NCERT solutions for class 12 physics chapter 2 Electrostatic Potential and Capacitance - This chapter is the continuation of chapter 1 electric charges and fields. Chapter 2 talks about potential due to different systems of charges and about capacitance and dielectrics. Solutions of NCERT for class 12 physics chapter 2 electrostatic potential and capacitance require certain concepts to be studied in the first chapter to solve questions. The CBSE NCERT solutions for class 12 physics chapter 2 electrostatic potential and capacitance are helpful in the preparation of board exam. The important derivations of the NCERT class 12 physics chapter 2 are potential and potential energy due to a dipole. Not only derivation and problems are important, the theory portion of dielectric and polarization is also important. The solutions for NCERT help students to perform well in exams. Some times in competitive exams like NEET and JEE Mains similar types of questions from NCERT solutions for class 12 physics NCERT chapter 2 electrostatic potential and capacitance are asked.
Given, two charge particles
The separation between two charged particle
Now, let's assume the point P between two charged particles where the electric potential is zero is x meter away from and meter away from
So,
The potential at point P :
Hence the point between two charged particles where the electric potential is zero lies 10cm away from and 6 cm away from
Now, Let's assume a point Q which is outside the line segment joining two charges and having zero electric potential .let the point Q lie r meter away from and (0.16+r) meter away from
So electric potential at point Q = 0
Hence the second point where the electric potential is zero is 24cm away from and 40cm away from
The electric potential at O due to one charge,
q = 5 × 10^{-6} C
r = distance between charge and O = 10 cm = 0.1 m
Using the superposition principle, each charge at corners contribute in the same direction to the total electric potential at the point O.
=
Therefore the required potential at the centre is
Given, 2 charges with charges and .
An equipotential plane is a plane where the electric potential is the same at every point on the plane. Here if we see the plane which is perpendicular to line AB and passes through the midpoint of the line segment joining A and B, we see that at every point the electric potential is zero because the distance of all the points from two charged particles is same. Since the magnitude of charges is the same they cancel out the electric potential by them.
Hence required plane is plane perpendicular to line AB and passing through the midpoint of AB which is 3cm away from both charges.
The direction of the electric field in this surface is normal to the plane and in the direction of line joining A and B. Since both charges have the same magnitude and different sign, they cancel out the component of the electric field which is parallel to the surface.
Since the charge is uniformly distributed and it always remains on the surface of the conductor, the electric field inside the sphere will be zero.
Given,
Charge on the conductor
The radius of a spherical conductor
Now,
the electric field outside the spherical conductor is given by:
Hence electric field just outside is .
Given,
charge on the conductor
The radius of the spherical conductor
Now,
the electric field at point 18cm away from the centre of the spherical conductor is given by:
Hence electric field at the point 18cm away from the centre of the sphere is
As we know,
where A= area of the plate
= permittivity of the free space
d = distance between the plates.
Now, Given
The capacitance between plates initially
Now, capacitance when the distance is reduced half and filled with the substance of dielectric 6
Hence new capacitance is 96pF.
Given, 3 capacitor of 9pF connected in series,
the equivalent capacitance when connected in series is given by
Hence total capacitance of the combination is 3pF
Given,
supply Voltage V = 120 V
The potential difference across each capacitor will be one-third of the total voltage
Hence potential difference across each capacitor is 40V.
Given, 3 capacitors with are connected in series,
the equivalent capacitance when connected in parallel is given by
Hence, the equivalent capacitance is 9pF.
Given, 3 capacitors connected in parallel with
Supply voltage
Since they are connected in parallel, the voltage across each capacitor is 100V.
So, charge on 2pf capacitor :
Charge on 3pF capacitor:
Charge on 4pF capacitor:
Hence charges on capacitors are 2pC,3pC and 4pC respectively
Given,
Area of the capacitor plate =
Distance between the plates
Now,
The capacitance of the parallel plate capacitor is
= permittivity of free space =
putting all know value we get,
Hence capacitance of the capacitor is 17.71pF.
Now,
Charge on the plate of the capacitor :
Hence charge on each plate of the capacitor is .
Given,
The dielectric constant of the inserted mica sheet = 6
The thickness of the sheet = 3mm
Supply voltage
Initial capacitance =
Final capacitance =
Final charge on the capacitor =
Hence on inserting the sheet charge on each plate changes to .
