Electrostatic Potential And Capacitance Class 12th Physics Part I CBSE Solution

Class 12th Physics Part I CBSE Solution
Exercises
  1. Two charges 5 108 C and 3 108 C are located 16 cm apart. At what point(s) on the line…
  2. A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the…
  3. Two charges 2 μC and -2 μC are placed at points A and B 6 cm apart. (a) Identify an…
  4. A spherical conductor of radius 12 cm has a charge of 1.6 × 10-7C distributed uniformly on…
  5. A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF =…
  6. Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total…
  7. Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. (a) What…
  8. In a parallel plate capacitor with air between the plates, each plate has an area of 6 ×…
  9. Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica…
  10. A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in…
  11. A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and…
Additional Exercises
  1. A charge of 8 m C is located at the origin. Calculate the work done in taking a small…
  2. A cube of side b has a charge q at each of its vertices. Determine the potential and…
  3. Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the…
  4. A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q. (a) A…
  5. Show that the normal component of electrostatic field has a discontinuity from one side of…
  6. Show that the tangential component of electrostatic field is continuous from one side of a…
  7. A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial…
  8. In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å: (a)…
  9. If one of the two electrons of a H2 molecule is removed, we get a hydrogen molecular ion…
  10. Two charged conducting spheres of radii a and b are connected to each other by a wire.…
  11. Two charges -q and + q are located at points (0, 0, -a) and (0, 0, a), respectively. (a)…
  12. Figure 2.34 shows a charge array known as an electric quadrupole. For a point on the axis…
  13. An electrical technician requires a capacitance of 2 μF in a circuit across a potential…
  14. What is the area of the plates of a 2 F parallel plate capacitor, given that the…
  15. Obtain the equivalent capacitance of the network in Fig. 2.35. For a 300 V supply,…
  16. The plates of a parallel plate capacitor have an area of 90 cm^2 each and are separated by…
  17. A 4 μF capacitor is charged by a 200 V supply. It is then disconnected from the supply,…
  18. Show that the force on each plate of a parallel plate capacitor has a magnitude equal to…
  19. A spherical capacitor consists of two concentric spherical conductors, held in position by…
  20. A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13…
  21. Answer carefully: (a) Two large conducting spheres carrying charges Q1 and Q2 are brought…
  22. A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and…
  23. A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material…
  24. Describe schematically the equipotential surfaces corresponding to (a) a constant electric…
  25. In a Van de Graff type generator a spherical metal shell is to be a 15 × 10^6 V electrode.…
  26. A small sphere of radius r1 and charge q1 is enclosed by a spherical shell of radius r2…
  27. Answer the following: (a) The top of the atmosphere is at about 400 kV with respect to the…

Exercises
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.


Answer:

Given,

Two charges qA = 5 x 10-8 C and qB = -3x10-8 C



Distance between two charges, r = 16 cm = 0.16 m




Consider a point O on the line joining two charges where the electric potential is zero due to two charges.





From the figure we can see that, x = distance of point O from charge qA




Electric potential at point O due to qA,





ϵ0 is the permittivity of free space. Its value is 8.85 x 10-12 F/m




Electric potential at point O due to qB







Since the total electric potential at O is zero,




⇒ VA + VB = 0



⇒ 


⇒ 


On cross multiplying we get,


5 × (0.16 -x) = 3x


⇒ 0.8 – 5x = 3x


⇒ 8x = 0.8




⇒x = 0.1 m = 10 cm(from charge qA)


∴ at a distance of 10 cm from the positive charge, the potential is zero between the two 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:

Let O be the center of the hexagon. It contains the charges at all its 6 vertices, each charge = + 5 μC = 5 × 10-6 C. The side of the hexagon is 10 cm = 0.1 m


It follows that the point O, when joined to the two ends of a side of the hexagon forms an equilateral triangle


Electric potential at O due to one charge,



Here, q = amount of charge = 5 × 10-6 C


r = distance between charge and O = 0.1 m


ε0 = absolute permittivity of free space


Since at the each of the hexagon, a charge of 5 μC (5 × 10-6 C) is placed, total electric potential at the point O due to the charges at the six corners,



⇒ 


⇒ 


⇒ V1 = 2.7 × 106 V



Question 3.

