Bihar Board - Class 12 Physics - Chapter 2: ELECTROSTATIC POTENTIAL AND CAPACITANCE Long Answer Question
1. Three identical capacitors C1 , C2 and C3 of capacitance 6 µF each are connected to a 12 V battery as shown. Find
(i) charge on each capacitor
(ii) equivalent capacitance of the network
(iii) energy stored in the network of capacitors.
Answer:
C1 and C2 in series, make C4 = 3µF
using
1C4 = 1C1 + 1C2
(i) 12V of potential is available in C4 and C3.
Charge in C3 = Q3 =C3V
= 6 × 10-6 × 12 = 72 µC
Charge in C4 = Q4 =C4V
= 3 × 10-6 × 12 = 36 µC
∴ Charge on C1 and C2 will also be 36 μC
(ii) C4 and C3 are in parallel to the source
∴ Ceq = 3 + 6 = 9 μF
(iii) Energy stored =
12 Ceq V2 = 12 (910-6) 122
= 648 × 10-6 = 6.48 × 10-4 joule
2. The equivalent capacitance of the combination between A and B in the given figure is 4 μF.
(i) Calculate capacitance of the capacitor C.
(ii) Calculate charge on each capacitor if a 12 V battery is connected across terminals A and B.
(iii) What will be the potential drop across each capacitor?
Answer:
Ceq= 4 μF
(i) Since 20 μF and C are in series, we have
14 = 120 + 1C
1C = 14 - 120 = 5-120 = 420
C = 204 = 5F
(ii) Charge drawn from 12 V battery is Q
= Ceq.V = 4 × 12 = 48 μC
So charge on each capacitor = 48 pC
(iii) Potential drop across
20 μF , V20= QC = 48F20F =2.4 Volt
And 5 μF , V5 = 48F5F = 9.6 Volt
3. A parallel plate capacitor, each with plate area A. and separation d, is charged to a potential difference V. The battery used to charge it remains connected. A dielectric slab of thickness d and dielectric constant k is now placed between the plates. What change, if any, will take place in :
(i) charge on plates?
(ii) electric field intensity between the plates?
(iii) capacitance of the capacitor?
Justify your answer in each case.
Answer:
Given : Plate area of either plate of parallel plate capacitor = A
Distance between the plates = d and
potential difference between the plates = V
∴ Initially capacitance, C = 0Ad ,
Charge on plate, Q = CV
As the battery remains connected throughout, the potential difference between the plates remains unchanged (V’ = V) on placing, a dielectric slab of thickness ‘d’ and dielectric constant ‘k’ between the plates.
(i) New charge on plates, Q’ = C’ V’ = kCV = kQ
Thus, charge changes to k times of its original value.
(ii) Electric field intensity between the plates, E’
= V 'd = V d = E
Thus, electric field intensity between the plates of capacitor remains unchanged.
(iii) New capacitance of the capacitor,
C ' = k0Ad = kC
4.Calculate the potential energy of a point charge -q placed along the axis due to a charge +Q uniformly distributed along a ring of radius R. Sketch P.E as a function of axial distance z from the centre of the ring. Looking at the graph, can you see what would happen if -q is displaced slightly from the centre of the ring (along the axis)?
Answer:
The potential energy (U) of a point charge q at a potential V is given by U=qV. In this case, a negatively charged particle is located on the axis of a ring with charge Q. The ring has a radius of a, and the electric potential at a distance x from the centre of the ring along the axis can be calculated.
The graph in Fig. shows how the potential energy changes as a function of axial distance z from the centre of the ring. If a charge of -q is slightly displaced from the centre of the ring along the axis and released, it will undergo oscillations. However, it is not possible to determine the nature of these oscillations simply by examining the graph.
5. A network of four capacitors each of 15 μF capacitance is connected cf to a 500 V supply as shown in the figure. Determine
(a) equivalent capacitance of the network and
(b) charge on each capacitor.
Answer:
Now CS and C4are in parallel
(a) ∴ Equivalent capacitance of the network,
C = CS + C4 = 15 + 5 = 20 μF
(b) Charge on capacitor, C4is
q = C4V = (15 × 10-6) × 500 = 7500 μC … [V =500
Charge on each capacitor, C1, C2 and C3 will be, q = CSV = (5 × 10-6) × 500 = 2500 μC
6. A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 360 μC. When potential across the capacitor is reduced by 120 V, the charge stored in it becomes 120 μC. Calculate:
(i) The potential V and the unknown capacitance C.
(ii) What will be the charge stored in the
capacitor, if the voltage applied had increased by 120 V?
