it behaves like a point charge. (Delhi 2007)
LA (5 marks)
5. Consider a system of n charges q 1 , q 2 .... qn with position vectors r 1 , r 2 , r 3 , .... rn relative to some origin ‘O’. Deduce the expression for the net electric field
E at a point P with position vector rp due to this system of charges. (3/5, Foreign 2015)
1.9 Electric Field Lines
VSA (1 mark)
6. Why do the electrostatic field lines not form closed loops? (AI 2014, AI 2012C) 7. Why do the electric field lines never cross each other? (AI 2014)
SA I (2 marks)
8. The electric field
E due to a point charge at any point near it is defined as
E = lim
q
F
→ 0 q
, where q
is the test charge and
F is the force acting on
it. What is the physical significance of lim
q → 0
in
this expression? Draw the electric field lines of a point charge Q when (i) Q > 0 and (ii) Q < 0. (Delhi 2007)
SA II (3 marks)
9. A point charge (+Q) is kept in the vicinity of an uncharged conducting plate. Sketch the electric field lines between the charge and the plate. (1/3, Foreign 2014)
1.10 Electric Flux
VSA (1 mark)
10. Write an expression for the flux Df, of the electric field
E through an area element D
S.
(Delhi 2010C)
SA I (2 marks)
11. (i) Define the term ‘electric flux’. Write its SI unit. (ii) What is the flux due to electric field (^) E = 3 × 103 iN/C through a square of side 10 cm, when it is held normal to
E?
(AI 2015C)
12. Given a uniform electric field
E = 5 × 103 iN/C. Find the flux of this field through a square of 10 cm on a side whose plane is parallel to the
Previous Years’ CBSE Board Questions^ PREVIOUS YEARS MCQS
Electric Charges and Fields 7
27. What is the electric flux through a cube of side 1 cm which encloses an electric dipole? (Delhi 2015) 28. A charge ‘q’ is placed at the centre of a cube of side l. What is the electric flux passing through each face of the cube? (AI 2012) 29. Figure shows three point charges, +2q, –q, +3q. Two charges +2q and – q are enclosed within a surface ‘S’. What is the electric flux due to this configuration through the surface ‘S’? (Delhi 2010) 30. A charge Q mC is placed at the centre of a cube. What is the electric flux coming out from any one surface? (AI 2010) 31. If the radius of the Gaussian surface enclosing a charge is halved, how does the electric flux through the Gaussian surface change? (AI 2008)
SA I (2 marks)
32. Show that the electric field at the surface of a
charged conductor is given by
E = n
s
e 0
, where
s is the surface charge density and n^ is a unit
vector normal to the surface in the outward direction. (AI 2010)
33. Define electric flux. Write its S.I. unit. A charge q is enclosed by a spherical surface of radius R. If the radius is reduced to half, how would the electric flux through the surface change? (AI 2009) 34. A sphere S 1 of radius r 1 encloses a charge Q, if there is another concentric sphere S 2 of radius r 2 (r 2 > r 1 ) and there are no additional charges between S 1 and S 2. Find the ratio of electric flux through S 1 and S 2. (AI 2009) 35. Define electric flux. Write its S.I. unit. A spherical rubber balloon carries a charge that is uniformly distributed over its surface. As the balloon is blown up and increases in size, how does the total electric flux coming out of the surface change? Give reason. (Delhi 2007)
S
+ 2 q q
+ 3 q
SA II (3 marks)
36. A hollow cylindrical box of length 1 m and area of cross-section 25 cm 2 is placed in a three dimensional coordinate system as shown in the figure. The electric field in the region is given by
E = 50 xi , where^ E^ is in
N C–1^ and x is in metres. Find (i) net flux through the cylinder. (ii) charge enclosed by the cylinder.
1 m
z
O
y
x
(Delhi 2013)
37. State Gauss’s law in electrostatic. A cube with each side ‘a’ is kept in an electric field given by (^) E = Cxi, (as is shown in the figure) where^ C^ is a positive dimensional constant. Find out
x
y
z
a a
(i) the electric flux through the cube (ii) the net charge inside the cube. (Foreign 2012)
LA (5 marks)
38. Given the electric field in the region
E = 2 xi , find the
electric flux through the cube and the charge enclosed by it.