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?


Answer:


(a) For the given system of two charges, the equipotential surface will be a plane normal to the line AB joining the two charges and passing through its mid-point O. On any point on this plane, the potential is zero.


(b) The electric field is in a direction from the point A to point B i.e. from the positive charge to the negative charge and normal to the equipotential surface.



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

(a) inside the sphere

(b) just outside the sphere

(c) at a point 18 cm from the centre of the sphere?


Answer:

Given, q = 1.6 x 10-7 C

Radius of the sphere, r = 12 cm = 0.12 m


(a) Inside the sphere: The charge on a conductor resides on its outer surface. Therefore, electric field inside the sphere is zero.


(b) Just outside the sphere: For a point on the charged spherical conductor or outside it, the charge may be assumed to be concentrated at its center.


Therefore, the electric field just outside the sphere is given as:



where, q is the charge due to which the field is being calculated



putting the values of “r” and “q” in the equation, we get,



∴E = 105 N/C


(c) At a point 18 cm from the center: Given, R = 18 cm


Electric field at a point 18 cm from the center




∴ E = 4.4 x 104 N/C



Question 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:

Given: Capacitance of capacitor when medium between two plates is air, C = 8 pF = 8 × 10–12 F

Now, the capacitance of a parallel plate capacitor is given as:



where, A be the area of each plate


d be the distance between the two plates of the parallel plate capacitor


k is the dielectric constant, (for air , k = 1)


Capacitance, C = 8 pF(Given)


⇒ 


Suppose that the capacitance of the capacitor becomes C’ when the distance between the plates is reduced to half (d’ = d/2) and the space between them is filled with a substance of dielectric constant K = 6.



From equation (i) and (ii),


C’ = 12 x 8 x 10-12 = 96 x 10-12 = 96 pF



Question 6.

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?


Answer:

Given, C1 = C2 = C3 = 9 pF = 9 x 10-12 F

V = 120 volt



(a) Total capacitance of the series combination is given by



⇒ 


⇒ 



Cnet = 3 x 10-12 F = 3 pico-Farad.


(b) Let q be the charge on each capacitor.


Sum of the potential difference across the plates of the three capacitors must be equal to 120 V


⇒ V1 + V2 + V3 = Vnet


⇒  (∵ Vnet = 120 V)


⇒ 


⇒ 


⇒ q = 360 x 10-12 C


Therefore, potential difference across a capacitor




Question 7.

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.


Answer:

Given, C1 = 2 pF

C2 = 3 pF


C3 = 4 pF , V = 100 volt



(a) Total capacitance of the parallel combination is


C = C1 + C2 + C3 = 2 + 3 + 4 = 9 pF


(b) Let q1 , q2 and q3 be that charges on the capacitor C1 , C2 and C3 respectively.


In the parallel combination the potential difference across each capacitor will be equal to the supply voltage i.e., 100 V


⇒ q1 = C1V = 2 x 10-12 × 100 = 2 × 10-10 C


⇒ q2 = C2V = 3 x 10-12 × 100 = 3 × 10-10 C


⇒ q3 = C3V = 4 x 10-12 × 100 = 4 × 10-10 C



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:

Given, A = 6 x 10-3 m2

D = 3 mm = 3 x 10-3 m


Capacitance of the capacitor,



⇒ C = 1.77 x 10-11 F = 17.7 pF


When the capacitor is connected to a 100 V supply, the charge on the each plate of the capacitor,


q = CV = 1.77 x 10-11 x 100 = 1.77 x 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,

(a) while the voltage supply remained connected.

(b) after the supply was disconnected.


Answer:

(a) When the voltage supply remains connected:

The capacitance of the capacitor will become K times.