Answer:
(i) Let the capacitance be C
∴ Charge on Q1 = CV or 360 μC = CV …(i)
In second case,
Q2 = C(V – 120)
⇒ 120 μC = C(V – 120) …(ii)
From equation (i) and (ii)
3 = V(V-120)
3V – 360 = V ⇒ 2V = 360
By putting this value of V in (ii)
120 × 10-6 = C(180 – 120)
C = 12010-660 = 2μF
(ii) Charge stored when voltage is increased by 120 V
Q’ = 2μF × (180 + 120) V = 600 μC
7. Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.
Solution:
Assuming a contradicting statement, the potential within a closed equipotential surface is not uniform. Let the potential just inside the surface be different to that on the surface having a potential gradient (dVdr).
Therefore, an electric field is created, which is represented by the following:
E=−dVdr
As a result, field lines either point inward or outward from the surface. These lines cannot originate from the surface itself, as it is equipotential. Field lines can only be formed when they have a charge at the other end within the surface. This conclusion contradicts the initial assumption. Therefore, it must be the case that the entire volume inside is equipotential.
8. A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 300 µC. When potential across the capacitor is reduced by 100 V, the charge stored in it becomes 100 V. Calculate the potential V and the unknown capacitance. What will be the charge stored in the capacitor if the voltage applied had increased by 100 V?
Answer:
(i) Charge stored, Q = CV
300 µC = C × V ,
potential is reduced by 100 V
100 µC = C(V – 100) = CV – 100 C
100 µC = 300 µC – 100 C
⇒ 100 C = 300 µC – 100 µC
⇒ 100 C = 200 µC
Therefore, capacitance C = 2µF
Potential , V = 300C2F = 150V
(ii) Charge stored when voltage applied is increased by 100 V
Q’ = 2µF × (150 + 100) = 500µC
9. Define an equipotential surface. Draw equipotential surfaces :
(i) in the case of a single point charge and
(ii) in a constant electric field in Z-direction. Why the equipotential surfaces about a single charge are not equidistant?
(iii) Can electric field exist tangentially to an equipotential surface?
Answer:
A surface with a constant value of potential at all points of the surface is defined as ‘equipotential surface’ „
(i) Equipotential surface for a single point charge.
(ii) Equipotential surface in a constant electric field as shown in the adjoining diagram.
Equipotential surface are not equidistant, because
V ∝ 1r
(iii) No, If the field lines are tangential, work will be done in moving a charge on the surface which goes against the definition of equipotential surface.
10. Define polarization and capacitor.
Answer:
Polarization :-
It is defined as the process of stretching of dielectric atoms due to the displacement of charges in the atoms under the action of applied electric field. This stretching of atoms continues till the restoring force becomes just equal and opposite to the force exerted by the applied electric field on the charges. Thus the C.G. of the nucleus does not coincide with the C.G. of electrons that are – vely charged due to this stretching and the atom thus acquires a dipole moment.
Capacitor :-
A capacitor is a device used for storing large quantities of charge. A given conductor cannot be charged to any extent. After a certain limit, change given to the conductor leakes away into the atmosphere. Capacity of a given conductor is a limit to which it can acquire charge. Capacitor or condenser is an arrangement of conductors by which capacity of a given conductor can be increased.
11. An electric dipole consists of two opposite charges each of 1 μC separated by 2 cm. The dipole is placed in an external uniform field of 105NC-1 Find the work done in rotating the dipole through 180° starting from its initially aligned position.
Answer:
Here, q = 1 μC = 1 x 10-6C
2a = dipole
length = 2 cm = 2 x10-2 m.
θ = 0°, θ2 = 180°, E = 105 NC-1.
.’. p = 2aq = 2 x 10-2 x 10-6 = 2 x 10-8 cm.
W = work done = ?
Using the relation
W- = pE (cosθ1 – cosθ2)
= 2 x 10-8 x 105 (cos 0 – cos 180)
= 2 x 10-3[1 – (- 1)]
= 2x 10-3 x 2 = 4 x 10-3J
= 0.004 J.
12 .(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 V m-1. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage, so there is no field inside).
(b) A man fixes outside the house one evening a two meter high insulating slab carrying on its top a large aluminum sheet of area 1 m2. Will he get an electric shock if he touches the metal sheet next morning?
(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?
(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning?
Answer:
(a) Our body and the ground are at the same electric potential. As we step out into the open, the original equipotential surfaces of open air change, keeping our head and the ground at the same potential.
(b) Yes, it is because the aluminum sheet gets charged due to discharging current and raises to the extent depending on the capacitor formed by the sheet and the ground.
(c) The atmosphere of earth gets continuously charged due to lightning, thunderstorms but simultaneously it gets discharged through normal weather zones. This keeps the system balanced.
(d) Light, sound and heat energy. Light energy in lightning and heat and sound energy in the accompanying thunder.
13. A long charged cylinder of linear charge density k is surrounded by a hollow coaxial conducting cylinder. What is the electric field in the space between the two cylinders?