If a 3mm mica sheet is inserted between plates of the capacitor after disconnecting it from the power supply, the voltage across the capacitor be changed. Since the charge on the capacitor can not go anywhere, if we change the capacitance (which we are doing by inserting mica sheet here), the voltage across the capacitor has to be adjusted accordingly.
As obtained from question number 8 charge on each plate of the capacitor is
As we know,
the electrostatic energy stored in the capacitor is
Here,
So,
Hence energy stored in the capacitor is
Given
Energy stored :
Now, when it is disconnected and connected from another capacitor of capacitance 600pF
New capacitance
New electrostatic energy
Hence loss in energy
Given,
The initial distance between two charges
The final distance between two charges
Hence total work is done
The path of the charge does not matter, only initial and final position matters.
As we know,
the distance between vertices and the centre of the cube
Where b is the side of the cube.
So potential at the centre of the cube:
Hence electric potential at the centre will be
The electric field will be zero at the centre due to symmetry i.e. every charge lying in the opposite vertices will cancel each other's field.
(a) at the mid-point of the line joining the two charges
As we know
outside the sphere, we can assume it like a point charge. so,
the electric potential at midpoint of the two-sphere
where q1 and q2 are charges and d is the distance between them
So,
The electric field
The distance of the point from both the charges :
Hence,
Electric potential:
Electric field due to q1
Electric field due to q2
Now,
Resultant Electric field :
Where is the angle between both electric field directions
Here,
Hence
2.15 (a) A spherical conducting shell of inner radius r_{1 }and outer radius r_{2} has a charge Q.
The charge placed on the centre is q, so -q will be the charge induced in the inner shell and + q will be induced in the outer shell
So,
charge density on the inner shell
charge Density on the outer shell
2.15 (b) A spherical conducting shell of inner radius r_{1} and outer radius r_{2} has a charge Q.
Yes, the electric field inside the cavity is zero even when the shape is irregular and not the sphere. Suppose a Gaussian surface inside the cavity, now since there is no charge inside it, the electric flux through it will be zero according to the guess law. Also, all of the charges will reside on the surface of the conductor so, net charge inside is zero. hence electric field inside cavity is zero.
The electric field on one side of Surface with charge density
The electric field on another side of Surface with charge density
Now, resultant of both surfaces:
As E1 and E2 are opposite in direction. we have
There has to be a discontinuity at the sheet of the charge since both electric fields are in the opposite direction.
Now,
Since the electric field is zero inside the conductor,
the electric field just outside the conductor is
Let's assume a rectangular loop of length l and small width b.
Now,
Line integral along the loop :
This implies
From here,
Since and are the tangential component of the electric field, the tangential component of the electric field is continuous across the surface
The charge density of the cylinder with length l and radius r =
The radius of another hollow cylinder with same length = R
Now, let our gaussian surface be a cylinder with the same length and different radius r
the electric flux through Gaussian surface
Hence electric field ar a distance r from the axis of the cylinder is
2.18 (a) In a hydrogen atom, the electron and proton are bound at a distance of about
As we know,
the distance between electron-proton of the hydrogen atom
The potential energy of the system = potential energy at infinity - potential energy at distance d
As we know,
Hence potential energy of the system is -27.2eV.
2.18 (b) In a hydrogen atom, the electron and proton are bound at a distance of about
the potential energy of the system is -27.2eV. (from the previous question)
Kinetic energy is half of the potential energy in magnitude. so kinetic energy = 27/2 = 13.6eV
so,
total energy = 13.6 - 27.2 = -13.6eV
Hence the minimum work required to free the electron is 13.6eV
2.18 (c) In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 :
When potential energy is zero at 1.06 away,
The potential energy of the system =potential energy at -potential energy at d
Hence potential energy, in this case, would be -13.6eV
Given,
Distance between proton 1 and 2
Distance between proton 1 and electron
Distance between proton 2 and electron
Now,
The potential energy of the system :
Substituting the values, we get
Since both spheres are connected through the wire, their potential will be the same
Let electric field at A and B be .
Now,
also
Also
Therefore,
Therefore the ratio of the electric field is b/a.
2.21 (a) Two charges –q and +q are located at points (0, 0, –a) and (0, 0, a), respectively.
(a) What is the electrostatic potential at the points (0, 0, z) and (x, y, 0) ?
