# NCERT Physics 12 Electrostatic Potential and Capacitance Chapter 2 Exercise

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Physics class 12 electrostatic potential and capacitance chapter 2 solution is here in this website. Also you can download sample paper from NCERT physics class 12.

## NCERT Physics 12 Electrostatic Potential and Capacitance Chapter 2 Exercise

Q.1. Two charges 5  10-8  C and -3  10-8 C are located 10  cm apart. At what points on the line joining the two charges is the electric potential zero ? Take the potential at infinity to be zero. Q.2. A regular hexagon of side 10 cm has a charge 5 μ C at each of its vertices. Calculate the potential at the centra of the hexagon.
Sol. In 0 is centre of hexagon ABCDEFA of each side 10 cm. As is clear from the figure, OAB,
OBC, etc. are equilateral triangles. Therefore, OA = OB = OC= OD = OE = OF
= r = 10cm = 10-1m,
As potential is scalar, therefore, potential at O is Q.3. Two charges + μ C and -2μ C 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 ?
Sol. (a) The plane normal to AB and passing through its middle point has zero potential everywhere. (b)
The direction of electric field every point on this surface is along normal to the plane in the
direction AB.
Q.4. spherical conductor of radius 12 cm has a charge of 1.6 10-7 C distributed uniformly an its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the centra of the sphere ?
Sol. Here, r = 12 cm = 12× 10-7 C.
(a) Inside the sphere, E = 0
(b) Just outside the sphere. (say, on the surface of the sphere ) Q.5. A parallel plate capacitor with air between the plats is reduced by half and the space 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 ? Q.6. Three capacitors each of capacitance 9 pF are connected in series. What is the total capacitance of the connection ? Q.7. Three capacitors of capacitances 2 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 100 V supply.
Sol. Cp = 2+3+4 = 9pF
For each capacitor, V is same = 100 volt Q.8. . In a parallel plate capacitor with air between the plates, each plate has 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, whit is the charge on each plate of the capacitor ? Q.9. Explain whit would happen if in the capacitor in Ǫ. 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.
Sol. (a) while the voltage supply remained connected, voltage remains constant.
Capacity increases to C’ = KC0 = 6×1.77×10-11 F 6
Charge increases to q’ = C’ V = 6×1.77×10-11 ×102 C.
(b) After the supply was disconnected, charge remains constant. New capacity,
C’ KC0 = 6×1.77×10-11 F Q.10. A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored In the capacitor ?
Sol. Here, C = 12 pF = 12 ×10-12 F, V = 50 volt, U = ? Q.11. A 600 pF capacitor is charged by a 200 V supply. It is them disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process ? Q.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 form point P (0,0,3cm) to a point Q (0,4,cm, 0) via a point R (0,6 cm,9 cm).

Sol.. In  a charge q = 8 mC = 8 is located at the origin O. charge to be carried is

q0 = -2 from P to Q,

where OP = rp = 3cm = 3

and    OQ = rQ = 4 cm = 4 10-2 m

As electrostatic forces are conservative forces, work done is independent of the path. Therefore,
there is no relevance of point R.  Q.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. Further, electric field intensity at the centre due to all the eight charges is zero, because the fields
due to individual charges cancel in pairs.
Q.14. Two tiny speres carrying charges 1-5 μ C and 2-5 μ C 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 mid point in a plane normal to the line and passing through the mid point.  Cleary. E is towards q1.
(b) Let P be the point in a plane normal to the line passing through the mid point, where OP =
10cm =0.1m, along AP produced
Electrc intensity at P due to q2 ;  Q.15. A spherical conducting shell of inner radius r1 and outer radius r2 has a charge. Ǫ. (a) A charge q placed at the centra of the shell. What is the surface charge density in the inner and outer surfaces of the shell ?
(b) Is the electric filed intensity inside a cavity (with no charge ) zero, even if the shell is not spherical, but has any irregular shape ? Explain.

Sol. (a) The charge + Ǫ resides on the outer surface of the shell. The charge q placed at the centra of the
shell induces charge – q on the inner surface and charge+ q on the outer surface of the shell, therefore. Total charge on inner surface of the shell is – q and total charge on the outer
surface of the shell is (Q + q).  (b) Electric field intensity inside a cavity with no charge is zero. Even when the shell has any irregular
shape. If we were to take a closed loop. Part of which is inside the cavity along a field line, and the
rest outside it. Then net work done by the field in carrying a test charge over the closed loop will
not ne zero. This is impossible for an electrostatic field. Hence electric field intensity inside a cavity
with no charge is always zero.

Q.16 show that the normal component of electrostatic field has a discontinuity form one  side of a charged surface to another given by ( 22).Where    is a unit vector normal to the surface at a point and  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 to electric field is

(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.

.

