Chapter 2 : Electrostatic Potential And Capacitance

Arivihan
 Electrostatic Potential And Capacitance

Arivihan
IMPORTANT TOPICS 3-39

PREVIOUS YEAR QUESTIONS 41-74

Numerical Question 75-98

 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

VERY SHORT SHORT LONG

CHAPTER OBJECTIVE
ANSWER TYPE ANSWER TYPE ANSWER TYPE
WEIGHTAGE (1 MARKS)

Arivihan
(2 MARKS) (3 MARKS) (4 MARKS)

3 1 - 1
9 2 - 1 1
- 1 1 1
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

Chapter at a Glance

Arivihan
1. Electrostatic potential and potential difference

2. Electric potential due to point charge.

3. Electric potential due to dipole in axial and coaxial position

4. Equipotential surface and its properties

5. Relationship between electric field and electric potential

 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

Chapter at a Glance

Arivihan
1. Electrostatic potential energy

2. Conductor – electrostatics

3. Dielectrics and their polarization

4. Dielectric constant , polarization vector and Electrical Susceptibility

5. Dielectric Strength
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

Chapter at a Glance

Arivihan
1. Electrical capacitance

2. Capacitance of spherical conductor and parallel plate capacitor

3. Capacitance of parallel plate capacitor with partially filled dielectric

medium

4. Net capacitance in series and parallel combination of capacitor

5. Energy stored in a capacitor & Energy loss in redistribution of charge

 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

Multiple Choice Questions

Arivihan
Q. The angle between electric lines of force and equipotential surface is :

A. 0°
B. 90°
C. 180°
D. None of these

Correct answer is option B

 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

Multiple Choice Questions

Arivihan
Q. The angle between electric lines of force and equipotential surface is :

A. 0°
B. 90°
C. 180°
D. None of these

Correct answer is option B

 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

Multiple Choice Questions

Arivihan
Q. The electric field inside a charged spherical shell is :

A. Zero
B. Equal to KQ/r2
C. Equal to KQ/r
D. Directly proportional to the distance

Correct answer is option A

 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

Multiple Choice Questions

Arivihan
Q. A dielectric when placed in external electric field induces an electric field
inside it which ____________ the external electric field.

A. Increases
B. Decreases
C. Has no effect
D. None of these

Correct answer is option B

 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

Multiple Choice Questions

Arivihan
Q. A long charged cylinder of linear charge density λ is surrounded by a
hollow coaxial conducting cylinder. The electric field in the space between
two cylinder is :

A. E=0
B. E = λ/(2πε₀r)
C. E = λr/(2πε₀)
D. E = λ/(4πε₀r²)

Correct answer is option B

 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

Fill in the blanks

Arivihan
1. One volt = 1/300 stat-volt.

2. 1 eV =1.6 X 10 ⁻¹⁹ Joule.

3. The dielectric strength of air is 3.0 X 106 Volt/meter.

4. There is always loss of energy in redistribution of charge.

 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

TRUE or FALSE

Arivihan
1. The potential due to point charge at infinite distance from it is
infinite.FALSE
2. Electric field is zero inside a conductor. TRUE

3. Electric potential is a scalar quantity. TRUE

4. The potential at a point on equatorial line due to dipole is always zero.

TRUE
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

One word / Sentence

Arivihan
1. Write the dimensional formula of electric potential. Ans : [M¹ L² T⁻³ A⁻¹].
2. Write an expression which gives relationship between electric field and
electric potential difference. Ans : E = -dV/dr
3. Write an expression for the potential energy of a system of two charge.Ans
: U= [1/(4πε₀)].[Q₁Q₂/r ]
4. What would be the radius of a spherical conductor whose capacitance is 1
Farad. Ans : 9 x 10 ⁶ km.
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (2 marks)

Arivihan
Q.1. What is an equipotential surface? Prove that no work is done in moving
a unit charge from one point to another point of equipotential surface.

Ans: Equipotential surface: An equipotential surface is the locus of all those points
at which the potential due to distribution of charge remains same.
Proof: Let a test charge q₀ is moved from point A to B on the equipotential surface.
In this process work done is W then the potential difference between them is
VB - VA= W/q₀
But for an equipotential surface :
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (2 marks)

Arivihan
Q.1. What is an equipotential surface? Prove that no work is done in moving
a unit charge from one point to another point of equipotential surface.

