NCERT Class 12 Physics Electric Charges and Fields Questions and Answers Important

BSEB Bihar Board Class 12 Physics Solutions Chapter 1 Electric Charges and Fields

NCERT Class 12 Physics Electric Charges and Fields

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NCERT  Class 12 Physics Electric Charges and Fields Questions and Answers

Question 1.
What is the force between two small charged spheres having charges of 2 x 10-7 C and 3 x 10-7 C placed 30 cm apart in air ?
Here., q1 = 2 x 10-7 C, q2 = 3 x 10-7 C, r = 30 cm = 0.30 m. F = ?
Using the relation,

= 6 x 10-3N.

Question 2.
The electrostatic force on a small sphere of charge 0.4 pC due to another small sphere of charge – 0.8 pC in air is 0.2 N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?

Here, q1 = 0.4 µC, q2 = – 0.8 µC = – 0.8 x 10-6 C = 0.4 x 10-6 C F = electrostatic force between q1 and q2 = 0.2 N.
(a) r = ?

= 16 x 9 x 10-4 = 144 x 10-4 m2
.’. r = 12 x 10-2 m = 0.12 m.

(b) Force on q2 due to q1 = ?
We know that electrostatic forces always, appear in priors and follow Newton’s 3rd law of motion.
I F21 ! = Force on q2 due to q1 = 0.2 N and is attractive in nature.

= 0.2 N.

Question 3.
Checks that the ratio ke2/Gmem is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
Dimensions of e2 are = [C2] Dimensions of k are = [Nm2 C-2 ] = [ML3T-2 C-2] Dimensions of G are = [M-1 L3 T-2] Dimensions of m2 = [M] ∴ Dimensions of \frac{\mathrm{Ke}^{2}}{\mathrm{Gm}_{\mathrm{e}} \mathrm{m}_{\mathrm{p}}} are obtamed as :

.’. \frac{\mathrm{Ke}^{2}}{\mathrm{Gm}_{\mathrm{e}} \mathrm{m}_{\mathrm{p}}} is a dimensionless quantity as it has no units.

Now using e = 1.6 x 10-19 C, k – 9 x 109 Nm2 C-2, G = 6.67 x 10-11 Nm2 kg-22, me = 9.1 x 1031 kg,
mp = 1.66 x 10-27kg, we get

= 2.29 x 1039
This factor represents the ratio of electrostatic force and the gravitational force between an electron and a proton.

Question 4.
(a) Explain the meaning of the statement ‘electric charge of a body is quantised’.
(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges ?
(a) The meaning of the statement ‘electric charge of a body is quantised’ is that the charge on it is always some integral multiple of elementary charge of an electron or a proton (= e in magnitude) i.e., charge on a body never varies continuously but it varies in the form of discrete packets called quanta or packets of charge. Mathematically, the charge on a body can be expressed as
q = ±ne
where n is an integer, e = magnitude of the charge of an electron or proton = 1.6 x 10-19 C. A fraction of the fundamental charge e has never been observed in free state.

(b) In practice, the charge on a charged body at macroscopic level is very large while the charge on an electron is very small. When electrons are added to or removed from a body, the change taking place in the total charge on the body is so small that the charge seems to be varying in a continuous manner. Thus quantisation of electric charge can be ignored at macroscopic level i.e., when dealing with a large scale charged body.

Question 5.
When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservaton of charge.
Initially i.e., before rubbing both the glass rod and silk cloth are electrically neutral. In other words, net charge On the glass rod and silk cloth is zero. When the glass rod is rubbed with silk cloth, a few electrons get transferred from the rod to the silk cloth, thus glass rod becomes positively charged and silk cloth negatively charged. The positive charge on the glass rod is exactly equal to the negative charge on the silk cloth, so net charge on the system is again zero.