1)electric potential at point (0,0,z)
distance from
distance from
Now,
Electric potential :
2) Since the point,(x,y,0) lies in the normal to the axis of the dipole, the electric potential at this point is zero.
2.21 (b) Two charges –q and +q are located at points (0, 0, –a) and (0, 0, a), respectively.
(b) Obtain the dependence of potential on the distance r of a point from the origin when
Here, since distance r is much greater than half the distance between charges, the potential V at a distance r is inversely proportional to the square of the distance
2.21 (c) Two charges –q and +q are located at points (0, 0, –a) and (0, 0, a), respectively
Since point (5,0,0) is equidistance from both charges, they both will cancel out each other potential and hence potential at this point is zero.
Similarly, point (–7,0,0) is also equidistance from both charges. and hence potential at this point is zero.
Since potential at both the point is zero, the work done in moving charge from one point to other is zero. Work done is independent of the path.
Here, As we can see
The electrostatic potential caused by the system of three charges at point P is given by
Since
From here we conclude that
Whereas we know that for a dipole,
And for a monopole,
Let's assume n capacitor connected in series and m number of such rows,
Now,
As given
The total voltage of the circuit = 1000V
and the total voltage a capacitor can withstand = 400
From here the total number of the capacitor in series
Since the number of capacitors can never be a fraction, we take n = 3.
Now,
Total capacitance required =
Number of rows we need
Hence capacitors should be connected in 6 parallel rows where each row contains 3 capacitors in series.
Given,
The capacitance of the parallel plate capacitor
Separation between plated
Now, As we know
Hence, to get capacitance in farads, the area of the plate should be of the order od kilometre which is not good practice, and so that is why ordinary capacitors are of range
Given.
Now,
Lets first calculate the equivalent capacitance of
Now let's calculate the equivalent of
Now let's calculate the equivalent of
Now,
The total charge on capacitors:
So,
The voltage across is given by
The charge on is given by
The potential difference across is
Hence Charge on
And Charge on :
(a) How much electrostatic energy is stored by the capacitor?
Here
The capacitance of the parallel plate capacitor :
The electrostatic energy stored in the capacitor is given by :
Hence, the electrostatic energy stored by the capacitor is .
The volume of the capacitor is:
Now,
Energy stored in the capacitor per unit volume :
Now, the relation between u and E.
Here,
The charge on the capacitance Initially
Total electrostatic energy initially
Now, when it is disconnected and connected to another capacitor
Total new capacitance =
Now, by conserving the charge on the capacitor:
Now,
New electrostatic energy :
Therefore,
Lost in electrostatic energy
Let
The surface charge density of the capacitor =
Area of the plate =
Now,
As we know,
When the separation is increased by ,
work done by external force=
Now,
Increase in potential energy :
By work-energy theorem,
putting the value of
origin of 1/2 lies in the fact that field is zero inside the conductor and field just outside is E, hence it is the average value of E/2 that contributes to the force.
Given
the radius of the outer shell =
the radius of the inner shell =
charge on Inner surface of outer shell =
Induced charge on the outer surface of inner shell =
Now,
The potential difference between the two shells
Now Capacitance is given by
Hence proved.
(a) Determine the capacitance of the capacitor.
The capacitance of the capacitor is given by:
Here,
Hence Capacitance of the capacitor is .
(b) what is the potential of the inner sphere?
Potential of the inner sphere is given by
Hence the potential of the inner sphere is .
The radius of the isolated sphere
Now, Capacitance of sphere:
On comparing it with the concentric sphere, it is evident that it has lesser capacitance.this is due to the fact that the concentric sphere is connected to the earth.
Hence the potential difference is less and capacitance is more than the isolated sphere.
Answer carefully:
2.31 (a)
The charge on the sphere is not exactly a point charge, we assume it when the distance between two bodies is large. when the two charged sphere is brought closer, the charge distribution on them will no longer remain uniform. Hence it is not true that electrostatic force between them exactly given by .
Answer carefully:
2.31 (b)
If Coulomb’s law involved dependence (instead of ), would Gauss’s law be still true?
Since the solid angle is proportional to and not proportional to ,
The guess law which is equivalent of coulombs law will not hold true.