Answer:
In Figure, A is a long charged cylinder of linear charge density k, length l and radius a. A hollow co-axial conducting cylinder B of length l and radius b surrounds A. The charge q = kl spreads uniformly on the outer surface of A. It induces – q charge on the cylinder B, which spreads on the inner surface of B. An electric field E is produced in the space between the two cylinders, which is directed radially outwards. Let us consider a coaxial cylindrical Gaussian surface of radius r. The electric flux through the cylindrical Gaussian surface is
E = E.dS = EdS cos0。= EdS = E(2rl)
The electric flux through the end faces of the cylindrical Gaussian surface is zero, as E is parallel to them. According to Gauss’s theorem ,
E =E(2rl)= q0 = l0
E = 20r
14. A spherical conducting shell of inner radius r, and outer radius r2 has a charge Q.
A charge q is placed at the center of the shell. What is the surface charge density on the inner and outer surfaces of the shell?
Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.
Answer:
1. Surface charge density on the inner surface of shell is
in = -q4r1 2
and on the outer surface of shell is
out = Q+q4r2 2
2. The electric flux linked with any closed surface S inside the conductor is zero as the electric field inside the conductor is zero.
So, by Gauss’s theorem net charge enclosed by closed surface S is also zero i.e.,
qnet = 0
So, if there is no charge inside the cavity, then there cannot be any charge on the inner surface of the shell and hence electric field inside the cavity will be zero, even though the shell may not be spherical.
15. Describe schematically the equipotential surfaces corresponding to
A constant electric field in the z-direction.
a field that uniformly increases in magnitude but remains in a constant (say, z) direction.
a single positive charge at the origin, and
a uniform grid consisting of long equally spaced parallel charged wires in a plane.
Answer:
Planes parallel to the x-y plane or normal to the electric field in z-direction.
Planes parallel to x-y plane or normal to the electric field in z-direction, but the planes having different fixed potentials will become closer with increase in electric field intensity.
Concentric spherical surfaces with their centers at origin.
A time dependent changing shape nearer to the grid, and at far off distances from the grid, it slowly becomes planar and parallel to the grid.
16. A charged cylinder is found to have a linear charge density of λ and is encompassed by a hollow coaxial conducting cylinder. Determine the electric field in the space between both cylinders.
Answer:
Assuming the length of the charged cylinder and the hollow coaxial conducting cylinder both to be L.
As is mentioned already, the charge density is λ
Now, Let E be the electric field in the space between the two cylinders
With regard to Gauss theorem, the electric flux via Gaussian surface is given as Φ = E (2πd)L
Therefore, d is the distance between the common axis of the cylinders, such that
θ = E (2πd)L = q0
q shows the charge on the inner surface of the outer cylinder
And, 0 is the permittivity of the free space
Thus, it can be said, E (2πd)L = l0
Henceforth, the electric field 2d0 between both cylinders.
17. Considering the separation between the plates is about 0.5 cm, determine the area of plates which have a 2F parallel plate capacitor.
Answer:
As per the equation,
Capacitance, C = 0Ad
The Capacitor has a capacitance of, C=2 F
Separation between the plates,
d = 0.5 cm
= 0.5 x 10-2 m
0 = free space permittivity = 8.85 x 10-12 C2 N-1 m-2
Plate area , A = Cd/0
A = [2 x 0.5 x 10-2]/8.85 x10-12
A = 1130 x 106 m2
18. (a) How can a positively charged body be at zero or negative potential?
(b) Why does water have a very high dielectric constant?
Answer:
(a) If a positively charged body B is moved gradually towards a small positively charged body A, at a particular distance, the -ve charge induced on A due to B will become equal to +ve charge initially present on body A. Hence the body A will be at zero potential. If body B is further moved towards A, -ve charge will be induced on body A. Hence body A will be +vely charged but at -ve potential. So a positively charged body can be at zero or negative potential.
(b) Due to unsymmetric placement of the atoms of water molecule, it possesses a permanent electric dipole moment which is about 0.6 x 10-20 Cm. This magnitude of dipole moment is about 10 times more than the induced dipole moment acquired by the molecule. Hence due to this large value of permanent dipole moment, the dielectric constant of water molecules is very high.
19. Give the physical meaning of electrostatic potential.
Answer:
Level of water determines the direction of flow of water. Similarly, temperature of a body determines the direction of flow of heat from or towards the body when it is brought in contact with another body. Exactly on these lines the electrostatic potential of a body represents its state of electrification/charging which determines the direction of flow of charge when this body is kept in contact with another charged body. The charge always flows from a body at higher potential to the one at lower potential. The flow continues till their potentials are equal. Earth is supposed to: be at zero potential. When a charged body is held in Contact with earth, and charge flows from body to earth, the potential of the body must be greater than the potential of earth. Hence the potential of the body must be positive (>zero). On the contrary, if charge flows from earth to body the potential of the body is said to be negative (< zero).