Therefore, C’ = kC


Where k = dielectric constant = 6 × 17.7pF = 106.2 pF


The potential difference across the two plates of the capacitor will remain equal to the supply voltage i.e. 100 V


The charge on the capacitor,


q’ = C’V = 160.2 x 10-12 x 100


= 1.602 x 10-8 C


(b) After the voltage supply is disconnected:


As calculated above, the capacitance of the capacitor, C’ = 106.2 pF


The potential difference will decrease on introducing mica sheet by a factor of K,


⇒ 


The charge on the capacitor,


q’ = C’V’ = 106.2 × 10-12 × 16.67


q’ = 1.77 x 10-9 C



Question 10.

A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?


Answer:

Given, C = 12 pF = 12 x 10-12 F

V = 50 V


The electrostatic energy stored in the capacitor,


W = (1/2) CV2 = (1/2) × 12 × 10-12 × (50)2 = 1.5 × 10-8 J



Question 11.

A 600pF capacitor is charged by a 200V 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:

Given, C1 = 600 pF = 600 x 10-12 F

V1 = 200 V


Energy stored in the capacitor,


U1 = (1/2) C1 (V1)2 = (1/2) × 600 × 10-12 × (200)2


= 12 × 10-6 J


When this charged capacitor is connected to another uncharged capacitor C2 ( = 600 pF) ,they will share charges, till potential differences across their plates become equal.


Total charge on the two capacitors,


q = C1V1 + C2V2 = 600 × 10-12 × 200 + 0


= 12 × 10-8 C


Total capacitance of the two capacitors,


C = C1 + C2 = 600 pF + 600 pF


= 1200 pF


= 1200 x 10-12 F


Therefore, common potential of the two capacitors,



Energy stored in the combination of the two capacitors,


U2 = (1/2)CV2 = (1/2) x 1200 x 10-12 x (100)2


= 6 x 10-6 J


Therefore, energy lost by the capacitor C1( in the form of heat and electromagnetic radiation),


U1 – U2 = 12 x 10-6 – 6 x 10-6 = 6 x 10-6 J




Additional Exercises
Question 1.

A charge of 8 m C 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:

Given,


Charge at origin (O), q0 = 8 × 10-3 C


The charge (q1 = -2 Nc) is moving through the given points, P(0, 0, 3)cm, R(0, 6, 9)cm and Q(0, 4, 0)cm respectively.


The picture below represents above situation,



Since, Work is independent on the path followed we concentrate on point P and Q only.


Potential at any point is given by, 


Where,


q = charge


d = distance from origin


ϵ0 = permittivity of space


and, 


Thus, Potential at point P, 


(∵ distance of point from origin = 3 cm = 0.03m)


Potential at point Q, 


Work done, W = q1 × (Change in potential difference)





= 1.27 J



Question 2.

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:

Given, side of the cube = b units


Charge at each vertex = q C



In the cube shown above,


d = diagonal of any face =  =  b units


l = diagonal of whole cube =  = √3 b units


Thus, the distance of centre of the cube from each vertices =  =  units


∴ Electric potential at the centre of the cube, V = 


(∵ 8 charges)


Where,


q = charge


d = distance from origin=


ϵ0 = permittivity of space


∴ V =  N units C-1


As the charges as symmetrically placed along the vertex of the cube, the net Electric field will be zero at the centre as the field due to 2 charges placed opposite to each other will cancel, so for all charges, net field will be 0.


Question 3.

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.


Answer:

Potential at any point is given by, V =  N m C-1


Electric field generated by a charge, E =  N C-1


Where,


q = charge


d = distance from origin


ϵ0 = permittivity of space


and,  = 9 × 109 N m2 C-2


(a) Figure below represents the given situation,



Let the potential at O (midpoint) be V and Electric field be E.


We have,


∴ V =  +  = 2.4 × 105 N m C-1


and,


E = electric field due to charge 2.5μC – electric field due to charge 1.5μC


∴E = 


⇒ E = 4 × 105 N C-1


(b) Figure below represents the given situation,



From the figure point Z is required location and its distance from both the charges is equal and equal to,


d = AZ = BZ = 


⇒ d = 0.18 m


and θ = 


∴ θ = 56.250


Thus, Potential and electric field at point Z is given by,


V = 


⇒ V = 2 × 105 N m C-1


and,


E = 


Where,


E1 = electric field generated by 1.5μC charge at Z


E2 = electric field generated by 2.5μC charge at Z


and, θ = angle between distance and perpendicular height (as shown above) and,



∴ E = 6.6 × 105 N C-1



Question 4.

