Class 12 physics chapter 1 NCERT solutions: Electrostatic Potential and Capacitance PDF: We have bought here the best NCERT solutions. Download the PDF or see the answer online that up to you. We have also brought video on this. Get all the part of electrostatic.
Question 1:
Two charges 5 × 10−8 C and −3 × 10−8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
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Answer
There are two charges,
Distance between the two charges, d = 16 cm = 0.16 m
Consider a point P on the line joining the two charges, as shown in the given figure.
r = Distance of point P from charge q1
Let the electric potential (V) at point P be zero.
Potential at point P is the sum of potentials caused by charges q1 and q2 respectively.
Where,
= Permittivity of free space
For V = 0, equation (i) reduces to
Therefore, the potential is zero at a distance of 10 cm from the positive charge between the charges.
Suppose point P is outside the system of two charges at a distance s from the negative charge, where potential is zero, as shown in the following figure.
For this arrangement, potential is given by,
For V = 0, equation (ii) reduces to
Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.
Question 2
A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon.
Answer
The given figure shows six equal amount of charges, q, at the vertices of a regular hexagon.
Where,
Charge, q = 5 µC = 5 × 10−6 C
Side of the hexagon, l = AB = BC = CD = DE = EF = FA = 10 cm
Distance of each vertex from centre O, d = 10 cm
Electric potential at point O,
Where,
= Permittivity of free space
Therefore, the potential at the centre of the hexagon is 2.7 × 106 V.
Question 3:
Two charges 2 μC and −2 µC are placed at points A and B 6 cm apart.
Identify an equipotential surface of the system. What is the direction of the electric field at every point on this surface?
Answer
The situation is represented in the given figure.
An equipotential surface is the plane on which total potential is zero everywhere. This plane is normal to line AB. The plane is located at the mid-point of line AB because the magnitude of charges is the same.
The direction of the electric field at every point on this surface is normal to the plane in the direction of AB.
Question 4:
A spherical conductor of radius 12 cm has a charge of 1.6 × 10−7C distributed uniformly on its surface. What is the electric field
Inside the sphere
Just outside the sphere
At a point 18 cm from the centre of the sphere?
Answer
Radius of the spherical conductor, r = 12 cm = 0.12 m
Charge is uniformly distributed over the conductor, q = 1.6 × 10−7 C
Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it.
Electric field E just outside the conductor is given by the relation,
Where,
= Permittivity of free space
Therefore, the electric field just outside the sphere is .
Electric field at a point 18 m from the centre of the sphere = E1
Distance of the point from the centre, d = 18 cm = 0.18 m
Therefore, the electric field at a point 18 cm from the centre of the sphere is
.
Question 2.5:
A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF =
10−12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
Answer
Capacitance between the parallel plates of the capacitor, C = 8 pF
Initially, distance between the parallel plates was d and it was filled with air. Dielectric constant of air, k = 1
Capacitance, C, is given by the formula,
Where,
A = Area of each plate
= Permittivity of free space
If distance between the plates is reduced to half, then new distance, d’ =
Dielectric constant of the substance filled in between the plates, k= 6
Hence, capacitance of the capacitor becomes
Taking ratios of equations (i) and (ii), we obtain
Therefore, the capacitance between the plates is 96 pF
Question 6:
Three capacitors each of capacitance 9 pF are connected in series.
What is the total capacitance of the combination?
What is the potential difference across each capacitor if the combination is connected to a 120 V supply?
Answer
Capacitance of each of the three capacitors, C = 9 pF
Equivalent capacitance (C’) of the combination of the capacitors is given by the relation,
Therefore, total capacitance of the combination is .
Supply voltage, V = 100 V
Potential difference (V’) across each capacitor is equal to one-third of the supply voltage.
Therefore, the potential difference across each capacitor is 40 V.
Question 7:
Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. What is the total capacitance of the combination? Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Answer
Capacitances of the given capacitors are
For the parallel combination of the capacitors, equivalent capacitor algebraic sum,is given by the
Therefore, total capacitance of the combination is 9 pF
Supply voltage, V = 100 V
The voltage through all the three capacitors is same = V = 100 V
Charge on a capacitor of capacitance C and potential difference V is given by the relation,
q = VC … (i) For C = 2 pF,
For C = 3 pF,
For C = 4 pF,
Question 8:
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
Answer
Area of each plate of the parallel plate capacitor, A = 6 × 10−3 m2
Distance between the plates, d = 3 mm = 3 × 10−3 m
Supply voltage, V = 100 V
Capacitance C of a parallel plate capacitor is given by,
Where,
= Permittivity of free space
= 8.854 × 10−12 N−1 m−2 C−2
Therefore, capacitance of the capacitor is 17.71 pF and charge on each plate is 1.771 × 10−9 C.
Question 9:
Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, While the voltage supply remained connected. After the supply was disconnected.