Thus the appearance of charge on the glass rod and silk cloth is in accordance with the law of conservation of charge as the total charge of the isolated system is constant. Similarly when ebonite rod is rubbed with fur, they acquire – ve and + ve charges respectively and net charge is zero again. Thus we conclude that charge is neither created nor destroyed but it is merely transferred from one body to another which is consistent with the law of conservation of charge.

Question 9.
A system has two charges qA = 2.5 x 10-7 C and qB = – 2.5 x 10-7 C located at points A (0, 0, – 15 cm) and B (0, 0, + 15 cm) respectively. Wftat is the total charge and electric dipole moment of the system?
The charges qA and qH are located at points A (0,0, – 15 cm) and B (0,0, + 15 cm) on z -axis as shown in the figure here. They forrp an electric dipole.
d – total chrage = ?
p = electric dipole moment of the system = ?
p = q.2a.
Now q = qA + qB = 2-5 x 10-7 + (-2.5 x 10-7) = 0
Also 2a = AB = dipole length= OA + OB
= 15 + 15 = 30 cm
= 0.30 m.
∴ p = either charge x dipole length
= qA x AB = 2.5 x 10-7 x 0.30
= 7.5 x 10-8 cm
The electric dipole moment is directed from B to A i.e., along negative Z-axis.

Question 10.
An electric dipole with dipole moment 4 x 10-9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 x 10-4 NC-1. Calculate the magnitude of the torque acting on the dipole.
Here, p = 4 x 10-9 Cm, E = 5 x 104NC-1,θ = 30°, τ = ?
Using the formula, x = pE sin θ, we get
τ = 4 x 10-9 x 5 x 104 x sin 30°
= 20 x 10-5 x \frac {1}{2}
= 10-4 Nm.

Question 11.
A polythene piece rubbed with wool is found to have a negative charge of 3 x 10-7 C.
(a) Estimate the number of electrons transferred (from which to which?)
(b) Is there a transfer of mass from wool to polythene?
(a) Here, q = Total charge transferred = – 3 x 10-7 C. Charge on an electron, e = -1.6 x 10-19 C.
n = no. of electrons transferred = ?
As the polythene piece rubbed with wool is found to attain -ve charge, so the electrons are transferred from wool to polvihene piece.
From quantisa Lion of charge, we know that q = ne.
n = \frac {q}{e} = \frac{-3 \times 10^{7}}{-1.6 \times 10^{-19}} = 1.875 x 1012
= 2 x 1012

(b) Yes, there is a transfer of mass from wool to polythene as electrons are material particles and are transferred from wool to polythene piece.
m = mass of each electron = 9.1 x 10-13 kg,
n = 1.875 x 1012
M = total mass transferred to polythene = ?
= m x n
= 9.1 x 1031 x 1.875 x 1012 = 1.71 x 10-18kg ≈ 2 x 10-18 kg.

Question 12.
(a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of eiectrosatic repulsion if the charge on each is 6.5 x 10-7C? The radii of A and B are negligible compared to the distance of seperation.
(b) What is the force of repulsion of each sphere is charged double the above amount, and the distance between them is halved?
(a) Charge on 1st sphere, A = qA = 6.5 x 10-7C
Charge on 2nd sphere, B = qB = 6.5 x 10-7C
Distance between sphere A and B = 50 cm = 0.5m = d
We know that,

(b) If each sphere is charged double the amount, then
qA = qB = 2 x6.5 x 10-7C
= 13 x 10-7C
and r = \frac {1}{2} x 50cm = 0.25m
We know that,

Question 13.
Suppose the spheres A and B of previous exercise have identical sizes. A thud sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the force of repulsion between A and B ?
Initial charges on spheres A and B
= qA = qB = 6.5 x 10-7C
r = Distance between the spheres A and B = 0.5 m.
All the spheres will have equal charges on being brought in contact because they are of the same size.
When 3rd sphere C having charge e3 = zero is brought in contact with 1st sphere A, then ;
Charge on A = Charge on C = q 1say ,

or q1 = 3.25 x 10-7 C
When 3rd sphere C having charge 3.25 x 10-7 C is brought in contact with 2nd sphere B, then; charge left on B say q2 is given by –
Charge on B = Charge on C

or q2 = 4.875 x 10-7C
If F be the force of repulsion between spheres A and B and when sphere C is removed, then according to Coulomb’s law of electrostatic forces.

