Bihar Board - Class 12 Physics - Chapter 1: Electric Charges and Fields Long Answer Question
For Class 12 Bihar Board, the chapter Electric Charges and Fields in Physics focuses on the properties of electric charges, Coulomb’s law, and the concept of the electric field. It also covers the electric potential and Gauss's law. You can also find the long-question answers that provide a clear understanding of these important concepts, aiding in better preparation for exams on Vidyakul.
Long Question Answer
1. (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = o) of the configuration. 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 at certain distance apart.
Answer:
(a) Let us assume that a radial electric field is present with origin as centre. The E is 0 at origin and a small test charge ‘q0’ is placed at origin. Force on test charge at origin is zero.
F=qE=0
But as the test charge is displaced a little, the electrostatic force will drift it further away so charge is in unstable equilibrium.
(b) A test charge placed at mid point of equal charges is in stable equilibrium, for lateral displacement. Initially at mid point position of test charge, two repulsion forces on test charge are equal and opposite.
FA = FB so q0 is in equilibrium.
But on displacing q0 close to one of the charge, one of the forces become stronger and pushes the charge towards another.
FA > FB so q0 moves towards B.
2. Which among the curves shown in Figure cannot possibly represent electrostatic field lines?
Answer:
(a) It is wrong, because electric field lines must be normal to the surface of conductor outside it.
(b) It is wrong because field lines cannot start or originate from negative charge, and also cannot end or submerge into positive charge.
(c) It is correct
(d) It is wrong because electric field lines never intersect each other.
(e) It is wrong because electric field lines cannot form closed loops.
3. Two metallic spheres A and B kept on insulating stands are in contact with each other. A positively charged rod P is brought near the sphere A as shown in the figure. The two spheres are separated from each other, and the rod P is removed. What will be the nature of charges on spheres A and B?
Answer:
Sphere A will be negatively charged.
Sphere B will be positively charged.
Explanation: If positively charged rod P is brought near metallic sphere A due to induction negative charge starts building up at the left surface of A and positive charge on the right surface of B.
If the two spheres are separated from each other, the two spheres are found to be oppositely
charged. If rod P is removed, the charges on spheres rearrange themselves and get uniformly
distributed over them.
4. (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. 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.
Answer:
(a) Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and displaced from its position in any direction, then it experiences a restoring force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards towards the null point. There is a net inward flux of electric field through a closed surface around the null point. According to Gauss’s law, the flux of electric field through a surface, which is not enclosing any charge, is zero. Hence, the equilibrium of the test charge can be stable.
(b) Two charges of same magnitude and same sign are placed at a certain distance. The mid-point of the joining line of the charges is the null point. When a test charged is displaced along the line, it experiences a restoring force. If it is displaced normal to the joining line, then the net force takes it away from the null point. Hence, the charge is unstable because stability of equilibrium requires restoring force in all directions.
5. (a) A conductor A with a cavity as shown in Fig. 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 inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q
(c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.
Answer:
(a) Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing the cavity. The electric field intensity E inside the charged conductor is zero.
Let q is the charge inside the conductor and 0is the permittivity of free space.
According to Gauss’s law,
Flux, = E.ds =q0
Here, E = 0
q0 = 0
∵ 00
∴ q =0
Therefore, charge inside the conductor is zero.
The entire charge Q appears on the outer surface of the conductor.
(b) The outer surface of conductor A has a charge of amount Q. Another conductor B having charge +q is kept inside conductor A and it is insulated from A. Hence, a charge of amount −q will be induced in the inner surface of conductor A and +q is induced on the outer surface of conductor A. Therefore, total charge on the outer surface of conductor A is Q + q.
(c) A sensitive instrument can be shielded from the strong electrostatic field in its environment by enclosing it fully inside a metallic surface. A closed metallic body acts as an electrostatic shield.
6. What is Electric Charge?
Answer:
Electric Charge :-
Electric Charge is a fundamental property of a matter which is responsible for electric forces between the bodies. Two electrons placed at a small separation are found to repeal each other, this repulsive force (Electric force) is only because of the electric charge on electrons.
