Electric Charges and Fields
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An arbitrary surface encloses a dipole. What is the electric flux through this surface?
An arbitrary surface encloses a dipole. What is the electric flux through this surface?
This is a Short Answer Type Questions as classified in NCERT Exemplar
Explanation- Electric flux through a dipole is always zero, because the positive and negative charges cancel each other out
Total charge -Q is uniformly spread along the length of a ring of radius R. A small test charge +q of mass m is kept at the centre of the ring and is given a gentle push along the axis of the ring.
(a) Show that the particle executes a simple harmonic oscillation.
(b) Obtain its time period.
Total charge -Q is uniformly spread along the length of a ring of radius R. A small test charge +q of mass m is kept at the centre of the ring and is given a gentle push along the axis of the ring.
(a) Show that the particle executes a simple harmonic oscillation.
(b) Obtain its time period.
This is a Long Answer Type Questions as classified in NCERT Exemplar
Explanation- electric field at the axis of the ring is E=KQy/ (R2+z2)3/2 where z is distance .

F=qE=KQqy/ (R2+z2)3/2
When z<
F= = -Kz
So force is directly proportional to – distance that is completely defined that is follows S.H.M
(b)w
Two charges –q each are fixed separated by distance 2d. A third charge q of mass m placed at the mid-point is displaced slightly by x (x<<d) perpendicular="" to="" the="" line="" joining="" two="" fixed="" charged="" as="" shown="" in="" fig.="" 1.14.="" show="" that="" q="" will="" perform="" simple="" harmonic="" oscillation="" of="" time="" period.<="" strong=""></d)>
T=[8π3ɛo md3/q2]1/2
Two charges –q each are fixed separated by distance 2d. A third charge q of mass m placed at the mid-point is displaced slightly by x (x<<d) perpendicular="" to="" the="" line="" joining="" two="" fixed="" charged="" as="" shown="" in="" fig.="" 1.14.="" show="" that="" q="" will="" perform="" simple="" harmonic="" oscillation="" of="" time="" period.<="" strong=""></d)>
T=[8π3ɛo md3/q2]1/2
Explanation- two charge -q at A and B
AB=AO+OB=2d and x= small distance perpendicular to O

When x
F=qq/4 where AP=BP=r but horizontal components gets cancel out each other and vertical components gets add .
If angle APO=O the net force
There is another useful system of units, besides the SI/mks A system, called the cgs (centimeter-gram-second) system. In this system Coloumb’s law is given by F=Qq/r2 vector r where the distance r is measured in cm (= 10–2 m), F in dynes (=10–5 N) and the charges in electrostatic units (es units), where 1es unit of charge=1/[3]x10-9C The number [3] actually arises from the speed of light in vaccum which is now taken to be exactly given by c = 2.99792458 × 108m/s. An approximate value of c then is c = [3] × 108 m/s. (i) Show that the coloumb law in cgs units yields 1 esu of charge = 1 (dyne)1/2 cm. Obtain the dimensions of units of charge in terms of mass M, length L and time T. Show that it is given in terms of fractional powers of M and L. (ii) Write 1 esu of charge = x C, where x is a dimensionless number. Show that this gives

There is another useful system of units, besides the SI/mks A system, called the cgs (centimeter-gram-second) system. In this system Coloumb’s law is given by F=Qq/r2 vector r where the distance r is measured in cm (= 10–2 m), F in dynes (=10–5 N) and the charges in electrostatic units (es units), where 1es unit of charge=1/[3]x10-9C The number [3] actually arises from the speed of light in vaccum which is now taken to be exactly given by c = 2.99792458 × 108m/s. An approximate value of c then is c = [3] × 108 m/s. (i) Show that the coloumb law in cgs units yields 1 esu of charge = 1 (dyne)1/2 cm. Obtain the dimensions of units of charge in terms of mass M, length L and time T. Show that it is given in terms of fractional powers of M and L. (ii) Write 1 esu of charge = x C, where x is a dimensionless number. Show that this gives

This is a Long Answer Type Questions as classified in NCERT Exemplar
Explanation- as we know F= Qq/r2= 1dyne = 1esu of charge2/1cm2
1 esu of charge = M1/2L3/2T—1
- Q=xC where x is dimensionless quantity
So F= =1 dyne = 10-5N
Taking x= 1/3 9
After solving we get 1/4
Two fixed, identical conducting plates (α &β ) , each of surface area S are charged to –Q and q, respectively, where Q > q > 0. A third identical plate (γ ), free to move is located on the other side of the plate with charge q at a distance d . The third plate is released and collides with the plate β . Assume the collision is elastic and the time of collision is sufficient to redistribute charge amongst β &γ . (a) Find the electric field acting on the plate γ before collision. (b) Find the charges on β and γ after the collision.