Question 14.
Figure below shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Charges on the two plates are shown in the figure. Since charged particles are deflected towards the oppositely charged plates, therefore (1) and (2) are -vely charged while particle (3) is +vely charged. Since all the three particles are crossing the same electric field with same speed, so they remain under the action of \overrightarrow{\mathrm{E}} for same time f (ray).
The deflection produced in the path of a charged particle along vertical
direction is given by, y = \frac {1}{2} at2 = \frac {1}{2} \frac{\mathrm{eE}}{\mathrm{m}} t2

Question 15.
Consider a uniform electric field E = 3 x 103 \hat { i } N/C.
(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane?
(b) What is the flux through the same square if the normal to its plane makes at 60° angle with the x-axis?
Here, E = 3 x 103 \hat { i } N/C-1i.e., the electric field acts along positive direction of x-axis.
Side of square = 10 cm
its surface area, ∆S = (10 cm)2 = 10-2 m3
or \frac{\mathrm{∆S}}{\mathrm{m}} = 10-2\hat { i }m2
as normal to the square is along x-axis.

(a) If Φ be the electric flux through the square, then
Φ = \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\Delta \mathrm{S}}
= (3 x 103 \hat { i }). (10-2 \hat { i })
= 3 x 103 x 10-2 \hat { i }. \hat { i } = 3x 10 = 30 Nm2C-1

(b) Here, angle between normal to the square i.e., area vector and the electric field is 60°.
∴ θ = 60°
∴ Φ = \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\Delta \mathrm{S}} = E. ∆S cos 60°
= 3 x 103 x 10-2 x \frac {1}{2}
= 15 Nm2 C-1.

Question 16.
What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?
Net flux over the cube is zero becuase the number of lines entering the cube of side 20 cm is same as the number of lines leaving the cube.

Question 17.
Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 x 103 Nm2/c
(i) What is the net charge inside the box?
(ii) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?
(i) Given, Φ = 8 x 103Nm2 C-1, ε0= 8.854 x 10-12 C2 N-1m-2
If the net charge inside the black box is q, then using formula
Φ = \frac{q}{\varepsilon_{0}} or we get, q = E0 Φ
or q = 8.854 x 10-12 x 8x 103C
q = 8.854 x 8 x 10-9 C
q = 70.832 x 10-9 C = 0.070832 x 10-6C = 0.071 pC

(ii) We cannot conclude that the net electric charge inside the box
is zero if the outward flux through the surface of black box is zero because there might be equal amounts of positive and negative charges cancelling each other and thus making the resultant charge equal to zero. Thus, we can only conclude that the net charge inside the box is zero.

Question 18.
A point charge +10 pC is a distance 5 cm directly above the centre of a square of side 10 cm as shown in figure. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)
The given square ABCD can be imagined as one of the side faces of a cube of side 0.10 m. The given charge can be imagined to be at the centre of this cube at a distance of 5 cm.
Here, q = + 10 pC = 10-5C.
Then according to Gauss’s Theorem, the total electric flux through all the 6 faces of the cube is given by
Φ = \frac{q}{\varepsilon_{0}}

If Φ be the electric flux through the square ABCD, then

Question 19.
A point charge of 2.0 qC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
Here, charge at the centre of the Gaussian surface,
q = 2μC = 2 x 10-6C.
E0 = 8.854 x 10-12 C2 N-1 m-2
Φ = electric flux through it =?
q = 2μC = 2 x 10-6C.
According to Gauss’s Theorem, the electric flux through the six faces of the cubes /.