When a glass rod is rubbed with silk, the rod acquires one kind of charge and the silk acquires the second kind of charge. This is true for any pair of objects that are rubbed to be electrified. Now if the electrified glass rod is brought in contact with silk, with which it was rubbed, they no longer attract each other.
Types of Electric Charge
There are two types of charges that exist in our nature.
Positive Charge
Negative Charge
If any object loses its electrons then they get a positive charge. It is denoted by (+q) sign. If any object gain electrons from another object then it gets a negative charge. It is denoted by (-q) sign. The charges were named as positive and negative by the American scientist Benjamin Franklin. If an object possesses an electric charge, it is said to be electrified or charged. When it has no charge it is said to be neutral.
7. What is Electric Flux and Solid Angle?
Answer:
Electric Flux :-
The number of electric field lines that are passing perpendicular through the unit surface of any plane is called electric flux. Consider an electric field that is uniform in both magnitude and direction, as in the figure.
We can write this as N ∝ EA, which means that the number of field lines is proportional to the product of E and A. This is a measure of electric flux and is represented by the symbol φ. In the above case, φ = EA cosθ. The SI unit of electric flux is N-m2/C or V-m(volt-metre).
Solid Angle :-
A solid angle is defined as an angle that is made at a point in place by an area. The SI unit of solid angle is steradian, and it is expressed as 'Sr'.
Sr = dA cosr2
8. State and proof Gauss’s Law ?
Answer:
Gauss’s Law :-
It states that the electric flux Φ through any closed surface is equal to (1/εo) times the net charged q enclosed by the surface. That is
Φ = E.dS= q0
Proof: Consider a point charge q surrounded by a spherical surface of radius r centered on the charge. The magnitude of the electric field everywhere on the surface of the sphere is
E = 140 . qr2 ......(1)
The electric field is perpendicular to the spherical surface at all points on the surface. The electric flux through the surface is
E = ∫EA cos .....(2)
Putting the value of E from eq. (1)
E = ∫140 . qr2 A cos
E = q40 .∫ 1r2 A cos
E = q40 .∫ d
E = q40 .4π
E = q0 .
This result says that the electric flux through a sphere that surrounds a charge q is equal to the charge divided by the constant ε0
9. A thin straight infinitely long conducting wire having charge density X is enclosed by a cylindrical surface of radius r and length l, its axis coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder.
Answer:
Since the field is everywhere radial, flux through the two ends of the cylindrical Gaussian surface is zero. At the cylindrical part of the surface, E is normal to the surface at every point, and its magnitude is constant, since it depends only on r. The surface area of the curved part is 2πrl, where l is the length of the cylinder.
Flux through the Gaussian surface = Flux through the curved cylindrical part of the surface is zero. At the cylindrical part of the surface, E is normal to the surface at every point, and its magnitude is constant, since at every point, and its magnitude is constant, since it depends only on r. The surface area of the cylinder.
Flux through the Gaussian surface = Flux through the curved cylindrical part of the surface
= E × 2πrl
(a) Electric field due to an infinitely long thin straight wire is radial.
(b) The Gaussian surface for a long thin wire of uniform linear charge density
The surface includes charge equal to λl.
Gauss’s law then gives
E × 2πrl = l0
i.e. E = 2πr0
10. Given a uniform electric field E = 510-3 i NC , find the flux of this field through a square of 10 cm on a side whose plane is parallel to the y-z plane. What would be the flux through the same square if the plane makes a 30° angle with the x-axis?
Answer:
Given :
E = 510-3 i NC
A = 10 × 10 × 10-4 m2,
Flux (ϕ) = EA cos θ
(i) For first case, θ = 0, cos 0 = 1
∴ Flux = (5 × 103) × (10 × 10 × 10-4)
(ii) Angle of square plane with x-axis = 30°
Hence the 0 will be 90° – 30° = 60°
EA cos θ = (5 × 103) × (10 × 10 × 10-4) × cos 60°
= 50 × 12
= 25 Nm2C-1
11. State Gauss’ law in electrostatics. Using this law derive an expression for the electric field due to a uniformly changed infinite plane sheet.
