Chapter 2 Electrostatic Potential And Capacitance Solutions
Question - 11 : - A 600 pF capacitor is charged by a 200 Vsupply. It is then disconnected from the supply and is connected to anotheruncharged 600 pF capacitor. How much electrostatic energy is lost in theprocess?
Answer - 11 : -
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 thecapacitor and another capacitor of capacitance C = 600 pF isconnected to it, then equivalent capacitance (C’) of thecombination is given by,
New electrostatic energy can be calculated as
Therefore, theelectrostatic energy lost in the process is
Question - 12 : - A charge of 8 mC is located at the origin.Calculate the work done in taking a small charge of −2 × 10−9 Cfrom 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 - 12 : -
Charge located at the origin, q =8 mC= 8 × 10−3 C
Magnitude of a small charge, which is takenfrom 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 electrostaticforce is independent of the path.
Therefore, work done during the process is1.27 J.
Question - 13 : - A cube of side b has acharge q at each of its vertices. Determine the potential andelectric field due to this charge array at the centre of the cube.
Answer - 13 : -
Length of the side of a cube = b
Charge at each of its vertices = q
A cube of side b is shownin 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 thecentre of the cube is due to the presence of eight charges at the vertices.
Therefore, the potential atthe centre of the cube is 
The electric field at the centre of thecube, due to the eight charges, gets cancelled. This is because the charges aredistributed symmetrically with respect to the centre of the cube. Hence, theelectric field is zero at the centre.
Question - 14 : - Two tiny spheres carrying charges 1.5 μC and 2.5 μC arelocated 30 cm apart. Find the potential and electric field:
(a) at the mid-point of the line joining the twocharges, and
(b) at a point 10 cm from this midpoint in a planenormal to the line and passing through the mid-point.
Answer - 14 : -
Two charges placed at points A and B are represented inthe 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 arethe electric potential and electric field respectively at O.
V1 = Potential due to charge at A + Potential due tocharge at B
Where,
∈0 = Permittivityof free space
E1 = Electric field due to q2 −Electric field due to q1
Therefore, the potential at mid-point is2.4 × 105 V and the electric field at mid-point is 4× 105 Vm−1. The field is directed from the larger charge to the smallercharge.
(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 electricpotential and electric field respectively at Z.
It can be observed from the figure that distance,
V2= Electric potential due to A + Electric Potential due toB
Electric field due to q atZ,
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 electricfield is 6.6 ×105 V m−1.
Question - 15 : - A spherical conducting shell of innerradius r1 and outer radius r2 has a charge Q.
(a) A charge q is placed at the centre ofthe shell. What is the surface charge density on the inner and outer surfacesof 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 - 15 : -
(a) Charge placed at the centre of a shell is +q.Hence, a charge of magnitude −q will be induced to the innersurface of the shell. Therefore, total charge on the inner surface of the shellis −q.
Surface charge density at the inner surface of the shellis given by the relation,
A charge of +q is induced onthe outer surface of the shell. A charge of magnitude Q isplaced on the outer surface of the shell. Therefore, total charge on the outersurface of the shell is Q + q. Surface chargedensity at the outer surface of the shell,
(b) Yes
The electric field intensity inside a cavityis zero, even if the shell is not spherical and has any irregular shape. Take aclosed loop such that a part of it is inside the cavity along a field linewhile the rest is inside the conductor. Net work done by the field in carryinga test charge over a closed loop is zero because the field inside the conductoris zero. Hence, electric field is zero, whatever is the shape.
Question - 16 : - (a) Show that the normal component of electrostaticfield has a discontinuity from one side of a charged surface to another givenby 
Where 
is a unit vector normal tothe surface at a point and σ is the surface charge density at that point. (Thedirection of is from side 1 to side 2.)Hence show that just outside a conductor, the electric field is σ 
(b) Show that the tangential component of electrostaticfield 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 electrostaticfield on a closed loop is zero.]
Answer - 16 : -
(a) Electric field on one side of a charged body is E1 andelectric field on the other side of the same body is E2.If infinite plane charged body has a uniform thickness, then electric field dueto 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 chargedbody,
Electric field at any point due to the two surfaces,
Since inside a closedconductor, 
= 0,
∴
Therefore, the electricfield just outside the conductor is 
(b) When a charged particle is moved from one point to theother on a closed loop, the work done by the electrostatic field is zero.Hence, the tangential component of electrostatic field is continuous from oneside of a charged surface to the other.
A long charged cylinder of linear chargeddensity λ is surrounded by a hollow co-axial conducting cylinder. What is theelectric field in the space between the two cylinders?
Answer - 17 : -
Charge density of the long charged cylinderof length L and radius r is λ.
Another cylinder of same length surroundsthe pervious cylinder. The radius of this cylinder is R.
Let E be the electricfield produced in the space between the two cylinders.
Electric flux through the Gaussian surface is given byGauss’s theorem as,
Where, d = Distance of apoint from the common axis of the cylinders
Let q be the total chargeon the cylinder.
It can be written as
Where,
q =Charge on the inner sphere of the outer cylinder
∈0 = Permittivityof free space
Therefore, the electricfield in the space between the two cylinders is 
Question - 18 : - In a hydrogen atom, the electron and proton are bound ata 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 electronfrom proton.
(b) What is the minimum work required to free theelectron, given that its kinetic energy in the orbit is half the magnitude ofpotential energy obtained in (a)?
(c) What are the answers to (a) and (b) above if thezero of potential energy is taken at 1.06 Å separation?
Answer - 18 : - The distance betweenelectron-proton of a hydrogen atom, 
Charge on an electron, q1 =−1.6 ×10−19 C
Charge on a proton, q2 =+1.6 ×10−19 C
(a) Potential at infinity is zero.
Potential energy of the system, =Potential energy at infinity − Potential energy at distance dwhere,
∈0 is thepermittivity of free space
14πε0=9×109 Nm2C-2∴ Potential energy=0-9×109×1.6×10-1920.53×10-10=-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 −27.2eV.
(b) Kinetic energy is half of the magnitude ofpotential energy.
Total energy = 13.6 − 27.2 = 13.6 eV
Therefore, the minimum work required to free the electronis 13.6 eV.
(c) When zero of potential energy is taken, 
∴Potential energy of the system = Potential energyat d1 − Potential energy at d
. In the ground state of an, the two protons areseparated 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 zeroof potential energy.
Answer - 19 : -
The system of two protons and one electron is representedin 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 andelectron, d2 = 1 ×10−10 m
Distance between proton 2 andelectron, d3 = 1 × 10−10 m
The potential energy at infinity is zero.
Potential energy of the system,
Therefore, the potential energy of thesystem is −19.2 eV.
Question - 20 : - Two charged conducting spheres ofradii a and b are connected to each other bya wire. What is the ratio of electric fields at the surfaces of the twospheres? Use the result obtained to explain why charge density on the sharp andpointed ends of a conductor is higher than on its flatter portions.
Answer - 20 : -
Let a be the radius of asphere A, QA be the charge on the sphere, and CA bethe capacitance of the sphere. Let b be the radius of a sphereB, QB be the charge on the sphere, and CB bethe capacitance of the sphere. Since the two spheres are connected with a wire,their potential (V) will become equal.
Let EAbe the electric field ofsphere 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 ofelectric fields at the surface is
A sharp and pointed end can be treated as asphere of very small radius and a flat portion behaves as a sphere of muchlarger radius.Therefore, charge density on sharp and pointed ends of theconductor is much higher than on its flatter portions.