Question 20.
A point charge causes an electric flux of – 1.0 x 103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge, (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?
Here, = electric flux through the spherical Gaussian surface
= -1.0 x 103 N m2 C-1.
r = radius of Gaussian spherical surfaces – 10 cm
Let q = charge enclosed at its centre.

(a) According to Gauss’s law, the electric flux through a Gaussian surface depends upon the charge enclosed inside the surface and not upon its size. Thus the electric flux will remain unchanged i.e., – 1.0 x 103 Nm2 C-1 through the spherical Gaussian surface of double radius i.e. of 20 cm as it also encloses the same amount of charge.

(b) q = point charge = ?, ε0 = 8.854 x 10-12 Nm-2 C2.
Using the formula,
Φ = \frac{q}{\varepsilon_{0}} we get
q = ε0 Φ = 8.854 x 10-12 x (-1.0 x 103) = – 8.854 x 10-9C = 8.8 nC.

Question 21.
A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 x 103 N/C and points radially inward, what is the net charge on the sphere?
Here R = radius of the conducting sphere = 0.10 m
r = distance of the point from centre of sphere = 20 cm = 0.2 m
Clearly r > R
E = electric field at a point at a distance of 20 cm from the sphere = 1.5 x 103 NC-1 acting inward.
q = net charge on the sphere = ?
Using the formula,

Also as E acts in the inward direction, so charge on the sphere is negative.
q = – 6.67 x 10-9 C = – 6.67 nC.

Question 22.
A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m2.
(a) Find the charge on the sphere, (b) What is the total electric flux leaving the surface of the sphere?
Here, a = surface charge density = 80.0 μ Cm2 = 80 x 10-6 Cm2
R = radius of the charged sphere = \frac {2.4}{2} = 1.2 m.
(a) q = charge on the sphere = ?
Using the relation, fa = σ \frac{q}{4 \pi R^{2}} , we get
q = 4πR2 x σ = 4 x \frac {22}{7} x (1-2)2 x 80 x 10-6
= 1.45 x 10-3 C.

(b) Φ = Total electric flux leaving the surface of the sphere = ?
Using Gauss’s Theorem, we get

Question 23.
An infinite line charge produces a field of 9 x 104 N/C at a distances of 2,cm. Calculate the linear charge density.
Here E = electric field produced by infinite line charge = 9 x 104 NC-1
r = distance of the point where E is produce = 2 cm = 0.02 m.
λ = linear charge density = ?
\frac{1}{4 \pi \varepsilon_{0}} = 9 x 109 N-1m-2C
Using the relation,
E = \frac{1}{2 \pi \varepsilon_{0}} \frac{λ}{r}
we get,

Question 24.
Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 x 10-22Cm2. What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?
Here a = surface charge density of plates
= 17.0 x 10-22 Cm-2
ε0 = 8.854 x 1012C2N1m-2

The arrangement of the plates is shown in the figure.
(a) Em the outerregion of the first plate E to the left of the first plate =?
Ther onlistotheeftof the fi(st plate thuselectric field E1 in this
region dueto the two plates is given by –
E1 = – E1 + (-E2)

E1 = 0

(b) E in the outer region of second plate = E to the right of second plate i.e., in region III = ?.
The region III, E is given by

(c) E between the plates = E in II region = EII = ? EII(I is given by
EII = EI+ (- E2)
(here E1 is +ve and E2 is – ve)
= E1 – E2