Ans: V A = VB
or V B - VA = 0
Work done to take unit charge from A to B
WAB = VB - VA = 0.
∴W=0
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (2 marks)

Arivihan
Q.2. What is electrostatic shielding write one example of it.

Ans : The phenomenon of making a region free from an electric field is called
electrostatic shielding, it is based on the fact that electric field vanishes inside the
cavity of a hollow conductor.

Example: In a thunder storm accompanied by lightning it is safe to sit inside the

car, rather than near a tree or the open ground. The metallic body of the car pro-
vides an electrostatic shielding from lightning.
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (2 marks)

Arivihan
Q.3. What are dielectrics ? What is polarization of dielectrics?
Ans : Dielectric :- The substances which does not allow current to flow through
them but shows electrical effect are called dielectrics.

When a dielectric is placed in an external electric field, then a dipole moment is

induced in it, in the direction of external field whose Value is proportional to the
intensity of external field. This phenomenon is called polarization of dielectrics.
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (3 marks)

Arivihan
Q.1. Derive relationship between electric susceptibility and dielectric
constant.
Ans: Electric susceptibility: If the net electric field E⃗ is not large, then the
polarization is proportional to the resultant field E⃗ existing in the dielectric
P⃗ ∝ E⃗
⇒ P⃗ = ε₀χE⃗
Where χ (chi) is a proportionality constant called electric susceptibility
∴ χ = P⃗/ε₀E⃗
The ratio of the polarization to ε₀ times the electric field is called electric
susceptibility of dielectric.
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (3 marks)

Arivihan
Q.1. Derive relationship between electric susceptibility and dielectric
constant.
Ans: Relation between K and χ
The net electric field in polarized dielectric is
E⃗ = E⃗₀ - E⃗ₚ
But E⃗ₚ = σₚ/ε₀ = P⃗/ε₀
∴ E⃗ = E⃗₀ - P⃗/ε₀
⇒ E = E₀ - ε₀χE/ε₀
E⃗ = E⃗₀ - χE⃗
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (3 marks)

Arivihan
Q.1. Derive relationship between electric susceptibility and dielectric
constant.

Ans: ⇒ E⃗ + χE⃗ = E⃗₀

⇒ E⃗(1 + χ) = E⃗₀
1 + χ = E₀/E
But E₀/E = K
∴ 1 + χ = K.
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (3 marks)

Arivihan
Q.2. Prove that the potential at a point on equatorial line due to dipole is
zero.
Ans: Let AB be an electric dipole consists of +q and -q
charges separated by a distance 2l. A point P lies on the
neutral axis at a distance r from its centre O. If a point P is
taken on the neutral axis then it will be equidistant from
both the charges of dipole as shown in Fig.
Now, AP = BP = √(r² + l²)
∴ Potential at P due to +q charge will be
V₁ = 1/(4πε₀) q/AP
or V₁ = 1/(4πε₀) q/√(r² + l²)
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (3 marks)

Arivihan
Q.2. Prove that the potential at a point on equatorial line due to dipole is
zero.
Ans: Similarly potential at P due to -q charge will be

V₂ = 1/(4πε₀) [-q/BP]

or V₂ = 1/(4πε₀) (-q)/√(r² + l²)

∴ Total potential due to dipole at point P

V = V₁ + V₂
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (3 marks)

Arivihan
Q.2. Prove that the potential at a point on equatorial line due to dipole is
zero.
Ans: = 1/(4πε₀) q/√(r² + l²) + 1/(4πε₀) (-q)/√(r² + l²)
= 1/(4πε₀) q/√(r² + l²) - 1/(4πε₀) q/√(r² + l²)
or V = 0.

Therefore the potential due to a dipole at a point in the equatorial position is

zero.
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (3 marks)

Arivihan
Q.3. Derive relation between intensity of electric field and potential
difference.
Ans: Suppose A and B are two points in the electric field of charge q. The direction
of electric field is radially outwards from A to B. Suppose the distance between A
and B is very small (i.e., dr) then the electric field between A and B can be taken as
uniform. As the potential is inversely proportional to distance hence potential at A
is more than that of B. Let the potential at B is V then that at A is V + dV.
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (3 marks)