Answer:
Gauss’ Law states that “the total flux through a closed surface is 10 times the net charge enclosed by
E = E.dS = q0
Let σ be the surface charge density (charge per unit area) of the given sheet and let P be a point at distance r from the sheet where we have to find
E
Choosing point P’, symmetrical with P on the other side of the sheet, let us draw a Gaussian cylindrical surface cutting through the sheet as shown in the diagram. As at the cylindrical part of the Gaussian surface, E and dS are at a right angle, the only surfaces having E and dS parallel are the plane ends
∴ E = E.dS + E.dS
E = EdS + EdS
= EA + EA =2EA
…[As E is outgoing from both plane ends, the flux is positive.
This is the total flux through the Gaussian surface.
Using Gauss’ law , E = q0
∴ 2EA = q0 = A0
∴ E = 20
This value is independent of r. Hence, the electric field intensity is same for all points near the charged sheet. This is called uniform electric field intensity.
12. Two thin concentric and coplanar spherical shells, of radii a and b (b > a) carry charges, q and Q, respectively. Find the magnitude of the electric field, at a point distant x, from their common centre for
(i) 0 < x < a
(ii) a ≤ x < b
(iii) b ≤ x < ∞
Answer:
Magnitude of Electric field : Two thin concentric and coplanar spherical shells of radii ‘a’ and ‘b’ (b > a) carry charges ‘q’ and ‘Q’ respectively.
(i) For 0 < x < a
Point lies inside both the spherical shells.
Hence, E(x) = 0
(ii) For a ≤ x < b
Point is outside the spherical shell of radius ‘a’ but inside the spherical shell of radius ‘b’.
∴ E(x) = 140 . qx2
(iii) For b ≤ x < ∞
Point is outside of both the spherical shells. Total effective charge at the centre equals (Q + q).
E(x) = 140 . (q+Q)x2
13. Four point charges qA = 2 μC, qB = −5 μC, qC = 2 μC, and qD = −5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?
Answer:
The given figure shows a square of side 10 cm with four charges placed at its corners. O is the centre of the square.
Where,
(Sides) AB = BC = CD = AD = 10 cm
(Diagonals) AC = BD = 102 cm
AO = OC = DO = OB = 1022 cm = 52cm
A charge of amount 1μC is placed at point O.
Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on 1 μC charge at centre O is zero.
14. (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. 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.
Answer:
(a) Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and displaced from its position in any direction, then it experiences a restoring force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards towards the null point. There is a net inward flux of electric field through a closed surface around the null point. According to Gauss’s law, the flux of electric field through a surface, which is not enclosing any charge, is zero. Hence, the equilibrium of the test charge can be stable.
(b) Two charges of same magnitude and same sign are placed at a certain distance. The mid-point of the joining line of the charges is the null point. When a test charged is displaced along the line, it experiences a restoring force. If it is displaced normal to the joining line, then the net force takes it away from the null point. Hence, the charge is unstable because stability of equilibrium requires restoring force in all directions.
15. Define Coulomb’s Law.
Answer:
Coulomb’s Law :- In 1785 Charles Coulomb (1736-1806) experimentally established the fundamental law of electric force between two stationary charged particles. He observed that An electric force between two charged particles has the following properties:
It is directed along a line joining the two particles and is inversely proportional to the square of the separation distance r, between them.
It is proportional to the product of the magnitudes of the charges, |q1| and |q2|, of the two particles.
It is attractive if the charges are of opposite sign and repulsive if the charges have the same sign.