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.


Answer:

(a) Charge placed at the centre of the sphere is + q. Thus, charge induced on the inner surface be –q.


Surface charge density at the inner surface of the sphere,


σ1 =  = 


Charge placed at the centre of the sphere is + q. Thus, charge induced on the inner surface be + q and surface charge distribution is Q. Therefore Total charge = Q + q.


Surface charge density at the outer surface of the sphere,


σ2 =  = 


(b) Yes


Whatever the shape of the shell, the electric field intensity in the cavity is zero. Take a closed loop such that it is inside the cavity and in conductor. Net work done in carrying charge along closed loop by electric field is zero. Thus, electric field is zero, whatever the shape.



Question 5.

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 .


Answer:

Electric field on one side of the charged is body is E1 and the electric field on the other side of the same body be E2. If infinite plane charged body has uniform thickness then the electric field due to one surface of the body is given by,


 = 


Where,


 = a unit vector normal to the surface at a point


σ = surface charge density at that point.


ϵ0 = permittivity of free space.


Electric field due to other surface of the charged body,


 = 


∴ Electric field due to both surfaces,


 -  =  +  = 


Since, Inside a closed conductor electric field s zero,


∴  =  = 


Hence, the electric field just outside the conductor = 



Question 6.

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:

Net work done in carrying charge along closed loop by electric field is zero. Hence, the tangential component of the electric field is continuous from one side of a charged surface to the other.



Question 7.

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:

Given,


Charge density of the long-charged cylinder of length L and radius r is λ. Another same type of cylinder with radius R surrounded it.


Let E is the electric field produced in the space between the two cylinders.


Electric flux through a Gaussian surface is given by the Gaussian theorem as,


Φ = E(2πd)L


Where, d = distance of a point from common axis of the cylinders.


Let q be the total charge on the cylinder,


∴ Φ = E(2πd)L = 


Where, q = charge of the inner sphere of the outer cylinder


ϵ0 = permittivity of space


Thus,


E(2πd)L = 


∴ 


Hence, the electric field between cylinders, 



Question 8.

In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:

(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.

(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?

(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?


Answer:

Given,


The distance between electron-proton of hydrogen atom = 0.53 Å = 0.53 × 10-10m


Charge on electron, q1 = - 1.6 × 10-19 C


Charge on proton, q2 = 1.6 × 10-19 C


(a) Potential energy at infinity is Zero.


Potential energy of a system,


= Potential energy at infinity – Potential energy at distance d


= 0 - 


Where, ϵ0 = permittivity of space


and,  =


∴ Potential energy, 


⇒ V = -43.7 × 10-19 J


On converting into electron volts we get,


⇒ 


⇒ V = -27.2 eV


(∵ 1eV = 1.6 × 10-19 J)


(b) Given, Kinetic energy is half of the potential energy in (a).


∴ Kinetic energy, K = 0.5 × (27.2) = 13.6 eV


Thus, Total energy of electron = -27.2 + 13.6 = -13.6 eV


Amount of work require to free the electron, A = Increase in energy the electron.


∴ A = 0-(-13.6) = 13.6 eV


(c) If we take potential energy Zero at 1.06 Å separation, then the potential energy of the system,


P = 


⇒P = -21.74 × 10-19 J


In electron volts, the potential energy is given as,


P = -13.585 eV.


(1eV = 1.6 × 10-19 J)


∴ The amount of work done to free the electron in this case,


W = -27.17- (-13.585) = -13.585 eV



Question 9.

If one of the two electrons of a H2 molecule is removed, we get a hydrogen molecular ion H+2. In the ground state of an H+2, the two protons are separated by roughly 1.5 Å, and the electron is roughly 1 Å from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.


Answer:

The system of two protons and electron is represented as below,



The total potential energy at infinity is Zero.


