Chapter 2 Electrostatic Potential And Capacitance Solutions
Question - 1 : - Two charges 5 × 10−8 C and−3 × 10−8 C are located 16 cm apart. At what point(s) on theline joining the two charges is the electric potential zero? Take the potentialat infinity to be zero.
Answer - 1 : -
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) atpoint P be zero.
Potential at point P is the sum ofpotentials 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 cmfrom the positive charge between the charges.
Suppose point P is outside the system oftwo charges at a distance s from the negative charge, wherepotential 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 adistance of 40 cm from the positive charge outside the system of charges.
Question - 2 : - A regular hexagon of side 10 cm has acharge 5 µC at each of its vertices. Calculate the potential at the centre ofthe hexagon.
Answer - 2 : -
The given figure shows six equal amount ofcharges, 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 centreO, d = 10 cm
Electric potential at point O,
Where,
= Permittivity of free space
Therefore, the potential at the centre ofthe hexagon is 2.7 × 106 V.
Question - 3 : - Two charges 2 μC and −2 µC are placed at points A and B 6cm apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field atevery point on this surface?
Answer - 3 : -
(a) The situation is represented in the given figure.
An equipotential surface is the plane on which totalpotential is zero everywhere. This plane is normal to line AB. The plane islocated at the mid-point of line AB because the magnitude of charges is thesame.
(b) The direction of the electric field at every point onthis surface is normal to the plane in the direction of AB.
Question - 4 : - A spherical conductor of radius 12 cm has acharge of 1.6 × 10−7C distributed uniformly on its surface. What isthe electric field
(a) Inside the sphere
(b) Just outside the sphere
(c) At a point 18 cm from the centre of the sphere?
Answer - 4 : -
(a) Radius of the spherical conductor, r =12 cm = 0.12 m
Charge is uniformly distributed over theconductor, q = 1.6 × 10−7 C
Electric field inside a spherical conductor is zero. Thisis because if there is field inside the conductor, then charges will move toneutralize it.
(b) Electric field E just outside theconductor is given by the relation,
Where,
= Permittivity of freespace
Therefore, the electricfield just outside the sphere is 
(c) Electric field at a point 18 m from the centre of thesphere = E1
Distance of the point from thecentre, d = 18 cm = 0.18 m
Therefore, the electric field at a point 18 cm from thecentre of the sphere is
Question - 5 : - A parallel plate capacitor with air betweenthe plates has a capacitance of 8 pF (1pF = 10−12 F). What willbe the capacitance if the distance between the plates is reduced by half, andthe space between them is filled with a substance of dielectric constant 6?
Answer - 5 : -
Capacitance between the parallel plates of the capacitor,C = 8 pF
Initially, distance between the parallelplates was d and it was filled with air. Dielectric constantof air, k = 1
Capacitance, C, is given by theformula,
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 thesubstance 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 theplates is 96 pF.
Three capacitors each of capacitance 9 pF are connectedin series.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across eachcapacitor if the combination is connected to a 120 V supply?
Answer - 6 : -
(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 pFTherefore,total capacitance of the combination is
3 pF.
(b) Supply voltage, V = 120 V
Potential difference (V‘) across each capacitoris equal to one-third of the supply voltage.
Therefore, the potential difference acrosseach capacitor is 40 V.
Three capacitors of capacitances 2 pF, 3 pF and 4 pF areconnected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if thecombination is connected to a 100 V supply.
Answer - 7 : - (a) Capacitances of the given capacitors are
For the parallelcombination 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 threecapacitors is same = V = 100 V
Charge on a capacitor of capacitance C andpotential difference V is given by the relation,
q = VC …(i)
For C = 2 pF,For C = 3 pF,
For C = 4 pF,
Question - 8 : - In a parallel plate capacitor with airbetween the plates, each plate has an area of 6 × 10−3 m2 andthe distance between the plates is 3 mm. Calculate the capacitance of thecapacitor. If this capacitor is connected to a 100 V supply, what is the chargeon each plate of the capacitor?
Answer - 8 : -
Area of each plate of the parallel platecapacitor, 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 platecapacitor is given by,Where,
= Permittivity of free space
= 8.854 × 10−12 N−1 m−2 C−2
Therefore, capacitance of the capacitor is17.71 pF and charge on each plate is 1.771 × 10−9 C.
Explain what would happen if in the capacitor given inExercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) wereinserted between the plates,
(a) While the voltage supply remained connected.
(b) After the supply was disconnected.
Answer - 9 : -
(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 noeffect on the amount of charge in the plates.
Charge = 1.771 × 10−9 C
Potential across the plates is given by,
Question - 10 : - A 12 pF capacitor is connected to a 50Vbattery. How much electrostatic energy is stored in the capacitor?
Answer - 10 : -
Capacitor of the capacitance, C =12 pF = 12 × 10−12 F
Potential difference, V =50 V
Electrostatic energy stored in the capacitor is given bythe relation,
Therefore, theelectrostatic energy stored in the capacitor is 