Chapter 2 Electrostatic Potential And Capacitance Solutions
Question - 21 : - Two charges −q and +q arelocated 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 thedistance r of a point from the origin when r/a >>1.
(c) How much work is done in moving a small testcharge 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 pointsis not along the x-axis?
Answer - 21 : -
(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 onthe axis of this dipole and point (x, y, 0) is normal to theaxis 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 thanhalf of the distance between the two charges. Hence, the potential (V)at a distance r is inversely proportional to square of thedistance i.e.
(c) Zero
The answer does not change if the path ofthe 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
Electrostatic potential, V2,at point (− 7, 0, 0) is given by,
Hence, no work is done in moving a smalltest charge from point (5, 0, 0) to point (−7, 0, 0) along the x-axis.
The answer does not change because workdone by the electrostatic field in moving a test charge between the two pointsis independent of the path connecting the two points.
Figure 2.34 shows a charge array known asan 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 anelectric monopole (i.e., a single charge).
Answer - 22 : -
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 distanceaway from point Y.
The system of charges forms an electric quadrupole.
It can be considered that the system of the electricquadrupole 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 threecharges at point P is given by,
Since
is taken as negligible.
It can be inferred thatpotential, 
However, it is known thatfor a dipole,
And, for a monopole, 
Question - 23 : - An electrical technician requires acapacitance of 2 µF in a circuit across a potential difference of 1 kV. A largenumber of 1 µF capacitors are available to him each of which can withstand apotential difference of not more than 400 V. Suggest a possible arrangementthat requires the minimum number of capacitors.
Answer - 23 : -
Total required capacitance, C =2 µF
Potential difference, V =1 kV = 1000 V
Capacitance of each capacitor, C1 =1µF
Each capacitor can withstand a potentialdifference, V1 = 400 V
Suppose a number of capacitors are connected in seriesand these series circuits are connected in parallel (row) to each other. Thepotential difference across each row must be 1000 V and potential differenceacross each capacitor must be 400 V. Hence, number of capacitors in each row isgiven as
Hence, there are three capacitors in each row.
Capacitance of each row
Let there are n rows, eachhaving three capacitors, which are connected in parallel. Hence, equivalentcapacitance of the circuit is given as
Hence, 6 rows of three capacitors arepresent in the circuit. A minimum of 6 × 3 i.e., 18 capacitors are required forthe given arrangement.
Question - 24 : - What is the area of the plates of a 2 Fparallel plate capacitor, given that the separation between the plates is 0.5cm? [You will realize from your answer why ordinary capacitors are in the rangeof µF or less. However, electrolytic capacitors do have a much largercapacitance (0.1 F) because of very minute separation between the conductors.]
Answer - 24 : -
Capacitance of a parallel capacitor, V =2 F
Distance between the two plates, d =0.5 cm = 0.5 × 10−2 m
Capacitance of a parallel plate capacitor is given by therelation,
Where
= Permittivity of free space = 8.85 ×10−12 C2 N−1 m−2
Hence, the area of the plates is too large.To avoid this situation, the capacitance is taken in the range of µF.
Question - 25 : - Obtain the equivalent capacitance of the network in Fig.2.35. For a 300 V supply, determine the charge and voltage across eachcapacitor.
Answer - 25 : -
Capacitance of capacitor C1 is100 pF.
Capacitance of capacitor C2 is200 pF.
Capacitance of capacitor C3 is200 pF.
Capacitance of capacitor C4 is100 pF.
Supply potential, V = 300V
Capacitors C2 and C3 areconnected in series. Let their equivalent capacitance be 
Capacitors C1 and C’ arein parallel. Let their equivalent capacitance be 
are connected in series. Let theirequivalent capacitance be C.Hence, the equivalentcapacitance of the circuit is 
Potential differenceacross 
= 
Potential difference across C4 = V4
Charge on 
Q4= CV
Hence, potential difference, V1,across C1 is 100 V.
Charge on C1 isgiven by,
C2 and C3 having samecapacitances have a potential difference of 100 V together. Since C2 and C3 arein series, the potential difference across C2 and C3 isgiven by,
V2 = V3 = 50 V
Therefore, charge on C2 isgiven by,
And charge on C3 isgiven by,
Hence, the equivalentcapacitance of the given circuit is
Question - 26 : - The plates of a parallel plate capacitorhave an area of 90 cm2 each and are separated by 2.5 mm. Thecapacitor is charged by connecting it to a 400 V supply.