From these observations, Coulomb proposed the following mathematical form for the electric force between two charges. The magnitude of the electric force F between charges q1 and q2 separated by a distance r is given by
F = k q1q2r2
where k is a constant called the Coulomb constant. The proportionality constant k in Coulomb’s law is similar to G in Newton’s law of gravitation. Instead of being a very small number like G (6.67 × 10–11), the electrical proportionality constant k is a very large number. It is approximately
k = 8.9875 × 109 N-m2C–2
The constant k is often written in terms of another constant, ε0, called the permittivity of free space. It is related to k by
K = 140
∴ F = 140 q1q2r2
0 = 14k = 8.85 × 10-12 C2 / Nm2
16. Define Electric Field. Find Electric Field due to a Point Charge.
Answer:
Electric Field :-
A charge produces something called an electric field in the space around it and this electric field exerts a force on any charge (except the source charge itself) placed in it. The electric field has its own existence and is present even if there is no additional charge to experience the force.
Intensity of Electric Field :-
The intensity of the electric field due to a charge configuration at a point is defined as the force acting on a unit positive charge at this point. Hence if a charge q experiences an electric force F at a point then the intensity of the electric field at this point is given as
E = F / q
It has S.I. units of newtons per coulomb (N/C).
Electric Field due to a Point Charge :-
To determine the direction of an electric field, consider a point charge q as a source charge. This charge creates an electric field at all points in the space surrounding it. A test charge q0 is placed at point P, a distance r from the source charge. According to Coulomb’s law, the force exerted by q on the test charge is
F = 140 qq0r2
This force is directed away from the source charge q, since the electric field at P, the position of the test charge, is defined by
E = Fq0
we find that at P, the electric field created by q is
E = 140 qq0r2
17. What is Electric Field Lines?
Answer:
Electric Field Lines:-
Electric field lines are a way of pictorially mapping the electric field around a configuration of charges. An electric field line is, in general, a curve drawn in such a way that the tangent to it at each point is in the direction of the net field at that point. The field lines follow some important general properties:
The tangent to electric field lines at any point gives the direction of the electric field at that point.
In free space, they are continuous curves that emerge from a positive charge and terminate at a negative charge
They do not intersect each other. If they do so, then it would mean two directions of the electric field at the point of intersection, which is not possible.
Electrostatic field lines do not form any closed loops. This follows from the conservative nature of the electric field
18.What is Electric Dipole ? find Torque of Electric Dipole in uniform Electric Field .
Answer:
Electric Dipole :-
A configuration of two charges of the same magnitude q, but of opposite sign, separated by a small distance (say 2a) is called an electric dipole.
The dipole moment for an electric dipole is a vector quantity directed from the negative charge to the positive charge and its magnitude is p = q × 2a (charge × separation). The SI unit of dipole moment is C-m (coulomb-metre).
Torque of Electric Dipole in uniform Electric Field :-
An electric dipole is placed in a uniform electric field E
The force experienced by the dipole is
F = qE
The two forces form a couple and it tries to turn the dipole. The torque due to the couple is given by
τ = either force × perpendicular distance between the forces
τ = qE × (2a sin θ)
τ = (2aq) E sin θ
τ = pE sin θ
τ = p × E
19. Write application of Gauss’s Law and find Electric field strength due to an infinitely long straight uniformly charged wire.
Answer:
Application of Gauss’s Law :-
It is used to calculate electric field due to an infinitely long straight uniformly charged wire.
It is also used to calculate electric field due to a uniformly charged infinite plane sheet.
It is also used to calculate electric field due to a uniformly charged thin spherical shell.
Electric field strength due to an infinitely long straight uniformly charged wire :-
Let a charged wire of infinite length be +q charge and its linear charge density λ be. To calculate the electric field due to this wire, let us assume a cylindrical Gaussian surface of radius r. Let the area of this Gaussian surface be dS1, dS2, dS3.