Two fixed, identical conducting plates (α &β ) , each of surface area S are charged to –Q and q, respectively, where Q > q > 0. A third identical plate (γ ), free to move is located on the other side of the plate with charge q at a distance d . The third plate is released and collides with the plate β . Assume the collision is elastic and the time of collision is sufficient to redistribute charge amongst β &γ . (a) Find the electric field acting on the plate γ before collision. (b) Find the charges on β and γ after the collision.

This is a Long Answer Type Questions as classified in NCERT Exemplar
Explanation- net electric field at plate γ due to two other plate
From plate 1 ,E1=
From plate 2 ,E2=
Total electric field E= E1 + E2 = to the left , if Q>q
electric field at o due to plate α =
electric field at o due to plate β
Consider a sphere of radius R with charge density distributed as ρ (r ) = kr for r ≤ R for r > R .

(a) Find the electric field at all points r.
(b) Suppose the total charge on the sphere is 2e where e is the electron charge. Where can two protons be embedded such that the force on each of them is zero. Assume that the introduction of the proton does not alter the negative charge distribution.
Consider a sphere of radius R with charge density distributed as ρ (r ) = kr for r ≤ R for r > R .

(a) Find the electric field at all points r.
(b) Suppose the total charge on the sphere is 2e where e is the electron charge. Where can two protons be embedded such that the force on each of them is zero. Assume that the introduction of the proton does not alter the negative charge distribution.
This is a Long Answer Type Questions as classified in NCERT Exemplar
Explanation-
(i) when r
= and we know that V =
= 4
=
E(4 ) = = so it is clear that E is radially outwards.
When r>R
=
&nb
In 1959 Lyttleton and Bondi suggested that the expansion of the Universe could be explained if matter carried a net charge. Suppose that the Universe is made up of hydrogen atoms with a number density N, which is maintained a constant. Let the charge on the proton be: ep=−(1+y)e where e is the electronic charge. Find the critical value of y such that expansion may start
In 1959 Lyttleton and Bondi suggested that the expansion of the Universe could be explained if matter carried a net charge. Suppose that the Universe is made up of hydrogen atoms with a number density N, which is maintained a constant. Let the charge on the proton be: ep=−(1+y)e where e is the electronic charge. Find the critical value of y such that expansion may start
This is a Long Answer Type Questions as classified in NCERT Exemplar
Explanation- let us consider that universe is of radius R
And we know that hydrogen is made up one electron and proton so net charge is
&
Five charges,q each are placed at the comers of a regular pentagon of side a.

i) What will be the electric field at O, the centre of the pentagon?
(ii) What will be the electric field at O if the charge from one of the comers (say A) is removed?
(iii) What will be the electric field at O if the charge q at A is replaced by – q?
(b) How would your answer to (a) be affected if pentagon is replaced by n-sided regular polygon with charge q at each of its comers?
Five charges,q each are placed at the comers of a regular pentagon of side a.

i) What will be the electric field at O, the centre of the pentagon?
(ii) What will be the electric field at O if the charge from one of the comers (say A) is removed?
(iii) What will be the electric field at O if the charge q at A is replaced by – q?
(b) How would your answer to (a) be affected if pentagon is replaced by n-sided regular polygon with charge q at each of its comers?
This is a Long Answer Type Questions as classified in NCERT Exemplar
Explanation- (a)
(i) The electric field at the center of pentagon is zero because the distance from the center is same.
(ii) The field through one charge is Kq/r2
(iii) When one charge is positive and other is negative then ne
1.34 Suppose that the particle in Exercise in 1.33 is an electron projected with velocity = 2.0 × 106 m s–1. If E between the plates separated by 0.5 cm is 9.1 × 102 N/C, where will the electron strike the upper plate? (|e|=1.6 × 10–19 C, me = 9.1 × 10–31 kg.)
1.34 Suppose that the particle in Exercise in 1.33 is an electron projected with velocity = 2.0 × 106 m s–1. If E between the plates separated by 0.5 cm is 9.1 × 102 N/C, where will the electron strike the upper plate? (|e|=1.6 × 10–19 C, me = 9.1 × 10–31 kg.)
1.34 Velocity of the particle, = 2.0 m/s
Separation of the two plates, d = 0.5 cm = 0.005 m
Electric field between two plates, E = 9.1 N/C
Charge of an electron, q = 1.6 C
Mass of an electron, = 9.1 kg
Let the electron strike the upper plate at the end of the plate L, when d