Sol. Proceeding as in Art. 1(c).20, normal component of electric field intensity due to a thin infinite plane
sheet of charge, on left side (side 1) (b) To show that the tangential component of electrostatic field continuous from one side of a
charged surface to another, we use the fact that work done by electrostatic field on a closed closed
loop is zero.
Q.17. A long charged cylinder of linear charge density ʎ is surrounded by a hollow co- axial conducting cylinder. What is the electric field in the space between the two cylinders ? Sol. In A is a long charged cylinder of linear charge density λ , length l and radius b surrounds
A. The charge q = λ l spreads uniformly on the outer surface of A . It induces – q charge on the
cylinder B, which spreads on the inner surface of B. An electric field E ⃗ is produced in the space
between the two cylinders, which is directed radially outwards. Let us consider a co – axial
cylindrical Gaussian surface of radius r. The electric flux through the cylindrical Gaussian surface is Q.18. 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 potential energy at infinite separation of electron from from proton.
(b) What is the minimum work required to free the electron, given that its KE in the in the orbit is half the magnitude of potential energy obtained in (a) ?
(c) What are the answers to (a) and (b) above, if zero of potential energy is taken at 1.06Å separation? Work required to free the electron = 13.58 eV
(c) Potential energy at a separation of r1 (=1.06Å) is Potential energy of the system, when zero of P.E. is taken at r1 = 1.06Å is
= P.E. at r1 – P.E. at r = 13.58 – 27.16 = -13.58eV
By shifting the zero of potential energy, work required to free the electron is not affected. It
continues to be the same, being equal to + 13.58 eV

Q.19.here, q1 = charge on electron (= -1.6 × 10-19C)

q2, q3 = charge on two protons, each = 1.6 × 10-19

r12 = distance between q1 and q2 = 1Å = 10-10 m

r23 = distance between q2 and q3 = 1.5 Å = 1.5× 10-10

r31 = distance between q3 and q1 = 1 Å = 10-10 m.

Taking zero of potential energy at infinity, we have Q.20. Two charged conducting spheres of radii a and b are connected to eachother by a wire. What is the ratio of electric fields at the surfaces of 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 ?
Sol. The charge flows from the surfaces at higher to the other at lower potential, till their potentials
become equal. After sharing, the charges on two spheres would be A sharp and pointed end can be treated of very small radius and a flat portion leaves as a sphere of
much larger radius. Therefore, charge density on sharp and pointed ends of conductor is much
higher than on its flatter portions.

Q.21.Two charges – q are located at points (0,0, – a) and (0,0, a ) respectively.

(i) What is the electrostatic potential at the points (0,0, z) and (x,y,0) ?

(ii) Obtain the dependence of potential on the distance r of the point from the origin, when

(iii) 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 path of test charge between the same point is not along the x-axis ?

(iv) If the above point charges are now placed in the same positions in a uniform external electric field  what would ne the potential energy of the charge system in its orientation of unstable equilibrium ? Justify your answer in each case.