Hence the total electric flux passing through the first surface dS1,
1 = E. dS1cos
1 = E. dS1cos90o
1 = 0 ….....(1)
Similarly, the total electric flux passing through the second surface dS2,
2 = E. dS2cos
2 = E. dS2cos0
2 = E. dS2
2 = E.dS2
2 = E(2πrl) ………….. .....(2)
Similarly, the total electric flux passing through the third surface dS3,
3 = E. dS3cos 90°
3 = 0 .....(3)
Hence, the total electric flux passing through the Gaussian surface,
Φ = 1 + 2 + 3
Φ = 0 + E.2πrl + 0
Φ = E.2πrl .....(4)
Putting Φ = q0 from Gauss's theorem,
q0 = E.2πrl
E = ql2r0
E = 2r0
20. Find deduction of Coulomb’s law from Gauss' Law .
Answer:
Deduction of Coulomb’s law from Gauss' Law :-
Consider a charge +q in place at origin in a vacuum. We want to calculate the electric field due to this charge at a distance r from the charge. Imagine that the charge is surrounded by an imaginary sphere of radius r as shown in the figure below. This sphere is called the Gaussian sphere.
Consider a small area element dS on the Gaussian sphere. We can calculate the flux through this area element due to charge as follows:
E. dS=E ∫dS
E. dS=E(4πr2)
Using this in Gauss theorem we get
E(4πr2) = q0
E = 140 . qr2
We know that
F = E q0
F = 140 . qq0r2
This is the required Coulomb’s law obtained from the Gauss theorem.
21. A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC /m2 . Find the charge on the sphere.
Answer:
The diameter of the sphere, d = 2.4 m
The radius of the sphere, r = 1.2 m
Surface charge density, σ = 80.0 μC /m2 = 80 × 10-6 C/m2
The total charge on the surface of the sphere can be calculated as follows,
Q = Charge density × Surface area
= σ × 4πr2
= 80 × 10-6 × 4 × 3.14 × ( 1.2 )2
= 1.447 × 10-3 C
Therefore, the charge on the sphere is 1.447 × 10-3 C.
22. (a) Why do the electrostatic field lines not form closed loops?
(b) Why do the electric field lines never cross each other?
Answer:
(a) Electric field lines do not form closed loops because the direction of an electric field is from positive to negative charge. So one can regard a line of force starting from a positive charge and ending on a negative charge. This indicates that electric field . lines do not form closed loops.
(b) The electric lines of force give the direction of the electric field. In case, two lines of force intersect, there will be two directions of the electric field at the point of intersection, which is not possible.
23. (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. 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.
Answer:
(a) Let us consider that the equilibrium is stable. Then if the test charge is displaced in any direction, it will experience a restoring force towards the null point. This means that all the field lines near the null point will be directed towards the null point. That is, there is a net inward flux of electric field through a closed surface around the null point. But according to Gauss’s law, the flux of an electric field through a surface, not enclosing any charge, must be zero. Hence, equilibrium cannot be stable.
(b) The null point is the mid-point of the line joining the two charges. If the test charge is displaced along the line from the null point, there is a restoring force. If the test charge is displaced normally to the line, the net force takes it away from the null point. Hence, the stability of equilibrium needs restoring force in all directions.
24.In a certain region of space, the electric field is along the z-direction throughout. The magnitude of the electric field is, however, not constant but increases uniformly along the positive z-direction at the rate of 105NC-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?
Answer:
The total dipole moment of the system, p=q×dl =−10-7 cm
The rate of increase of the magnitude of the electric field along the positive z-direction = 105 NC-1per metre.
The force experienced by the system is given as
F = qE
F = q(dE/dl) x dl
= p x (dE/dl)
=- 10-7 ×10-5
=−10-2N
The force experienced by the system is – 10-2 N. This is in the negative z-direction and opposite to the direction of the electric field. Therefore, the angle between the dipole moment and the electric field is 180°
Torque (τ) =pEsin180°=0
Therefore, the torque experienced by the system is zero.
25. (a) Depict the direction of the magnetic field lines due to a circular current carrying loop.
(b) Why do the electric field lines not form closed loops?
Answer:
(a) Direction of the magnetic field lines is given by right hand thumb rule.
(b) Electric field lines do not form closed loops because the direction of an electric field is from positive to negative charge. So one can regard a line of force starting from a positive charge and ending on a negative charge. This indicates that electric field lines do not form closed loops.
हिंदी के सभी अध्याय के महत्वपूर्ण प्रशन उत्तर के लिए अभी Download करें Vidyakul App - Free Download Click Here