(2/5, Delhi 2015)
39. Define electric flux. Write its S.I. unit. “Gauss’s law in electrostatics is true for any closed surface, no matter what its shape or size is”. Justify this statement with the help of a suitable example. (AI 2015)
a
x
y
z
8 CBSE Chapterwise-Topicwise Physics
40. Consider two hollow concentric spheres S 1 and S 2 , enclosing charges 2Q and 4Q respectively as shown in figure. (i) Find out the ratio of the electric flux through them. (ii) How will the electric flux through the sphere S 1 change if a medium of dielectric constant ‘er’ is introduced in the space inside S 1 in place of air? Deduce the necessary expression.
2 Q
S 1
4 Q S 2
(AI 2014)
41. (a) Define electric flux. Write its SI units. (b) The electric field components due to a charge inside the cube of side 0.1m are as shown: E (^) x = a x, where a = 500 N/C-m Ey = 0, Ez = 0. Calculate (i) the flux through the cube, and (ii) the charge inside the cube.
0.1 m
X
Y
Z
O 0.1 m
(AI 2008)
1.15 Applications of Gauss’s Law
VSA (1 mark)
42. Two charges of magnitudes –2Q and +Q are located at points (a, 0) and (4a, 0) respectively. What is the electric flux due to these charges through a sphere of radius ‘3a’ with its centre at the origin? (AI 2013)
SA I (2 marks)
43. A small metal sphere carrying charge +Q is located at the centre of a spherical cavity inside a large uncharged metallic spherical shell as
shown in the figure. Use Gauss’s law to find the expressions for the electric field at points P 1 and P 2.
P 2
(AI 2014)
44. Two concentric metallic spherical shells of radii R and 2R are given charges Q 1 and Q 2 respectively. The surface charge densities on the outer surfaces of the shells are equal. Determine the ratio Q 1 : Q 2. (Foreign 2013) 45. A spherical conducting shell of inner radius r 1 and outer radius r 2 has a charge Q. A charge q is placed at the centre of the shell. (a) What is the surface charge density on the (i) inner surface, (ii) outer surface of the shell? (b) Write the expression for the electric field at a point x > r 2 from the centre of the shell. (AI 2010)
SA II (3 marks)
46. Two infinitely large plane thin parallel sheets having surface charge densities s 1 and s 2 (s 1 > s 2 ) are shown in the figure. Write the magnitudes and directions of the net fields in the regions marked II and III.
σ 1
I
II A
B
III
σ 2
(Foreign 2014)
47. (i) State Gauss’s law. (ii) A thin straight infinitely long conducting wire of linear charge density ‘l’ is enclosed by a cylindrical surface of radius ‘r’ and length ‘l’. Its axis coinciding with the length of the
10 CBSE Chapterwise-Topicwise Physics
1. As in air, F q
r
2
pe^2
In medium, F ′ =
K
q
r
2
pe^2
\ F ′ =F
K
where K is dielectric constant of material and
K > 1 for insulators. Hence, the force is reduced, when a plastic sheet is inserted.
2.
(i) Pair (1mC, 2mC) : From upper graph it is clear that the force of repulsion increases with the reducing distance between two charges.
(ii) Pair (2mC, –3mC) : From lower graph it is clear that the force of attraction increases as the distance between two charges reduces.
3. Here, r 1 = 1m, r 2 = 3m, r 3 = 9m
Electric .field,
E = q + + + + ∞
pe 0 ( )^2 ( )^2 ( ) 92
= ×
q q 4
pe 0 pe 0 8 using^ S^
a ∞ (^) r
{ (^1) − }
1 pe
q (^) NC
4. Suppose total charge on ring of radius a is q. Charge q is uniformly distributed. We want to find electric field at point P on the axis of the charged ring. Consider a small element of the ring carrying charge dq. Electric field due to this small element is
dE
dE
can be resolved into two components as (i) dE cosq along PX and (ii) dE sinq along PY. Due to symmetry of ring all components of electric fields of small elements along y-axis cancel out. Resultant electric field at point P, E = (^) ∫ dEcos q
Here, dE
dq
r
dq
x a
pe 0 2 4 pe 0 ( 2 2 )
cos q = =
x
r
x
x 2 a^2
\ E dq
x a
x
x a
= ×
×
∫
4 pe 0 ( 2 2 ) ( 2 2 )
4 pe 0 2 2 3 2
x
x a
dq
E qx
x a
4 pe 0 ( 2 2 3 2) /
For large x as x >> a, so a^2 can be neglected,
\ E qx
x
q
x
pe 0 3 4 pe 0 2
which is the electric intensity due to a point charge at a distance x. Hence charged ring behaves as a point charge for points at large distances from it.