Question 25.
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 x 104 NC-1in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm-3. Estimate the radius of the drop, (g = 9.81 ms-2; e = 1.60 x 10-19 C).
Here, E = constant electric field = 2.55 x 104 Vm-1
e = charge of an e = 1.6 x 10’i9 C
n= no. of electrons = 12
If q = cha rge On the drop, then
q = ne = 12 x 1.6 x 10-19 C = 19.2 x 10-19 C.
If Fe be the electrostatic force on the oil drop due to electric field,
Then Fe – q E = 19.2 x 10-19 x 2.55 x 10-4 ……(1)
Also let Fg = Force on the drop due to gravity, then
Fg = mg = \frac {4}{3} πr3pg …….(2)
Here p = density of oil = 1.26 g cm3
= 1.26 x 10-3kg (10-2 m)-3
= 1.26 x 103 kg m-3
g = 9.81 ms-2
r = radius of the drop = ?
Putting these values in equation (2), we get –
Fg = \frac {4}{3} πr3 x 1.26 x 103 x 9.81 ……..(3)
As the drop remains stationary,
Fe = Fg
or electros tafic force = weight of the drop due to electric field
or 19.2 x 10-19 x 2.55 x 104 = \frac {4}{3} πr3 1.26 x 103 x 9.81

Question 26.
Which among the curves shown in figure cannot possibly represent electrostatic field lines?

Only (c) represents electric field lines.
(a) As the electrostatic field lines start or end only normally to the surface of the conductor, so figure (a) cannot represent electrostatic field lines as they are not normal to the surface.
(b) The electrostatic lines of force cannot start from negative charge and cannot end on positive charge, so fi^ufe ifb) cannot represent such lines.
(c) This figure represents electrostatic lines of force due to two isolated positive charges separated by some distance.
(d) As no two electrostatic lines of force intersect each other, so this figure cannot represent such lines.
(e) This figure does not represent electrostatic lines of force because they can’t form closed loop:

Question 27.
In a certian region of space, electric field is along the z- direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 105 NC-1 per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10-7 cm in the negative z-direction?
Let \overrightarrow{\mathrm{p}} = 2aq be the electric dipole moment of an electric dipole consisting of charges – q at A and + q at B and placed along z- axis. ,
\overrightarrow{\mathrm{p}} acts is negative z-direction
∴ its dipolemoment pz along z-direction is
pz = – 10-7cm.
The electric field is applied along the positive direction of z-axis, such that
\frac {dE}{dz} = 105 NC-1 m-1 ,F = ?, τ = Torque = ?
In a non-uniform electric field, the force on dipole is given by

The negative sign shows that force on the dipole is along negative z-axis.
Calculating of Torque (τ) – Both \overrightarrow{\mathrm{p}} and \overrightarrow{\mathrm{E}} act along negative z and + z axes respectively.
∴ 0 = 180°
Using the relation,
τ = pE sin 0, we get
τ = pE sin 180° = pE x 0
τ = 0
∴Torque on the dipole is zero.

Question 28.
(a) A conductor A with a cavity as shown in figure (a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor, (b) Another conductor B with charge q is i inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [figure (b)]. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

(a) Consider a Gaussian surface shown by dotted lines so as to enclose the cavity but lying wholly inside the conductor A shown in fig. a. Also we know that the electric field inside a conductor is zero, so the part of the conductor within the cavity does not contain any net charge. By extending the same argument, we can say that no charge can be present inside the Gaussian surface, which lies just within the conductor. Thus according to Gauss’s law

inside the conductor as E = 0 inside the Gaussian surface
∴ Q = 0 inside the Gaussian surface.
Hence, the entire charge Q must appear on the outer surface of the conductor A.

(b) Let us again consider a Gaussian surface shown by dotted lines in fig. b so that it encloses the cavity and the conductor B with charge q inserted in the cavity. As such, the electric flux will cross the Gaussian surface which means that the electric field exists inside the conductor A. But the electric field inside the conductor A must be zero. It will be true only if the charge q on the conductor B induces charge – q on the inner surface of the cavity and + q charge induced will move the outer surface of theconductor A- As there is already a charge + q on the outer surface of the conductor A, thus total charge on its outer surface becomes Q + q. Hence proved

(c) The electric field inside a hollowmetallic conductor is zero and it resides only on its outer surface. Therefore, the instrument can be shielded from the strong electrostatic fields by enclosing it with a hollow metallic structure or envelope.