Answer carefully:
2.31 (c)
when a small test charge is released at rest at a point in an electrostatic field configuration it travels along the field line passing through that point only if the field lines are straight because electric field lines give the direction of acceleration, not the velocity
Answer carefully:
2.31 (d)
The initial and final position will be the same for any orbit whether it is circular or elliptical. Hence work done will always be zero.
Answer carefully:
2.31 (e)
Since the electric potential is not a vector quantity unlike the electric field, it can never be discontinuous.
Answer carefully:
2.31 (f) What meaning would you give to the capacitance of a single conductor?
There is no meaning in the capacitor with a single plate factually. but we give it meaning by assuming the second plate at infinity. Hence capacitance of a single conductor is the amount of change required to raise the potential of the conductor by one unit amount.
Answer carefully:
2.31 (g)
Water has a much greater dielectric constant than mica because it posses a permanent dipole moment and has an unsymmetrical shape.
Given
Length of cylinder
inner radius
outer radius
Charge on the inner cylinder
Now as we know,
The capacitance of this system is given by
Now
Since the outer cylinder is earthed the potential at the inner cylinder is equal to the potential difference between two cylinders.
So
Potential of inner cylinder:
Given
Voltage rating in designing capacitor
The dielectric constant of the material
Dielectric strength of material =
Safety Condition:
The capacitance of the plate
Now, As we know,
Now,
Hence the minimum required area is
2.34 (a) Describe schematically the equipotential surfaces corresponding to a constant electric field in the z-direction
When the electric field is in the z-direction is constant, the potential in a direction perpendicular to z-axis remains constant. In other words, every plane parallel to the x-y plane is an equipotential plane.
2.34 (b) Describe schematically the equipotential surfaces corresponding to a field that uniformly increases in magnitude but remains in a constant (say, z) direction
The potential in a direction perpendicular to the direction of the field is always gonna be same irrespective of the magnitude of the electric field. Hence equipotential surface will be the plane, normal of which is the direction of the field.
2.34 (c) Describe schematically the equipotential surfaces corresponding to a single positive charge at the origin, and
For a single positive charge, the equipotential surface will be the sphere with centre at position of the charge which is origin in this case.
2.34 (d) Describe schematically the equipotential surfaces corresponding to a uniform grid consisting of long equally spaced parallel charged wires in a plane.
The equipotential surface near the grid is periodically varying.and after long distance it becomes parallel to the grid.
The potential difference between the inner sphere and shell;
So, the potential difference is independent of . And hence whenever q1 is positive, the charge will flow from sphere to the shell
Answer the following:
2.36 (a)
The surface of the earth and our body, both are good conductors. So our body and the ground both have the same equipotential surface as we are connected from the ground. When we move outside the house, the equipotential surfaces in the air changes so that our body and ground is kept at the same potential. Therefore we do not get an electric shock.
Answer the following:
2.36 (b)
Answer:
Yes, the man will get an electric shock. the aluminium sheet is gradually charged up by discharging current of atmosphere.eventually the voltage will increase up to a certain point depending on the capacitance of the capacitor formed by aluminium sheet, insulating slab and the ground. When the man touches the that charged metal, he will get a shock.
Answer the following:
2.36 (c)
Thunderstorm and lightning across the globe keep the atmosphere charged by releasing the light energy, heat energy, and sound energy in the atmosphere. In a way or other, the atmosphere is discharged through regions of ordinary weather. on an average, the two opposing currents are in equilibrium. Hence the atmosphere perpetually remains charged.
Answer the following:
2.36 (d)
Electrical energy, of the atmosphere, is dissipated as light energy which comes from lightning, heat energy and sound energy which comes from the thunderstorm.
The CBSE NCERT solutions for class 12 physics chapter 2 electrostatic potential and capacitance will help in securing a good score in board exam. On an average 4 to 6 % questions are asked from this chapter for the board exam. For JEE Main on an average 6% of questions are asked from electrostatics and for NEET exam 4 to 6% questions are asked from the unit electrostatics. The unit electrostatics contains the first two chapters of class 12 NCERT physics.
Will NCERT solve all the doubts pertaining to competitive exams?
NCERT can solve up to 70% of the doubts pertaining to competitive exams. Along with NCERT we can solve previous questions and use any one good reference books for more questions and concepts. For conceptual clarity NCERT is the best. Also solving NCERT questions will help in clearing the concept.