5. Let us consider a system of n charges q 1 , q 2 , q 3 ,... qn with position vectors r 1 , r 2 , r 3 , .... r (^) n relative to origin O.
Detailed Solutions
Electric Charges and Fields 11
Let
Fi be the force due to ith^ charge qi on q 0 Then, (^) F
q q r i i r i
= (^) i
0 pe^2
Here, r (^) i is the distance of the test charge q 0 from qi The electric field at the observation point P is given by
E F^
q q
q q r i r q
i q
i i
= = ⋅ i
lim 0 → 0 lim 0 → 0 0 0 0
0 2
4 pe (^) E
q r
i ir i
= ⋅ i
4 pe 0 2 ...(i) If
E is the electric field at point P due to the system of charges, then by the principle of superposition of electric fields, E E E E E (^) n Ei i
n = + + + + = =
1 2 3 ∑ 1
Using (i), we get (^) E
q r
ir i
i i
n = ⋅ =
∑
1 4 pe^0
=
∑
4 pe 0 12
q r
ir i
i i
n
6. (^) Electrostatic field lines do not form closed loops
due to conservative nature of electric field.
7. At the point of intersection of two field lines, there will be two directions for the resultant electric field. This is not acceptable.
8. lim
q → 0
represents that the charge q in
E F q q
→
lim 0
, is a test charge of infinitely small is
magnitude, so that it may not alter the electric field of the source charge.
(i) Q > 0
(ii) Q < 0
10. Electric flux Df =
E ⋅ D
S = E DS cos q.
11. (i) Electric flux : Total number of electric field lines crossing a surface normally is called electric flux. SI unit of electric flux is N m^2 C–1. (ii) The area of a surface can be represented as a vector along normal to the surface. Here
E = 3 × 103 i^ NC−^1 Area of the square DS = 10 × 10 cm^2 DS = 100 cm^2 = 10–2^ m^2 Since normal to the square is along x-axis, we have DS = 10–2^ i^ m^2 Electric flux through the square, f = ⋅ = × ⋅ −
^
E DS ( 3 10 3 i^ ) ( 10 2 i) f = 30 N m^2 C–
12. Here,
E = 5 × 10 3 i^ N/C Side of square = a = 10 cm = 0.1 m Area of square, S = a^2 = (0.1)^2 = 0.01 m^2 Case I : Area vector is along x-axis, S = 0 01. i 2
^ m Required flux, f = ⋅
E S
⇒ f = ( × ) ( .⋅ )
^ ^ 5 10 3 i 0 01i ⇒ f = 50 N m^2 /C Case II : Plane of the square makes a 30° angle with the x-axis. Here, angle between area vector and the electric field is 60°. So, required flux f′ = E ⋅S cos q = (5 × 10^3 )(10–2) cos60° = 25 N m^2 /C
13. Given electric field
E = 3 × 10^3 i^ NC – Magnitude of area, S = 10 cm^2 = 1 × 10–3^ m^2 (i) (^) When the surface is parallel to y-z plane, the normal to plane is along x-axis. normal to plane is along x-axis. In this case q = 0; so electric flux, f = ⋅ = × ⋅ × −
E S ( 3 10 3 i^ ) ( 1 103 i^ )= 3 N m^2 C–
S
E S
S
=60°
y
O
z
S x
Electric Charges and Fields 13
19. When q = 0° between p and
E , the dipole is in
the most stable equilibrium state in uniform external field.
20. Dipole in a uniform external field
Consider an electric dipole consisting of charges –q and +q and of length 2a placed in a uniform electric field
E making an angle q with electric field.
Force on charge –q at A = −q E
(opposite to
E )
Force on charge +q at B = qE
(along
E )
Electric dipole is under the action of two equal and unlike parallel forces, which give rise to a torque on the dipole. t = Force × Perpendicular distance between the two forces t = qE(AN) = qE(2a sinq) t = q(2a)E sinq t = pE sinq t = (^) p^ ×E Pairs of perpendicular vectors (a) ( ,^ t p) (b) (^) ( ,^ t E)
21. Refer to answer 20.
(i) When q = 0; t = 0 and
p (^) and
E are parallel and the dipole is in a position of stable equilibrium. (ii) When q = 180°, t = 0 and
p and
E are antiparallel and the dipole is in a position of unstable equilibrium.