(a) How much electrostatic energy is stored by thecapacitor?
(b) View this energy as stored in the electrostaticfield between the plates, and obtain the energy per unit volume u.Hence arrive at a relation between u and the magnitude ofelectric field E between the plates.
Answer - 26 : -
Area of the plates of a parallel platecapacitor, A = 90 cm2 = 90 × 10−4 m2
Distance between the plates, d =2.5 mm = 2.5 × 10−3 m
Potential difference across theplates, V = 400 V
(a) Capacitance of the capacitor is given by the relation,Electrostatic energy storedin the capacitor is given by the relation, 
Where,
= Permittivity of free space = 8.85 ×10−12 C2 N−1 m−2
Hence, the electrostaticenergy stored by the capacitor is 
(b) Volume of the given capacitor,
Energy stored in the capacitor per unit volume is givenby,
Where,
= Electric intensity = E
A 4 µF capacitor is charged by a 200 Vsupply. It is then disconnected from the supply, and is connected to anotheruncharged 2 µF capacitor. How much electrostatic energy of the first capacitoris lost in the form of heat and electromagnetic radiation?
Answer - 27 : - Capacitance of a chargedcapacitor, 
Supply voltage, V1 =200 V
Electrostatic energy stored in C1 isgiven by,
Capacitance of an unchargedcapacitor, 
When C2 isconnected to the circuit, the potential acquired by it is V2.
According to the conservation of charge,initial charge on capacitor C1 is equal to thefinal charge on capacitors, C1 and C2.
Electrostatic energy for the combination of twocapacitors is given by,
Hence, amount of electrostatic energy lostby capacitor C1
= E1 − E2
= 0.08 − 0.0533 = 0.0267
= 2.67 × 10−2 J
Show that the force on each plate of aparallel plate capacitor has a magnitude equal to (½) QE,where Q is the charge on the capacitor, and E isthe magnitude of electric field between the plates. Explain the origin of thefactor ½.
Answer - 28 : -
Let F be the force appliedto separate the plates of a parallel plate capacitor by a distance of Δx.Hence, work done by the force to do so = FΔx
As a result, the potential energy of thecapacitor increases by an amount given as uAΔx.
Where,
u =Energy density
A =Area of each plate
d =Distance between the plates
V =Potential difference across the plates
The work done will be equal to the increase in thepotential energy i.e.,
Electric intensity is given by,
However, capacitance, 
Charge on the capacitor is given by,
Q = CV
The physical origin of thefactor, 
, in the force formula liesin the fact that just outside the conductor, field is E andinside it is zero. Hence, it is the average value, 
, of the field thatcontributes to the force.
Question - 29 : - A spherical capacitor has an inner sphere of radius 12 cmand an outer sphere of radius 13 cm. The outer sphere is earthed and the innersphere is given a charge of 2.5 µC. The space between the concentric spheres isfilled 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 thatof an isolated sphere of radius 12 cm. Explain why the latter is much smaller.
Answer - 29 : - Radius of the innersphere, 
= 12 cm = 0.12 m
Radius of the outer sphere, 
= 13 cm = 0.13 m
Charge on the inner sphere,
Dielectric constant of aliquid, 
(a) 
Where,
= Permittivity of freespace = 
Hence, the capacitance ofthe capacitor is approximately 
(b) Potential of the inner sphere is given by,
Hence, the potential of theinner sphere is 
(c) Radius of an isolated sphere, r = 12 ×10−2 m
Capacitance of the sphere is given by the relation,
The capacitance of the isolated sphere isless in comparison to the concentric spheres. This is because the outer sphereof the concentric spheres is earthed. Hence, the potential difference is lessand the capacitance is more than the isolated sphere.
A cylindrical capacitor has two co-axialcylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder isearthed and the inner cylinder is given a charge of 3.5 µC. Determine thecapacitance of the system and the potential of the inner cylinder. Neglect endeffects (i.e., bending of field lines at the ends).
Answer - 30 : -
Length of a co-axial cylinder, l =15 cm = 0.15 m
Radius of outer cylinder, r1 =1.5 cm = 0.015 m
Radius of inner cylinder, r2 =1.4 cm = 0.014 m
Charge on the inner cylinder, q =3.5 µC = 3.5 × 10−6 C
Where,
= Permittivity offree space = 
Potential difference of the inner cylinder is given by,
