| Board | SCERT, Kerala |
| Text Book | NCERT Based |
| Class | Plus Two |
| Subject | Physics Textbook Solution |
| Chapter | Chapter 2 |
| Chapter Name | Electric Charges and Fields |
| Category | Plus Two Kerala |
Kerala Syllabus Plus Two Physics Textbook Solution Chapter 1 Electrostatic Potential and Capacitance
Plus Two Textbook Solutions
- Chapter 1: Electric Charges and Fields
- Chapter 2: Electrostatic Potential and Capacitance
- Chapter 3: Current Electricity
- Chapter 4: Moving Charges and Magnetism
- Chapter 5: Magnetism and Matter
- Chapter 6: Electromagnetic Induction
- Chapter 7: Alternating Current
- Chapter 8: Electromagnetic Waves
- Chapter 9: Ray Optics and Optical Instruments
- Chapter 10: Wave Optics
- Chapter 11: Dual Nature of Radiation and Matter
- Chapter 12: Atoms
- Chapter 13: Nuclei
- Chapter 14: Semiconductor Electronics: Materials, Devices, and Simple Circuits
- Chapter 15: Communication Systems
Plus Two Physics Part I
Plus Two Physics Part II
Chapter 2 Electrostatic Potential and Capacitance Solution
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?
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.
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.
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.
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.
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.
Two charges 2 μC and −2 µC are placed at points A and B 6 cm apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface?
(a) 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.
(b) The direction of the electric field at every point on this surface is normal to the plane in the direction of AB.
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
(a) Inside the sphere
(b) Just outside the sphere
(c) At a point 18 cm from the centre of the sphere?
(a) 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.
(b) 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 
.
(c) 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
.
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?
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, 
= 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.
Three capacitors each of capacitance 9 pF are connected in series.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?
(a) Capacitance of each of the three capacitors, C = 9 pF
Equivalent capacitance (C’) of the combination of the capacitors is given by the relation,
1C'=1C+1C+1C⇒1C'=19+19+19=13⇒C'=3 pF
Therefore, total capacitance of the combination is 3 pF
.
(b) Supply voltage, V = 120 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.
Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
(a) Capacitances of the given capacitors are
For the parallel combination of the capacitors, equivalent capacitor
is given by the algebraic sum,
Therefore, total capacitance of the combination is 9 pF.
(b) 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,
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,
(a) While the voltage supply remained connected.
(b) After the supply was disconnected.
(a) 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.
(b) 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,
A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
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
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
.
Additional Exercise
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).
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.
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.
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 centre of the cube, due to the eight charges, gets cancelled. This is because the charges are distributed symmetrically with respect to the centre of the cube. Hence, the electric field is zero at the centre.
Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:
(a) at the mid-point of the line joining the two charges, and
(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.
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
(a) 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.
(b) Consider a point Z such that normal distanceOZ = 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.
A spherical conducting shell of inner radius r1 and outer radius r2 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.
(a) 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,
(b) 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.
(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by
Where 
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 
(a) Electric field on one side of a charged body is E1 and electric field on the other side of the same body is E2. If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,
Where,
= Unit vector normal to the surface at a point
σ = Surface charge density at that point
Electric field due to the other surface of the charged body,
Electric field at any point due to the two surfaces,
Since inside a closed conductor, 
= 0,
∴
Therefore, the electric field just outside the conductor is 
.
(b) When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.
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?
Charge density of the long charged cylinder of length L and radius r is λ.
Another cylinder of same length surrounds the pervious cylinder. The radius of this cylinder is R.
Let E be the electric field produced in the space between the two cylinders.
Electric flux through the Gaussian surface is given by Gauss’s theorem as,
Where, d = Distance of a point from the common axis of the cylinders
Let q be the total charge on the cylinder.
It can be written as
Where,
q = Charge on the inner sphere of the outer cylinder
∈0 = Permittivity of free space
Therefore, the electric field in the space between the two cylinders is
.
The distance between electron-proton of a hydrogen atom, 
Charge on an electron, q1= −1.6 ×10−19C
Charge on a proton, q2= +1.6 ×10−19C
(a) Potential at infinity is zero.
Potential energy of the system, = Potential energy at infinity − Potential energy at distance d
where,
∈0 is the permittivity of free space
14πε0=9×109 Nm2C−2∴ Potential energy=0−9×109×(1.6×10−19)20.53×10−10=−43.47×10−19 J∵1.6×10−19 J=1 eV∴Potential energy=−43.7×10−19=−43.7×10−191.6×10−19=−27.2 eV
Therefore, the potential energy of the system is −27.2 eV.
(b) Kinetic energy is half of the magnitude of potential energy.
Total energy = 13.6 − 27.2 = 13.6 eV
Therefore, the minimum work required to free the electron is 13.6 eV.
(c) When zero of potential energy is taken, 
∴Potential energy of the system = Potential energy at d1 − Potential energy at d
If one of the two electrons of a H2 molecule is removed, we get a hydrogen molecular ion
. In the ground state of an, the two protons are separated by roughly 1.5 Ã…, and the electron is roughly 1 Ã… from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.