22. Refer to answer 20.
23. (a) Suppose an electric dipole of dipole moment
p is placed along a direction, making an angle q
with the direction of an external uniform electric field
E. Then, the torque acting on the dipole is
defined as pE sinq or
t = p ×E.
Its direction will be perpendicular to both
p and
E.
(b) If the field is non-uniform there would be a net force on the dipole in addition to the torque and the resulting motion would be a combination of translation and rotation.
(b) If the field is non-uniform there would be a net force on the dipole in addition to the torque and the resulting motion would be a combination of translation and rotation. (c) (i)
E is increasing parallel to^
p then q = 0°.
So torque becomes zero but the net force on the dipole will be in the direction of increasing electric field and hence it will have linear motion along the dipole moment. (ii)
E is increasing anti-parallel to^
p. So, the torque
still remains zero but the net force on the dipole will be in the direction of increasing electric field which is opposite to the dipole moment, hence it will have linear motion opposite to the dipole moment.
24. Refer to answer 20. 25. Electric field due to a system of charges : Consider a system of charges q 1 and q 2 with position vectors
r 1 and
r 2 relative to common origin O. Let P be any point with position vector r at which electric field is to be determined.
Electric field
E 1 due to q 1 is given by (^) E q r
r p
1 p 0
1 1
2 1
pe where r^ (^1) p is a unit vector in the direction from q 1 to P and r 1 p is the distance between q 1 and P. Similarly, electric field
E 2 due to q 2 is (^) E
q r
r p
2 p 0
2 2
2 2
pe where r^ (^2) pis a unit vector in the direction from q 2 to P and r 2 p is the distance between q 2 and P. By the superposition principle, the electric field
E at r^ due to the system of charges is E (r) = E^ r E r 1 ( )^ + 2 ( ) = 1 ⋅ ⋅ + ⋅ ⋅ 4
1 1
(^2 1 )
2 2 pe pe^2
q r
r q r
r p
p p
p
\
E r q r
r q r
r p
p p
( ) = + p
1 1
2 1
2 2 pe^2
14 CBSE Chapterwise-Topicwise Physics
26. According to Gauss’s law, the electric flux passing through a closed surface is given by
E ds
q
∫. =^
enclosed
e 0
When radius of spherical Gaussian surface is increased, its surface area will be increased but point charge enclosed in the sphere remains same. Hence there will be no change in the electric flux.
27. According to Gauss’s law, net flux through a
closed surface, f^ E E ds^ een
q
= (^) ∫ ⋅ =
(^0) Total charge enclosed, qen = 0 as net charge on dipole is zero.
\ fE =
e 0
28. By Gauss’s theorem, total flux through whole of
the cube, f
e
q
0 where, q is the total charge enclosed by the cube. As, charge is at centre, therefore, electric flux is symmetrically distributed on all 6 faces. Therefore,
Flux through each face of the cube, f′ =^ =
f
q
29. Total charge within a surface S = + 2q + (– q) = + q
\ Electric flux f = q e 0
30. Refer to answer 28. 31. The electric flux remains the same, as the charge enclosed remains the same. 32. Consider an elementary area dS on the surface of the charged conductor. Enclose this area element with a cylindrical gaussian surface as shown in figure. Now electric field inside a charged conductor is zero. Therefore, direction of field, just outside dS will be
normally outward i.e. in direction of n.
According to Gauss’s theorem, total electric flux coming out is
E S
S
⋅ d =
s d
e 0
[
E is electric field at the surface]
R
S
Gaussian surface
⇒ E dS s dS
e
cos
0
° = ⇒^ E^ =
s
e 0
33. Electric flux linked with a surface is the number of electric lines of force cutting through the surface normally. It’s SI unit is N m 2 C –1^ or V m. On decreasing the radius of spherical surface to half there will be no effect on the electric flux. 34. By Gauss's theorem, total flux passing through a closed surface, f e
q 0 \ f f e
1 e 2
1 0
0 2
1 2
= × = = = 1
q q
q q
Q
Q
\ f 1 : f 2 = 1 : 1.