Question 29.
A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (al2ej n, where n is the unit vector in the outward normal direction, and o is the surface charge density near the hole.
Let + σ be the surface charge density of the charged conductor, near the hole. Let A = area of the hole. The electric field is perpendicular to the plane sheet of charge and is directed in outward direction shown by \hat { i }
To find’ \overrightarrow{\mathrm{E}} in the hole, draw a Gaussian cylinder through, the hole.
Since no lines of force cross the side walls of the cylinder, so the component of \overrightarrow{\mathrm{E}} normal to the walls iz zero. At the ends of the cylinder, the normal component of \overrightarrow{\mathrm{E}} is equal to E.
Thus if Φ be the total electric flux through the Gaussian surface,

Also let q = charge enclosed in the Gaussian surface
∴ According to Gauss’s law,

From (1) and (2), we get

or in vector form E = \frac{\sigma}{2 \varepsilon_{0}} \hat{n}Hence proved.
Aliter: For a hollow conductor, the electric field at a point on its surface is given by
E = \frac{\sigma}{\varepsilon_{0}} \hat{n} and is directed outwards. and the \overrightarrow{\mathrm{E}} inside it is zero. If there is a hole on the surface of a conductor, then we say that the electric field at a point P is the sum of the E at P due to the field up hole and rest of the conductor.

Inside the conductor, the \overrightarrow{\mathrm{E}} due to filled up hole and the rest of the conductor is directed opposite so as to cancel each other, where as outside the conductor, the two fields act in the same direction and have the same magnitude. Let \overrightarrow{\mathrm{E}} = electric field in the hole.

2 x Electric field in the Hole = \overrightarrow{\mathrm{E}} due to the hollow conductor with filled in hole.
or 2\overrightarrow{\mathrm{E}} = \frac{\sigma}{\varepsilon_{0}} \hat{n}
or \overrightarrow{\mathrm{E}} = \frac{\sigma}{\varepsilon_{0}} \hat{n}

Question 30.
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density X without using Gauss’s law. [Hint: Use coulomb’s law directly and evaluate the necessary integral.] Answer:
Let AB be an infinite line charge having linear charge density X and centre O. + Let a = perpendicular distance of the point P from the line charge at which the electric field is to be calculated. Consider the long thin wire to be divided into a large number elementary segments and consider one such segment of length dl at point C and having charge dq at a distance r from P.

Also let ∠CPO = θ
∴ λ = \frac {dq}{dl} or dq = A.dl ……..(1)
If dE be the electric field produced at point P by the dq, then according to Coulomb’s law, dE is given by
dE = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{dq}}{\mathrm{r}^{2}} ………..(2)

The rectangular components of \overrightarrow{\mathrm{dE}} are shown in the figure. It is clear from the figure that dE sin 9 components due the symmetrical elements are equal in magnitude and act in opposite directions, thus cancel each other. While dE cos θ components due to the whole length of the wire act in the same direction and add upto given the net electric field \overrightarrow{\mathrm{E}} given by

Now in rt. ∠d ∆ COP,
cos θ = \frac {OP}{CP} = \frac {a}{r} …………(4)
∴ From-(1), (2), (3) and (4), we get

Now in rt. ∠d ∆ COP, cos θ = \frac {a}{r} or \frac {1}{r} = \frac {cos θ}{a}…(6)
and \frac {1}{a} = tan θ
or I = a tan θ ………..(7)
differentiating (7) w.r.t. O, we get
di = a sec2θ dθ ………….(8)
Also when x = – α, θ = \frac {π}{2} and when x = + α, θ = π/2 …………(9)
from (5), (6), (8) and (9), we get

which is the required expression.