20. Define conductors and insulators. Why were conductors called non-electrics and insulators were called dielectrics? Describe two types of dielectrics.
Answer: Conductors and Insulators :- A substance, which allows an electric current to flow through itself, is called a conductor. It has free electrons inside itself.
Examples are metals, human body, earth etc.
A substance, which does not allow an electric current to flow through itself, is called an insulator. It does not have free electrons inside itself.
Examples are ebonite, glass, quartz, rubber, plastic etc.
Non-electric and Dielectric:- When a conductor was rubbed, electric charge produced on it was conducted away through the body to earth. As no charge was found present on the rubbed conductor, it was supposed to have no charge. Hence it was called non-electric.
On the other hand, charge stays in an insulator when it is rubbed. It was supposed to have a charge. Hence it was called dielectric.
Types of dielectric:- Dielectrics are of two types- polar and non-polar.
Polar dielectrics. Their atoms have separate centers of positive and negative charges. These separated centers are called poles and the atoms are called dipoles.
Non-polar dielectrics. Their atoms do not have separate centers of positive and negative charges in normal conditions. But inside an electric field, they develop poles.
21. Define the term 'electric field intensity. Electric field inside a conductor is zero. Explain.
Answer: The electric field intensity at any point in the electric field is the force experienced by a unit positive charge placed at that point.
Electric field inside a conductor is zero. When a conductor is placed inside an electric field, the free electrons in the conductor start drifting under the influence of electric field. The free electrons move towards the positive plate. The redistribution of charges will create its own electric field due to induced charges till the electric field due to induced charges become equal to external field. Hence net electric field inside the conductor is zero.
22. What is a capacitor? Explain its principle.
Answer: Capacitor:- A capacitor is a device used for storing electric charge and electrical energy. It is an arrangement of conductors in which the capacitance of the system can be increased without increasing the size.
Principle of a Capacitor:- It is based on the principle that when an earthed conductor is based in the neighborhood of a charged conductor, the capacitance of the system increases considerably.
Explanation :- When an uncharged conductor B is placed near a charged conductor A, due to induction, negative charge will be induced on the nearer face of B and positive charge on its farther side as shown in Fig. The induced negative charge on B lowers the potential of A, while the induced positive charge raises its potential. As the induced negative charge is nearer to conductor A than the induced positive charge, the potential of conductor A on the whole is lowered. Therefore, capacitance of a conductor increases by a small amount, when another uncharged conductor is placed near it.
Now connect the conductor B to the earth as shown in Fig. The induced positive charge on B will immediately flow to the earth. However, the induced negative charge on B will stay on it. The potential of A will get lowered by a large amount. Thus, it follows that the capacitance of a conductor is greatly increased, when an earth-connected conductor is placed near it. It forms the principle of a capacitor.
23. Draw the electric field lines due to a uniformly charged thin spherical shell when charge on the shell is (i) positive and (ii) negative.
Answer: The electric field lines due to a uniformly charged thin spherical shell when the shell is positive are all directed radially in outward direction as shown in Fig. and all directed radically in inward direction as shown in Fig.
24. (a) Two large conducting spheres carrying charges Q1 and Q2 are brought close to each other, is the magnitude of electrostatic force between them exactly given by Q1Q240R2 r is the distance between their centers?
(b) If Coulomb's law involved 1r3 dependence (instead of 1r2), would Gauss’s law still be true?
(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?
(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?
(e) We know that the electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?
(f) What meaning would you give to the capacitance of a single conductor?
(g) Guess a possible reason why water has a much greater dielectric constant {- 80) then say, mica (= 6).
Answer:
(a) No, because Coulomb's law holds good only for point charges. ‘
(b) No, because in that case electric flux linked with the closed surface will also become dependent on V other than the charge enclosed by it.
(c)No, it will travel along the field line only if it is a straight line.
(d) Zero, whatever may be the shape of orbit may be. It is because work done in moving a charge in a closed path in an electric field is zero, as the electric field is a conservative field.
(e) No, electric potential is continuous
there. As E = 0, so dVdr = 0 or V = constant.
(f) It means that a single conductor is a capacitor whose other plate can be considered to be at infinity.
(g) A water molecule is a polar molecule with non-zero electric dipole moment, however mica does not have polar molecules. So, the dielectric constant of water is high.
25. Define equipotential surface. Write its properties.
Answer: Equipotential surface:- An equipotential surface is the locus of all those points at which the potential due to distribution of charge remains the same.
Properties:
Potentials on every point are equal
No work is done in moving a positive charge from one point to another
The electrical lines of force are normal to the equipotential surface
Two equipotential surfaces do not intersect each other.
All the points on the surface of a conductor are in electric contact. If the potentials are not same then the
charge will flow from higher potential to lower potential till the potential of both the points become the same.
Thus the surface of a conductor is always equipotential.
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