Sol. Hare, – q is at (0,0,-a) and + q is at (0,0,a) As work done = charge (V2-V1) ∴ w = zero
As work done by electrostatic field is independent of the path connecting the two points, therefore,
work done will continue to be zero along every path. Q.22. shows a charge array known as an ‘electric quadrupole’. For a point on the axis of the quadruple, 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 sigle charge.).
Sol. As is clear from an electric quadruple may be regarded as a system of three charges, + q,
– 2q and + q at A,B and C respectively.
Let AC = 2a, we have to calculate electric potential at any point p where BP = r using superposition
principle, potential at P is given by  Q.23. An electrical technician requires a capacitance of 2 μF in a circuit across a potential difference of 1 K V. A large number of 1μF capacitors are available to him. Each of which can withstand a potential difference of not more then 400V. Suggest a possible arrangement that requires minimum number of capacitors.   Q.24. What is the area f the plates 2 F Parallel plate capacitor, given that the separation between the plates is 0.5cm.
[You will realise from your answer why ordinary capacitors are in the range of μ F or less. However, electrolytic capacitors do have a much larger capacitance (0-1 F) because of very minute separation between the conductors]. Q.25. Obtain equivalent capacitance of the following net work, for a 300 V supply, determine the charge and voltage across each capacitor.
Sol. The equivalate circuits of network in are shown in
In as 100 pF and 200 pF capacitors are in series, pot. Diff. across C4 is in the ratio 2 : 1.   Q.26. The plates of a parallel plate have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V 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. Q.27. A 4 μ F capacitor is charged by a 200 V supply. It is them disconnected from the supply and is connected to another uncharged 2 μ F capacitor. How much electrostatic energy of the first capacitor is dissipated in the from of heat and electromagnetic radiation ?  Q.28. 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 factor 1/2 . The origin of factor 1/2 in force can be explained by the fact that inside the conductor, field is zero
and outside the conductor, the fiend is E. Therefore, the average value of the field ( i. e. E/2)
contributes to the force.
Q.29. A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports, show that the capacitance of this spherical capacitor is given by
C=(4∈_(0 r_1 r_2 ))/(r_1 r_2 ),
where r1 and r2 are the radii of outer and inner spheres respectively.
Sol. As is clear from, + Q charge spreads uniformly on inner surface of outer sphere of radius
r1 The induced charge – Q spreads uniformly on the outer surface of inner sphere of radius r2 . the
outer surface of outer sphere is earthed. Due to electrostatic shielding E = 0 for r < r2 and E = 0, for r> r1.
In the space between the two spheres., electric intensity E exists as shown. Potential difference
between the two spheres.  Q.30. A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. the other 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 the capacitor with that of radius 12 cm. Explain why the latter is much smaller ? The capacity of an isolated sphere is much smaller because in a capacitor, outer sphere is earthed,
potential difference decreases and capacitance increases.
Q.31. Answer carefully : (a) Two large conducting spheres carrying charges Q1 and Q2 are brought close to each other. Is the magnitude of electrostatic force between them exactly given by  where  r is distance between their centres ?
(b) If coulomb 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 line of force passing through that point ?
(d) What is the work done by the field of a nucleus in a complete circular orbit of 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 capacity of a single conductor ?
(g) Guess a possible reason why water has a much greater dielectric constant K = 80 than, say mica (K=6).
Sol. (a) When the charged spheres are brought close together, the charge distributions on them become
non- uniform. Therefore, coulomb’s law is not valid. Hence the magnitude of force is not given
exactly by this formula.
(b) No, Gauss’s law will not be true if coulomb’s law involved 1/r3 dependence instead of 1/r2
dependence .
(c) The line of force gives the direction of acceleration of charge. If the electric line of force is linear,
the test charge will move along the line. If the line of force is non linear, the charge will not go along
the line.
(d) force due to the field is directed towards the nucleus, and the electron does not move in the
direction of this force, therefore work done is zero when the orbit is circular. This is true even when
orbit is ellipitical as electric forces are conservative forces.
(e) No, electric potential is continuous.
(f) The capacity a single conductor implies that the second conductor is at infinity.
(g) This is because a molecule of water in its normal has an unsymmetrical shape and, therefore it
has a permanent dipole moment.
Q.32. A cylindrical capacitor has two co -axial cylinders of length 15 cm and radii 1.5 and 1.4 cm. The outer cylinder is earthed and inner cylinder is given a charge of 3.5 μ C. 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). Q.33. A parallel plate capacitor is to be designed with a voltage rating 1 k V using a material of dielectric constant 3 and dielectric strength about 107 V m-1. [Dielectric strength to conduct electricity through field a material can tolerate without break down, i.e., without starting to conduct electricity through partial ionization, 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 ? Q.34. Describe schematically, the equipotential surfaces corresponding to
(a) a constant electric field in the Z direction
(b) a fiend that uniformly increases in magnitude but remains in a constant, say z direction
(c) a single positive charge at the origin.
(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.
Sol. By definition an equipotential surface is that at every point of which potential is the same, In the
four cases given above :
(a) Equipotential surfaces are planes parallel to X – Y plane. These are equidistant.
(b) Equipotential surfaces are planes parallel to X – Y plane. As the field increases uniformly, distance
between the planes (differing by fixed potential) decreases.
(c)Equipotential surfaces are concentric spheres with origin at the Centre.
(d) Equipotential surfaces have the shape which changes periodically. At far off distances from the
grid, the shape of equipotential surfaces becomes parallel to the grid itself.
Q.35.In a Van de Graaf type generator, a spherical metal shell is to be a 15 166 volt 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.]
Sol. Here ,V = 15 × 106 volt., Dielectric strength = 5× 107 vm-1, Minimum radius, r = ?
Max. electric field, E = 10% dielectric strength Obviously. We cannot build an electrostatic generator, using a very small shell.
Q.36. A small of radius r1 and charge q2 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
Sol. As the charge resides always on the outer surface of the shell, therefore, when the sphere and shell
are connected by a wire, charge will flow essentially from the sphere to the sell. Whatever be the
magnitude and sign of charge q2. Q.37. Answer the following : (a) The top of the atmosphere is at about 400 KV with respect to the surface of earth, corresponding to an electric field that decreases with altitude, Near the surface of the earth, the field is about 100 Vm-1 . Why them do we not get an electric shock, as we step out of the house into the open ?(Assume the house to be a steel cage so that there is no field inside).
(b) A man fixes outside his house one evening a two Metra high insulating slab carrying on its top, a large aluminium sheet of area 1 m2. Will he get an electric shock If he ouches 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 is dissipated ?
(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning ?
Sol. (a) Since our body and the surface of earth, both are conducting, therefore, our body and the
ground from an equipotential surface. As we step out into the open from our house, the original
equipotential surfaces of open air change, keeping our body and the ground at the same potential.
That is why we do not get an electric shock.
(b) Yes, the man will get a shock. This is because the steady discharging current of the atmosphere
charges up the aluminium sheet gradually and raises its voltage to an extent depending on the
capacitance of the condenser formed by the aluminium sheet, the ground and the insulating slab.(c)
The atmosphere is being charged continuously by thunderstorms and lightning all over the globe. It
is also discharging due to the small conductivity of air. The two opposing processes, on an average,
are in equilibrium. Therefore, the atmosphere keeps charged.
(d) During a lightning, the electrical energy of the atmosphere is dissipated in the form of light, heat
and sound. ## NCERT Class 12 Physics other chapter Solutions

Updated: December 4, 2021 — 3:05 pm