Thus, Potential energy of the system is,


V = 


Where,


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 proton 1 and proton 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


ϵ0 = Permittivity of free space


and, 


∴ V = 



⇒ V = -30.7 × 10-19 J


Writing in electron volts, we get,


⇒ V = -19.2 eV


(∵ 1eV = 1.6 × 10-19 J)


Therefore, the potential energy of the system is -19.2 eV.



Question 10.

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:

Let a be the radius of the sphere A, QA be the charge on the sphere, and CA be the capacitance of the sphere. Let b be the radius of the sphere B, QB be the charge on the sphere, and CB be the capacitance of the sphere. Since, the two spheres are connected with wire, their potential V is equal.


Electric field generated by a charge, E = 


Where,


q = charge


d = distance from origin


ϵ0 = permittivity of space


Let EA be the electric field for sphere A and EB be the electric field for sphere B. Therefore their ratio,


 = 


However,  =  and  = 


⇒  = 


∴  =  = 


Hence, the ration of electric fields at surface is 



Question 11.

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)?

(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?


Answer:

The charges –q and + q at (0,0,-a) and (0,0,a) respectively forms a dipole.



(a) The point (0,0,z) lies on the axis of dipole, while the point (x,y,0) lies on the equatorial plane of the dipole.


The electric potential of dipole on any point on its axis is given by the formula–



where,  = Absolute Permittivity of free space


p = Dipole moment of the system of two charges


r = Distance of point from the centre of the dipole


here, p = 2qa and r = z.


The electric potential of dipole at point (0,0,z) = 


The electric potential of dipole on any equatorial point is zero


The electric potential of dipole at point (x,y,0) = 0.


(b) Given r/a>>1, which implies r>>a


the distance of point where potential is to be obtained 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. V  1/r2


(c) 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 = 



= 0


Electrostatic potential(V2) at point (-7, 0, 0) is given by,


V2 = 



= 0


The potential difference between points (5,0,0) and (-7,0,0,) is V1-V2 = 0.


Thus,


The work done in moving a small test charge from the point (5,0,0) to (–7,0,0) along the x-axis = 0.


The answer does not change if the path of the test charge between the same points is not along the x-axis 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.



Question 12.

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).



Answer:

The diagram is given as:


X Y Z



The given charges of same magnitude placed at points X, Y, and Z respectively, forms an electric quadrupole.


Where, charge + q is at point X, charge -2q is at point Y, and charge + q is at point Z.


The point P is at a distance r from point Y.


Here, XY = YZ = a


So, YP = r, PX = r + a, PZ = r-a.


The electrostatic potential due to the system of three charges at point P is given by,


∴ 


⇒ 


⇒ 


⇒ 


⇒ 


⇒ 


Since, r/a>>1


Then a/r<<1 which implies a2/r2 is negligible so it can be ignored,


Thus, potential becomes,


V = 


Relation of potential due to quadrupole with r is V1/r3


However, for a dipole it is V1/r2


And for a monopole it is V1/r



Question 13.

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.


Answer:

Potential difference across the circuit = 1kV = 1000V


Capacitance of each capacitor = 1 μF


Potential difference each capacitor can withstand = 400V


Capacitance required across the circuit = 2 μF


Assume n number of capacitors are connected in series and further m number of such series circuits are connected in parallel to each other.


As the potential difference in the circuit is 1000V so the potential difference across each row of n capacitors is 1000V, as the potential difference each capacitor can withstand is 400V,


Therefore, 400V × n = 1000V


⇒ n = 1000V/400V = 2.5~3capacitors in each row.


Now,


Capacitance of each row =  μF


As there are m column of three capacitors, which are connected in parallel.


Hence, equivalent capacitance of the circuit is


 + ……….m times = 


Since, Required capacitance = 2 μF



⇒ m = 6


Total number of capacitors = 3 × 6 = 18.



Question 14.

What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realize from your answer why ordinary capacitors are in the range of μF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]


Answer:

Let the area of the plates of capacitor be A.