Answer
Dielectric constant of the mica sheet, k = 6
Initial capacitance, C = 1.771 × 10−11 F
Supply voltage, V = 100 V
Potential across the plates remains 100 V.
Dielectric constant, k = 6
Initial capacitance, C = 1.771 × 10−11 F
If supply voltage is removed, then there will be no effect on the amount of charge in the plates.
Charge = 1.771 × 10−9 C
Potential across the plates is given by,
Question 10:
A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
Answer
Capacitor of the capacitance, C = 12 pF = 12 × 10−12 F
Potential difference, V = 50 V
Electrostatic energy stored in the capacitor is given by the relation,
Therefore, the electrostatic energy stored in the capacitor is
Question 11:
A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
Answer
Capacitance of the capacitor, C = 600 pF Potential difference, V = 200 V
Electrostatic energy stored in the capacitor is given by,
If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance (C’) of the combination is given by,
New electrostatic energy can be calculated as
Therefore, the electrostatic energy lost in the process is . .
Question 12:
A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of −2 × 10−9 C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).
Answer
Charge located at the origin, q = 8 mC= 8 × 10−3 C
Magnitude of a small charge, which is taken from a point P to point R to point Q, q1 = − 2 × 10−9 C
All the points are represented in the given figure.
Point P is at a distance, d1 = 3 cm, from the origin along z-axis.
Point Q is at a distance, d2 = 4 cm, from the origin along y-axis.
Potential at point P,
Potential at point Q,
Work done (W) by the electrostatic force is independent of the path.
Therefore, work done during the process is 1.27 J.
Question 13:
A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.
Answer
Length of the side of a cube = b
Charge at each of its vertices = q
A cube of side b is shown in the following figure.
d = Diagonal of one of the six faces of the cube
l = Length of the diagonal of the cube
The electric potential (V) at the centre of the cube is due to the presence of eight charges at the vertices.
Therefore, the potential at the centre of the cube is .
The electric field at the center of the cube, due to the eight charges, gets cancelled. This is because the charges are distributed symmetrically with respect to the center of the cube. Hence, the electric field is zero at the center.
Question 14:
Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the
potential and electric field: at the mid-point of the line joining the two charges, and at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.
Answer
Two charges placed at points A and B are represented in the given figure. O is the mid- point of the line joining the two charges.
Magnitude of charge located at A,
q1 = 1.5 μC Magnitude of charge located at B,
q2 = 2.5 μC Distance between the two charges,
d = 30 cm = 0.3 m
Let V1 and E1 are the electric potential and electric field respectively at O.
V1 = Potential due to charge at A + Potential due to charge at B
Where,
∈0 = Permittivity of free space
E1 = Electric field due to q2 − Electric field due to q1
Therefore, the potential at mid-point is 2.4 × 105 V and the electric field at mid-point is 4× 105 V m−1. The field is directed from the larger charge to the smaller charge.
Consider a point Z such that normal distance OZ = 10 cm = 0.1 m, as shown in the following figure.
V2 and E2 are the electric potential and electric field respectively at Z.
It can be observed from the figure that distance,
V2= Electric potential due to A + Electric Potential due to B
Electric field due to q at Z,
Electric field due to q2 at Z,
The resultant field intensity at Z,
Where, 2θis the angle, ∠AZ B
From the figure, we obtain
Therefore, the potential at a point 10 cm (perpendicular to the mid-point) is 2.0 × 105 V
and electric field is 6.6 ×105 V m−1.
Question 15:
A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q.
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? 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
Charge placed at the centre of a shell is +q. Hence, a charge of magnitude −q will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is −q.
Surface charge density at the inner surface of the shell is given by the relation,
A charge of +q is induced on the outer surface of the shell. A charge of magnitude Q is placed on the outer surface of the shell. Therefore, total charge on the outer surface of the shell is Q + q. Surface charge density at the outer surface of the shell,
Yes
The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape. Take a closed loop such that a part of it is inside the cavity along a field line while the rest is inside the conductor. Net work done by the field in carrying a test charge over a closed loop is zero because the field inside the conductor is zero. Hence, electric field is zero, whatever is the shape.
Question 36:
A small sphere of radius r1 and charge q1 is enclosed by a spherical shell of radius r2 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 electric field between a sphere and a shell is determined by the charge q1 on a small sphere. Hence, the potential difference, V, between the sphere
and the shell is independent of charge q2. For positive charge q1, potential difference V is
always positive.
Question 37:
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!)
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?
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?
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
We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential.
Yes, the man will get an electric shock if he touches the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminium sheet. As a result, its voltage rises gradually. The raise in the voltage depends on the capacitance of the capacitor formed by the aluminium slab and the ground.
The occurrence of thunderstorms and lightning charges the atmosphere continuously. Hence, even with the presence of discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains electrically neutral.
During lightning and thunderstorm, light energy, heat energy, and sound energy are dissipated in the atmosphere.
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