NCERT Solution Physics Class 12 Chapter 2 Electrostatic Potential And Capacitance
NCERT Solution Physics Class 12 Chapter 2 Electrostatic Potential And Capacitance all questions and answers. Physics Class 12 2nd Chapter Electrostatic Potential And Capacitance exercise solution and experts answer. As one of online learning platforms, we (netex.) are excited to offer the NCERT Solution Physics Class 12 Chapter 2. This solution is designed to help students who are looking to brush up on their physics concepts on Chapter 2 Electrostatic Potential And Capacitance.
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2.12)
2.13.) A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due o this charge array at the centre of the cube.
ANSWER-
Given cube of side b has charge q at each side of vertices is given below
Let,
b = diagonal of faces
l = length of cube diagonal
The electric potential (V) at the centre of the cube is due to the presence of eight charges at the vertices and given by
Where r is the Distance between centre and the vertices is half of the diagonal of the cube
From △ABF,
Using pythogorous theorem,
AB2 + BF2 = AF2
d2 = b2 + b2
d2 = 2b2
d = √2b
From △AFG,
Using pythogorous theorem,
Distance between centre and the vertices is half of the diagonal of the cube hence r = L/2 = √3/2b
From equation 1 we get,
2.14)
2.15.) A spherical conducting shell of inner radius r 1 and outer radius r 2 has a charge Q. (a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell? (b) 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-
(a) Charge placed at the centre of a shell is q which is positive Hence the inner surface of the shell will acquire the negative charge of same magnitude. Therefore, total charge on the inner surface of the shell is −q. induced charges on outer surface will be positive q charge as centre charge will repel the positive charges hence induced charge on outer surface will be +q but initially outer surface has Q charge hence total charge will become Q + q on outer surface.
As we know that surface charge density is charge per unit area
Hence,
For inner surface surface charge density is given by,
2.16 (a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by (E2 – E1). = σ/ε0 where n is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of is from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is σ/ε0.
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]
ANSWER-
Let E1 And E2 electric field on one side and other side of the same body then both electric fields are given by
Where n ̂ is unit vector normal to the surface and s is surface charge density.
Resultant electric field at any point due to both sides is given by
As electric field inside a conductor is zero (E1 =0) hence electric field outside the field is
(b) we know the fact that work done by electrostatic field on a closed loop is zero Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.
2.17 A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?
ANSWER-
A long charged cylinder of length L and linear charge density λ is surrounded by a hollow co-axial conducting cylinder of radius R.
Electric flux through the surface between them is given by
Φ = E x ds
where E is the electric field between the surfaces and ds be the surface area
Φ = E x (2πd) x L
Where d is distance of a point from common axis of the cylinder it is middle of the two cylinders.
Using gauss law if the Gaussian surface is drawn between the surfaces then this flux is equal to the total charge enclosed by the surface divided by permittivity of free space (ε0).
∴ Φ = q/ε0
E x (2πd) x L = q/ε0
If linear charge density λ then
∴ λ = Charge/length
∴ λ = q/L
∴ E x (2πd) x L = λL/ε0
∴ E = λ/2ε0 d
Therefore electric field between two surfaces is E = λ/2ε0 d
2.18)
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2.20.) Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.
ANSWER-
Consider two spheres A and B .
For Sphere A,
a = radius
QA = charge on sphere A
CA = capacitance of the sphere A
EA = Electric field of sphere A
For Sphere B,
b = radius
QB = charge on sphere B
CB = capacitance of the sphere B
EB = Electric field of sphere B
When two spheres are connected to each other by the wire both will get equal potential V.
We know that electric field due to a charged sphere of radius R and given by
2.21)
2.22.) Figure 2.32 shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for r/a >> 1, and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).
Answer-
From given figure
AB =BC = a and BP =r , AP = and CP = r-a .
Electrostatic potential at point P due to three charges is given by,
r +a
2.23)
2.26)
2.28) Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor ½.
ANSWER-
Let F be the force on the parallel plates when they are displaced by factor Dx. then work Is done on the plates due to this force and this work is stored in the plate in the form of the potential energy hence
Work done by force = increase in potential energy of capacitor
F x △X = △ (P.E) ……….(1)
But the potential energy is given by
△ (P.E) = uA△X
WHERE C is capacitor of the parallel plate capacitor = Aε0/d
And we also know that CV = Q = charge on capacitor
∴ F = 1/2 x Q x E
Hence proved.
Here factor 1/2 is came and for this we can give explanation that outside the conductor field is E but inside the conductor it is zero. Hence it has average value of E/2 which contributes to the value of force hence factor 1/2 is present in the formula of force.
2.29 A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig. 2.34). Show that the capacitance of a spherical capacitor is given by C = 4πεr1r2/r1-r2 Where r 1 and r 2 are the radii of outer and inner spheres, respectively.
ANSWER-
From given figure we can observe that,
The inner surface of the outer shell has charge +Q. The outer surface of the inner shell has induced charge −Q.
Potential difference between the two shells is given by
HENCE PROVED.
2.30)
2.31 Answer carefully: (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 Q1 Q2 /4πε0r2 , where r is the distance between their centres?
(b) If Coulomb’s law involved 1/r 3 dependence (instead of 1/r 2), would Gauss’s law be still 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 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) than say, mica (= 6).
ANSWER-
(a) As given spheres are very large in size there may be a non-uniform charge distribution on the spheres hence the magnitude of electrostatic force between them is not exactly given by Q1 Q2 /4πε0 r 2 , where r is the distance between their centre.
(b)Gauss’s law would not hold good if Coulomb’s law involved 1/r 3 dependence.
(c)we know that field lines represents the force and the direction of acceleration and tangent to the field lines represents direction of velocity hence A small test charge is released at rest at a point in an electrostatic field configuration. it will travel along the field line passing through that pointonly when field lines are straight so that tangent to this straight line is also along that straight line
(d)the force between nucleus and the electron is electrostatic force which is conservative force . and we know that work done by conservative force is independent of the path followed and for circular path the work done is zero. Hence whether the path followed by the electron is circular or elliptical the work done is zero.
2.32)
2.33)
2.35.) A small sphere of radius r 1 and charge q1 is enclosed by a spherical shell of radius r 2 and charge q2. Show that if q1 is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q2 on the shell is.
ANSWER-
According to Gauss’s law,the charge q1 is responsible for the electric field between a sphere and a shell. Hence, the potential difference V between the sphere and the shell is independent of Charge Q2 hence if the q1 is positive then potential s also positive charge flows from small sphere to the shell.
2.36 Answer the following:
(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 Vm–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 his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m2 . 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? (Hint: The earth has an electric field of about 100 Vm–1 at its surface in the downward direction, corresponding to a surface charge density = –10–9 C m–2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)
ANSWER-
(a) as soon as we step out of the house of steel cage the the surfaces having same potential of open air changes while the body and ground remains at same potential . this will prevent a body from getting electric shock.
(b) Here combination of ground and aluminium sheet with insulating slab in between will form a capacitor. The steady discharging current in atmosphere will charge the aluminium sheet thereby increasing the potential of the capacitor formed hence on the next morning a man gets shock when he touches the metal sheet.
(c)The atmosphere remains charges even though it has discharging current of 1800A because of the natural phenomena like lightning and thunderstorms.
(d)When lightning happens bright light can be seen with large intensity sound . Hence light energy , sound energy along with heat energy gets dissipated into the atmosphere due to the phenomenon like lightning.
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