Capacitance of a parallel capacitor (C) = 2 F


Distance between the two plates (d) = 0.5cm = 0.005m


Capacitance of a parallel plate capacitor is given by the relation,




Where,


 = Absolute Permittivity of free space = 8.8510-12 C2N-1m-2


 = 0.00112991012m2 ~ 1130km2


As the area of the plates required becomes too large so the capacitance is taken in the range of μF .



Question 15.

Obtain the equivalent capacitance of the network in Fig. 2.35. For a 300 V supply, determine the charge and voltage across each capacitor.



Answer:

Capacitance of capacitor C1 = 100pF, C2 = 200pF, C3 = 200 pF, C4 = 100 pF.


Supply potential (V) = 300 V


Let the equivalent capacitance of capacitors C2 and C3(connected in series) be C'.



⇒ C’ = 100pF


Let the equivalent capacitance of capacitors C1 and C’(connected in parallel) be C’’.


 C’’ = C’ + C1 = 100 + 100 = 200pF


Let the equivalent capacitance of C’’ and C4 (connected in series) be C’’’.



⇒ C’’’ = 200/3pF


Thus, the equivalent capacitance of the circuit = 200/3 pF.


Now, Supply potential (V) = 300 V


Charge on C4 is given by,


Q4 = C’’’V =  = 2 × 10-8C


So, V4 = Q4/C4 = (210-8)/(10010-12) = 200C


Potential difference across C1 is given by, V1 = V-V4 = 300-200 = 100V


Charge on C1 is given by, Q1 = C1V1 = 100 × 10-12 × 100 = 10-8C


C2 and C3 having same capacitances have a potential difference of 100 V together.


Since C1 and C3 are in series, the potential difference across C2 and C3 is given by,V2 = V3 = 50V


Therefore charge on C2 is given by,Q2 = C2V2 = 20010-1250 = 10-8C


And charge on C3 is given by,Q3 = C3V3 = 20010-1250 = 10-8C.


The charge and voltage across each capacitor are as follows:




Question 16.

The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400V supply.

(a) How much electrostatic energy is stored by the capacitor?

(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.


Answer:

Area of the plates of a parallel capacitor (A) = 90cm2 = 910-3m

Distance between the plates (d) = 2.5mm = 2.510-3m


Potential difference across the plates (V) = 400V


(a) Electrostatic energy stored in the capacitor(E) = CV2


Since, Capacitance (C) = 


∴ 


Where,  = Absolute Permittivity of free space = 8.8510-12C2N-1m-2



⇒ E = 2.5510-6J


The electrostatic energy stored by the capacitor = 2.5510-6J


(b) Volume between plates of capacitor (v) = A × d


= 910-3m2 × 2.510-3m


= 2.2510-5m3


Therefore, Energy per unit volume (u) = E/v




The energy per unit volume (u) = 0.113 Jm-3


Relation between u and the magnitude of electric field E between the plates


Since,


Energy per unit volume (u) = E/v


CV2 /v


Here, V = voltage across capacitor


v = volume between plates of capacitor,




Here, V/d is the Electric intensity or Electric field so, 



Question 17.

A 4 μF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?


Answer:

Capacitance of a charged capacitor(C) = 4 μF = 410-6F

Voltage supplied to the capacitor (V) = 200 V


Electrostatic energy of the capacitor (E) =  CV2 =  = 810-2 J


Capacitance of an uncharged capacitor (C’) = 2 μF = 210-6F


When both the capacitors are connected in a circuit, then initial charge on charged capacitor is equal to the final charge on both the capacitors in circuit. (According to law of conservation of charges)


Since, Charge = Voltage Capacitence


Therefore, CV = (C + C’) V’, where V’ is the voltage in the circuit when both capacitors are connected.


410-6F200V = (410-6F + 210-6F) V’



⇒ V =  V


Now, Electrostatic energy for the combination of two capacitors = 1/2 (C + C’)V’2



∴ U = 5.3310-2 J


Thus, the electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation = 810-2 J-5.3310-2 J = 6.6710-2J



Question 18.

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1/2) 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 1/2.


Answer:

Let the force applied to separate the plates of a parallel plate capacitor by a distance of x be F. Hence, work done = Fx.