35. Refer to answer 33. The total electric flux linked with surface of balloon remains the same because the charge on its surface remains the same on blowing it up. 36. (i)
1 m z
O
y
A B x
Given,
E = 50 x iand^ A^ = 25 cm^2 = 25 × 10–4^ m^2
As the electric field is only along the x-axis, so, flux will pass only through the cross-section of cylinder. Magnitude of electric field at cross-section A, EA = 50 × 1 = 50 N C– Magnitude of electric field at cross-section B, EB = 50 × 2 = 100 N C– The corresponding electric fluxes are fA =
E A ⋅ A= 50 × 25 × 10–4^ cos 180°
= – 0.125 N m^2 C– fB =
E B ⋅ A= 100 × 25 × 10–4^ cos 0°
= 0.25 N m^2 C– So, the net flux through the cylinder, f = fA + fB = –0.125 + 0.25 = 0.125 N m^2 C– (ii) Using Gauss’s law
E dA^
⋅ =q
∫ (^) e 0 ⇒^ 0 125.^ =^ 8 85. × 10 − 12
q
⇒ q = 8.85 × 0.125 × 10–12^ = 1.1 × 10–12^ C
37. (i) Gauss’s law in electrostatics states that the total electric flux through a closed surface enclosing
16 CBSE Chapterwise-Topicwise Physics
(a) (i) Surface charge density on the inner surface
of the shell is s
in p
= −q
4 r 12
(ii) Surface charge density on the outer surface of
shell is s
out p
Q +q
4 r 22
(b) Using, Gauss’s law, E x
Q q
x
4 pe 0 2
B
σ 1
I A
B
III
σ 2
II
A
B
A
In region II : The electric field due to the sheet of charge A will be from left to right (along the positive direction) and that due to the sheet of charge B will be from right to left (along the negative direction). Therefore, in region II, we have
E = + − ^
s e
s e
1 0
2 0
⇒
E = 1 −
0 e (s^1 s^2 )^ along positive direction
In region III : The electric fields due to both the charged sheets will be from left to right, i.e., along the positive direction. Therefore, in region III, we have
E = s + e
s e
1 0
2 0
⇒ E
0 e (s^1 s^2 )^ along positive direction
47. (i) According to Gauss’s law, total flux over a
closed surface S in vacuum is 1
e 0
times the total
charge enclosed by closed surface S
f
e
= (^) ∫ ⋅ =
E ds^
q
s
enclosed 0 (ii) Electric field intensity due to line charge or infinite long uniformly charged wire at point P at
distance r from it is obtained as : Assume a cylindrical Gaussian surface S with charged wire on its axis and point P on its surface, then net electric flux through surface S is
f = ⋅ = (^) ∫ ° + (^) ∫ °
E ds^ Eds^ Eds s
cos 90 cos 0
upper plane face
curved surface
∫∫
- ∫ Eds^ cos 90° lower plane face or f = 0 + EA + 0 or f = E ⋅ 2 prl
But by Gauss’s theorem, f =
q l
e
l
0 e 0
where q is the charge on length l of wire enclosed by cylindrical surface S, and l is uniform linear charge density of wire.
\ E × 2prl =
l
e
l
0
or E =
l
2 pe 0 r
directed normal to the surface of charged wire.
48. Consider a thin spherical shell of radius R carrying charge q. To find the electric field outside the shell, we consider a spherical Gaussian surface of radius r (> R), concentric with given shell. The electric field
E is same at every point of
Gaussian surface and directed radially outwards
(as is unit vector n^ so that q = 0°)
According to Gauss’s theorem,
^
E ds^ E n ds^
q
s s ∫ ⋅^ =^ ∫ ⋅^ =
e 0
or E ds
q
E r
q
s
\ =
∫ (^) e
p
e
0 2 0
R
ds
O
r
P
E^
q
Gaussian surface n ^ r
Electric Charges and Fields 17
E
q
r
4 pe 02
Vectorially,
E
q r
= r
4 pe 0 2 Special cases (i) At the point on the surface of the shell, r = R
\ (^) E q R
4 pe 0 2
(ii) If s is the surface charge density on the shell then q = 4pR^2 s
\ (^) E R R
0
2 pe^2
p s s e
(iii) If the point P lies inside the spherical shell then the Gaussian surface encloses no charge i.e., r < R ... (^) q = 0, hence E = 0
O
E
r = R
r < R
r > R
Distance from centre ( ) r
49. (i) Refer to answer 47 (i). (ii) Assume a cylindrical Gaussian surface S cutting through plane sheet of charge, such that point P lies on its plane face, then net electric flux through surface S is
f = ⋅ = (^) ∫ ⋅ + (^) ∫ ⋅ + ⋅
E ds E ds E ds E ds
left plane face
curved surface
riight plane face
∫ ∫ S
or left plane face
curved surface
f = (^) ∫ Eds ° + (^) ∫ Eds ° +
Eds
cos cos
cos
∫^00 ° right plane face
or f = EA + 0 + EA = 2EA But by Gauss’s theorem f =
q A
e
s
0 e 0
where q is the charge in area A of sheet enclosed by cylindrical surface S and s is uniform surface charge density of sheet.