Question 31.
It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge + (2/3) e, and the ‘down’ quark (denoted by d) of charge (-1/3) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron. .
Two types of quarks are denoted by u and d.
The charge on up quark (u) = + \frac {2}{3} e.
and the charge on down quark (d) = – \frac {2}{3} e.
The charge on the proton is + e and it is made of three quarks.
Therefore, the possible quark composition of a proton ¡s uud.
∴ Totaicharge = \frac {2}{3} e + \frac {2}{3}e – \frac {1}{3}e = \frac {4-1}{3}e = e
On the other hand, neutron is a neutral particle but it is aLso made of three quarks. For this, the possible composition of the neutron is udd.
∴ total charge on neutron = + \frac {2}{3}e + (\frac {-e}{3}) + (\frac {-e}{3}) = 0.

Question 32.
(a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the cofiguration. Show that the equilibrium of the test charge is necessarily unstable.
(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.
(a) An arbitrary electrostatic configuration consists of two charges of unequal charges of unequal magnitude and of same sign, e.g., consider a system of two fixed charges + ve and + e separated by a distance r placed at point A and B respectively. Let a test charge q0 be placed at a point C at a distance x from + 4e. C is the point is resultant field or force on 90 is zero.

i.e., If \overrightarrow{\mathrm{F}_{1}} and \overrightarrow{\mathrm{F}_{2}} be the forces acting on q0 due to + 4e and + e respectively.
\left|\overrightarrow{F_{1}}\right| = \left|\overrightarrow{F_{2}}\right|

For equilibrium, the charge q0 can be either +ve or – ve.

Case I. If q0 is – ve, then it experiences equal attractive force due to both the charges. When it is displaced on either side along the line joining the two charges from its equilibrium position, the attractive force due to one charge gets increased while due’to the other, it is decreased. As a result of this, charge – q0 no longer returns to its equilibrium position i.e., equilibrium of – ve charge is necessarily unstable.

Case II. If q0 is the + ve but displaced perpendicular to line joining the two charges, then resultant force tends to displace it further more i.e., it will not come-back to its equilibrium position i.e., equilibrium will be unstable.

(b) Let the simple configuration consists of two equal charges + q at point A and B. As now the two charges are of same nature and have same magnitude hence their resultant \overrightarrow{\mathrm{E}} will be zero at the mid-point O of the line joining them being equal and opposite and system will be unstable if the charge is slightly displaced, it executes S.H.M.

Question 33.
A particle of mass m and charge (-q) enters the region between the two charged plates initially moving along x-axis with speed vx (like particle 1 in Fig.). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the verticle deflection of the particle at the far edge of the plateis qEL2/(2m vx2).
Compare this motion with motion of a proj ectile in gravitational field discussed in Section 4.10 of Class XITextbook of physics.
Consider two charged plates Q, P and let E be the electric field acting between the two plates.

E will be acting from Q to P.
Let t be the time taken by the particle to cross electric field region having vertical deflection y.
t = \frac{L}{v_{x}}
Charge – q will follow a parabolic path between the charged plates Q and P.
Let a be the acceleration produced in the particle along y axis: Initial velocity along y axis uy = 0
Also we know that \overrightarrow{\mathrm{F}} = m \overrightarrow{\mathrm{a}}

Where (-ve) sign shows that \overrightarrow{\mathrm{a}} is opposite in direction to \overrightarrow{\mathrm{F}}
Using formula

NCERT Class 12 Physics Electric Charges and Fields

Question 34.
Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vx = 2.0 x 106 m s-1. If E between the plates separated by 0.5 cm is 9.1 x 102 N/C, where will the electron strike the upper plate? (| e | = 1.6 x 10-19 C, me = 9.1 x 1031 kg.)
ux= 2 x 106 ms-1
d = separation between the plates = 0.5 cm = 0.5 x 10-2 m. .
E = electric field between the plates = 9.1 x 102NC-1.
q = 1.6 x 10-19 C, me = mass of electron = 9.1 x 10-31 kg
y = defection of electron towards upper plate = ?
y is given by = \frac{\mathrm{q} \mathrm{EL}^{2}}{2 \mathrm{~m} v_{\mathrm{x}}^{2}}
Sustituting the values of q, E, m, V but the value of L is not given.
.’. Insufficient Data.

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