The increase in potential energy of the capacitor = uv = uAx


Where, u = Energy density, A = area of each plate and v = volume between plates of capacitors.


Since, work done will be equal to the increase in the potential energy i.e.,


work done = increase in potential energy of the capacitor


Fx = uAx


F = uA



Since, u = and  = E (Electric intensity or Electric field)





and 


Therefore, F = 1/2 (CV)E and Q = CV


Thus, F = QE


The physical origin of the factor in the force formula lies in the fact that just outside the conductor, electric field is E and inside it is zero. Hence, it is the average value  of the field that contributes to the force.



Question 19.

A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig. 2.36). Show that the capacitance of a spherical capacitor is given by



Where r1 and r2 are the radii of outer and inner spheres, respectively.



Answer:

Radius of the outer shell = r1


Radius of the inner shell = r2


Charge on the inner surface of the outer shell = Q


Charge on the outer surface of the inner shell = -Q


Potential difference between the two shells, 



Where,  = Absolute Permittivity of free space = 8.8510-12C2N-1m-2


Since, Capacitance, 




Hence, proved.



Question 20.

A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 μC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.

(a) Determine the capacitance of the capacitor.

(b) What is the potential of the inner sphere?

(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.


Answer:

Radius of the outer shell (r1) = 13cm = 0.13m


Radius of the inner shell (r2) = 12cm = 0.12m


Charge on the outer surface of the inner shell = 2.5 μC = 2.510-6C


Dielectric constant of liquid () = 32


Since, Potential difference between the two shells,



∴ Δ


Where,  = Absolute Permittivity of free space = 8.8510-12C2N-1m-2


Therefore 




(a) Capacitance, 



⇒ C = 5.5510-9F


Hence, the capacitance of the capacitor = 5.5510-9F


(b) The potential of a capacitor (V), is given as:



⇒ 


⇒ V = 450 V


Hence, the potential of the inner sphere = 450V


(c) Radius of the isolated sphere(R) = 12 cm = 0.12m


Capacitance of an isolated sphere = 4πε0R


∴ C = 4 × 3.14 × 8.85 × 10-12 × 0.12


C = 1.3310-11F


Where, ϵ0 is the Absolute Permittivity of free space = 8.8510-12C2N-1m-2The capacitance of the isolated sphere is less than that of the concentric spheres because the outer sphere of the concentric spheres is earthed. Hence, the potential difference is less and the capacitance is more than the isolated sphere.



Question 21.

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/r3 dependence (instead of 1/r2), 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) According to question, two large conducting spheres with charge Q1 and Q2 are placed such that distance between their centre is r. Then, the magnitude of electrostatic force between the two spheres is not exactly equal to 


Since, the magnitude of electrostatic force between two-point charges Q1 and Q2 placed at a distance r is given by the relation,


.


Here both the spheres have non-uniform distribution of charges Q1 and Q2 on their surface.


(b) Gauss’s law states that net electric flux through a surface is the electric field of the charge enclosed by the surface multiplied by the area of the surface projected in a plane perpendicular to the field and this product is a constant equal to charge enclosed within the surface divided by ϵ0


i.e. 


or 


Where, electric flux,


E = electric field


and dA is the elementary area vector of the surface, Q = charge enclosed by the surface, ϵ0 = permittivity of free space.


Since, Electric field involves 1/r2 and area vector involves the term of r2, therefore their product becomes constant.


But, If Coulomb’s law involved 1/r3 dependence (instead of 1/r2), product of electric field and area vector Gauss’s law will not be true.


(c) If a small test charge is released at rest at a point in an electrostatic field configuration, then it will travel along the field lines passing through that point, as the tangent drawn at a point on the field line gives the direction of acceleration at that point.


(d) When the electron completes an orbit, either circular or elliptical, the displacement becomes zero, so the work done by the field of a nucleus is zero. ∵ Work = force × displacement


(e) Electric field is discontinuous across the surface of a charged conductor, but the electric potential is continuous.


(f) The capacitance of a single conductor means a capacitance of a parallel plate capacitor with one plate as the conductor while the other placed at infinity.