\ 2 EA = s
e
A
0
or E =
s
directed normal to surface of charged sheet (i) away from it, if it is positively charged and (ii) towards it, if it is negatively charged.
50. Refer to answer 47. 51. + q
Refer to answer 49(ii).
52. Consider two infinite plane parallel sheets of charge A and B. Let s 1 = +s and s 2 = –s be the uniform surface densities of charge on A and B respectively.
A
(^)
B The electric field between two plates is given by
E = E − E = − = −
^
1 2 1 0
2 (^2 2 0 2 0 2 0 )
s e
s e
s e
s e
s s e \ (^) E = 2 ⇒ E= (^2 0 )
s e
s e
53. Refer to answer 49. 54. Refer to answer 48. 55. Consider a sphere of radius r with centre O surrounded by a large concentric conducting shell of radius R.
q r^ P^ Q x R
To calculate the electric field intensity at any point P, where OP = x, imagine a Gaussian surface with
2.2 Electrostatic Potential 2.3 Potential due to a Point Charge 2.4 Potential due to an Electric Dipole 2.5 Potential due to a System of Charges 2.6 Equipotential Surfaces 2.7 Potential Energy of a System of Charges 2.8 Potential Energy in an External Field 2.9 Electrostatics of Conductors
2.10 Dielectics and Polarisation 2.11 Capacitors and Capacitance 2.12 The Parallel Plate Capacitor 2.13 Effect of Dielectric on Capacitance 2.14 Combination of Capacitors 2.15 Energy Stored in a Capacitor 2.16 Van de Graaff Generator
Topicwise Analysis of Last 10 Years’ CBSE Board Questions (2016-2007)
8 Maximum weightage is of Energy Stored in a Capacitor. 8 Maximum VSA and SA I type questions were asked from Equipotential surfaces.
8 Maximum SA II type questions were asked from Energy Stored in a Capacitor.
Electrostatic Potential
and Capacitance
7 6 5 4 3 2 1 0
VSA SA I SA II VBQ (^) LA
Topic
Number of questions 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14 2.15 2.
8
9
20 CBSE Chapterwise-Topicwise Physics
(^8) Electric potential : Electric potential at a point is defined as amount of work done in bringing a unit positive charge from infinity to that point. It is denoted by symbol V.
V W q
X Electric potential is a scalar quantity. The SI unit of potential is volt and its dimensional formula is [ML^2 T–3A–1]. X Electric potential at a point distantr from a point charge q is
V q r
4 pe 0
X Electric potential due to group of charges : The electric potential at a point due to a group of charges is equal to the algebraic sum of the electric potentials due to individual charges at that point.
V
q r
q r
q r
q r q r
n n i i i
n
^
=
∑
0
1 2 2
3
(^0 )
pe
pe
1 3
X Electric potential at any point due to an electric dipole
The electric potential at point P due to an electric dipole
V
p r
p r r
pe 0 2 4 0 2
q pe
cos ^
- At axial point : When the pointP lies on the axial line of dipole i.e., q = 0°.
V
p r
4 pe 0 2
- At equatorial point : When the pointP lies
on the equatorial line of the dipole, i.e., q = 90° \ V = 0. X Electric potential due to a uniformly charged spherical shell of uniform surface charge density s and radius R at a distance r from the centre the shell is given as follows :
- At a point outside the shelli.e., r > R
V R r
q r
= s = e pe
2
0 0
- At a point on the shelli.e., r = R
V R^ q R
= s = e 0 pe 0
- At a point inside the shelli.e., r < R
V R^ q R
= s = e 0 pe 0
Here, q = 4pR^2 s The variation of V with r for a uniformly charged thin spherical shell is shown in the figure.
X Electric potential due to a non-conducting solid sphere of uniform volume charge density r and radius R at distant r from the sphere is given as follows : X At a point outside the spherei.e., r > R
V
R
r
q r
r e pe
3 (^30 )
X At a point on the spherei.e., r = R
V R^ q R
= r = e pe
2 (^30 )
X At a point inside the spherei.e., r < R
V
R r (^) q R r R
r − 3 e pe
2 2
0
2 2 3
Here q^ =^ R
p 3 r
QUICK RECAP