The system of two protons and one electron is represented in the given figure.
Charge on proton 1, q1 = 1.6 ×10−19 C
Charge on proton 2, q2 = 1.6 ×10−19 C
Charge on electron, q3 = −1.6 ×10−19 C
Distance between protons 1 and 2, d1 = 1.5 ×10−10 m
Distance between proton 1 and electron, d2 = 1 ×10−10 m
Distance between proton 2 and electron, d3 = 1 × 10−10 m
The potential energy at infinity is zero.
Potential energy of the system,
Therefore, the potential energy of the system is −19.2 eV.
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.
Let a be the radius of a sphere A, QAbe the charge on the sphere, and CA be the capacitance of the sphere. Let b be the radius of a sphere B, QB be the charge on the sphere, and CB be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (V) will become equal.
Let EAbe the electric field of sphere A and EB be the electric field of sphere B. Therefore, their ratio,
Putting the value of (2) in (1), we obtain
Therefore, the ratio of electric fields at the surface is
.
A sharp and pointed end can be treated as a sphere of very small radius and a flat portion behaves as a sphere of much larger radius.Therefore, charge density on sharp and pointed ends of the conductor is much higher than on its flatter portions.
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?
(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1.
(c) How much work is done in moving a small test charge from the point (5, 0, 0) to (−7, 0, 0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?
(a) Zero at both the points
Charge − q is located at (0, 0, − a) and charge + q is located at (0, 0, a). Hence, they form a dipole. Point (0, 0, z) is on the axis of this dipole and point (x, y, 0) is normal to the axis of the dipole. Hence, electrostatic potential at point (x, y, 0) is zero. Electrostatic potential at point (0, 0, z) is given by,
Where,
= Permittivity of free space
p = Dipole moment of the system of two charges = 2qa
(b) Distance r is much greater than half of the distance between the two charges. Hence, the potential (V) at a distance r is inversely proportional to square of the distance i.e., 
(c) Zero
The answer does not change if the path of the test is not along the x-axis.
A test charge is moved from point (5, 0, 0) to point (−7, 0, 0) along the x-axis. Electrostatic potential (V1) at point (5, 0, 0) is given by,
V1=−q4Ï€∈01(5−0)2√+(−a)2+ q4Ï€∈01(5−0)2+√(a)2 =−q4Ï€∈025+a2√+q4Ï€∈025+a2√ =0
Electrostatic potential, V2, at point (− 7, 0, 0) is given by,
Hence, no work is done in moving a small test charge from point (5, 0, 0) to point (−7, 0, 0) along the x-axis.
The answer does not change because work done by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.
Figure 2.34 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).
Four charges of same magnitude are placed at points X, Y, Y, and Z respectively, as shown in the following figure.
A point is located at P, which is r distance away from point Y.
The system of charges forms an electric quadrupole.
It can be considered that the system of the electric quadrupole has three charges.
Charge +q placed at point X
Charge −2q placed at point Y
Charge +q placed at point Z
XY = YZ = a
YP = r
PX = r + a
PZ = r − a
Electrostatic potential caused by the system of three charges at point P is given by,
Since
,
is taken as negligible.
It can be inferred that potential, 
However, it is known that for a dipole, 
And, for a monopole, 
An electrical technician requires a capacitance of 2 µF in a circuit across a potential difference of 1 kV. A large number of 1 µF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.
Total required capacitance, C = 2 µF
Potential difference, V = 1 kV = 1000 V
Capacitance of each capacitor, C1 = 1µF
Each capacitor can withstand a potential difference, V1 = 400 V
Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to each other. The potential difference across each row must be 1000 V and potential difference across each capacitor must be 400 V. Hence, number of capacitors in each row is given as
Hence, there are three capacitors in each row.
Capacitance of each row 
Let there are n rows, each having three capacitors, which are connected in parallel. Hence, equivalent capacitance of the circuit is given as
Hence, 6 rows of three capacitors are present in the circuit. A minimum of 6 × 3 i.e., 18 capacitors are required for the given arrangement.
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Chapter 2: Electrostatic Potential and Capacitance Textbook Solution
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Plus Two Physics Chapter Wise Textbook Solutions
- Chapter 1: Electric Charges and Fields
- Chapter 2: Electrostatic Potential and Capacitance
- Chapter 3: Current Electricity
- Chapter 4: Moving Charges and Magnetism
- Chapter 5: Magnetism and Matter
- Chapter 6: Electromagnetic Induction
- Chapter 7: Alternating Current
- Chapter 8: Electromagnetic Waves
- Chapter 9: Ray Optics and Optical Instruments
- Chapter 10: Wave Optics
- Chapter 11: Dual Nature of Radiation and Matter
- Chapter 12: Atoms
- Chapter 13: Nuclei
- Chapter 14: Semiconductor Electronics: Materials, Devices, and Simple Circuits
- Chapter 15: Communication Systems
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