(g) Water has a permanent dipole moment while mica doesn’t, because water has an unsymmetrical space as compared to mica. So, water has a greater dielectric constant than mica.



Question 22.

A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 μC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).


Answer:

Radius of outer cylinder(R) = 1.5cm = 0.015m


Radius of inner cylinder(r) = 1.4cm = 0.014m


Charge on the inner cylinder(q) = 3.5m = 3.5 × 10-6C


Length of a co-axial cylinder(l) = 15cm = 0.15m


Capacitance of a co-axial cylinder of radii R and r, is given as:



where ϵ0 = Permittivity of free space = 8.8510-12N-1 m-2 C2


Therefore, 


(Trick: Multiply by 2.303 to convert loge to log10)


∴ C = 1.210-10F


Potential difference of the inner cylinder,



∴ 


∴ V = 2.29104V


Therefore, Capacitance = 1.210-10F


Potential difference = 2.29104V



Question 23.

A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm-1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength.

What minimum area of the plates is required to have a capacitance of 50 pF?


Answer:

Potential difference of a parallel plate capacitor(V) = 1kV = 1000V


Dielectric constant of a material(ϵr) = 3


Dielectric strength = 107V/m


Electric field intensity(E) = 10%of 107


⇒ E = 106V/m


(since, the field intensity never exceeds 10% of the dielectric strength)


Capacitance of the parallel plate capacitor(C) = 50pF = 50 × 10-12F


Distance between the plates (d), 


⇒ 


⇒ d = 10-3m


∵ Capacitance = 


where, ϵ0 = Permittivity of free space = 8.85 × 10-12N-1 m-2 C2


A = area of plates


∴ 


⇒ A = 50 × 10-12F × 10-3m/8.85 × 10-12N-1 m-2 C2 × 3


⇒ A = 1.88 × 10-3m2


∴ Minimum area of the plates required = 1.88 × 10-3m2



Question 24.

Describe schematically the equipotential surfaces corresponding to

(a) a constant electric field in the z-direction,

(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction,

(c) a single positive charge at the origin, and

(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.


Answer:

(a) The equipotential surfaces for a constant electric field in the z-direction are the equidistant planes parallel to the x-y plane.


(b) The equipotential surfaces for a field that uniformly increases in magnitude but remains in a constant in z direction are the planes parallel to the x-y plane and this field increases when the planes get closer.


For both cases A and B the Equipotential Surface is same as below



(c) The equipotential surfaces for a single positive charge at the origin are the concentric spheres centered at the origin.



D. The equipotential surface for a uniform grid consisting of long equally spaced parallel charged wires in a plane, near the grid is shaped as a periodically alternating curve. At a larger distance from the grid, the surface becomes parallel to the grid.



Question 25.

In a Van de Graff type generator a spherical metal shell is to be a 15 × 106 V electrode. The dielectric strength of the gas surrounding the electrode is 5 × 107 Vm-1. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)


Answer:

Potential Difference of the spherical metal shell (V) = 15 × 106 V

Dielectric strength of the gas surrounding the electrode = 5 × 107Vm-1


Dielectric strength = Electric field intensity (E)


The minimum radius of the spherical shell (r),



∴ r = 15 × 106 V/5 × 107 Vm-1 = 0.3m



Question 26.

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 on the smaller sphere.


∴ The potential difference (V) between the sphere and the shell depends on q1 and is independent of charge q2.


∵ q1 is positive so V is positive which implies charge will necessarily flow from the sphere to the shell.



Question 27.

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) We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, and our body being a good conductor suitably comes in equipotential with the earth surface.


(b) Yes, the man will get an electric shock if he comes in contact with the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminium sheet. As a result a substantial potential is built gradually. The raise in the voltage depends on the capacitance of the capacitor formed by the aluminium slab and the ground.


(c) The occurrence of thunderstorms and lightning charges the atmosphere continuously. So, during lightning even with the presence of a discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains neutralized.


(d) During lightning and thunderstorm, heat energy, light energy and sound energy are dissipated in the atmosphere.


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