Arivihan
Q.3. Derive relation between intensity of electric field and potential
difference.
Ans: Work done in bringing the test charge q₀ from B to A is
dW = q₀dV ...(1)
Force acting on q₀ will be
F = q₀E
Work done in bringing the test charge against the repulsion force will be
dW = -q₀Edr, (∵ Work = Force × Displacement) ...(2)
The negative sign shows that the direction of displacement and direction of force
are opposite to each other.
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (3 marks)

Arivihan
Q.3. Derive relation between intensity of electric field and potential
difference.
Ans: From eqns. (1) and (2), we get
q₀dV = -q₀Edr
or dV = -Edr
or E = -dV/dr
This is required relation between intensity of electric field and potential difference.
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (4 marks)

Arivihan
Q.1. Derive an expression for the electric potential at a point due to an
electric dipole in axial position of an electric dipole.
Ans: Let us consider about an electric dipole consists of two charges +q and -q which
are located at positions A and B respectively and separated by a small distance 2l in
air or Vacuum as shown in Fig. Let us take a point P on the axis of dipole at which the
potential due to dipole is to be obtained.
The potential at P due to +q charge will be
V₁ = [1/(4πε₀)] ·[ q/BP]
Or = [1/(4πε₀)] · [q/(r-l)]
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (4 marks)

Arivihan
Q.1. Derive an expression for the electric potential at a point due to an
electric dipole in axial position of an electric dipole.

Ans: And potential at P due to -q charge will be

V₂ = [1/(4πε₀)] · [(-q)/AP]
Or = [-1/(4πε₀)] · [q/(r+l)]
By the principle of superposition, total potential at point P will be
V = V₁ + V₂
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (4 marks)

Arivihan
Q.1. Derive an expression for the electric potential at a point due to an
electric dipole in axial position of an electric dipole.

Ans: = [1/(4πε₀)] [q/(r-l)] + [(1/(4πε₀)] [q/(r+l)]

= [1/(4πε₀)] q[1/(r-l) - 1/(r+l)]
= [1/(4πε₀)] q[(r+l-r+l)/(r²-l²)]
or V = [1/(4πε₀)] [(2ql)/(r²-l²)]
But 2ql = p (dipole moment)
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (4 marks)

Arivihan
Q.1. Derive an expression for the electric potential at a point due to an
electric dipole in axial position of an electric dipole.

Ans: ∴ V = [1/(4πε₀)] [p/(r²-l²)] volt

If the dipole is small then l<<r

∴ V = [1/(4πε₀)] [p/r²] volt

This is required expression.

 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (4 marks)

Arivihan
Q.2. Derive an expression for the capacitance of parallel plate capacitor
when dielectric medium is partially filled between plates of capacitor. [IMP]

Ans: Let A and B are parallel plates of a capacitor. The distance between the plates is
d and plate of thickness t and dielectric constant εᵣ
is introduced.
Now, plate A is given charge +Q.
Let the charge density be σ.
∴ σ = Q/A
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (4 marks)

Arivihan
Q.2. Derive an expression for the capacitance of parallel plate capacitor
when dielectric medium is partially filled between plates of capacitor. [IMP]

Ans: Intensity of field in air,

E₀ = σ/ε₀ = Q/ε₀A ...(1)
If the intensity of field inside the dielectric medium be E, then
Dielectric constant = Electric field in Vacuum/Electric field in medium
or εᵣ = E₀/E
or E = E₀/εᵣ = Q/ε₀kA
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (4 marks)

Arivihan
Q.2. Derive an expression for the capacitance of parallel plate capacitor
when dielectric medium is partially filled between plates of capacitor. [IMP]

Ans: Now, potential difference between A and B,

V = E₀(d-t) + Et, [(d-t) is Vacuum distance]
= [ Q/ε₀A (d-t)] + [Q/ε₀kA t]
= Q/ε₀A[d-t + t/k]
But, we have C = Q/V
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (4 marks)

Arivihan
Q.2. Derive an expression for the capacitance of parallel plate capacitor
when dielectric medium is partially filled between plates of capacitor. [IMP]

Ans: or C = Q/(Q/ε₀A[d-t + t/k])

∴ C = ε₀A/(d-t + t/k) = ε₀A/[d-t{1-(1/k)}]
This is the required expression.
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (4 marks)

Arivihan
Q.3. Prove that there is always loss of energy in redistribution of charge .
Ans: Let A and B are two conductors of capacitors C₁ and C₂ respectively. When
charges Q₁ and Q₂ are given separately, the potential of A and B becomes V₁ and V₂
respectively.
∴ Total charge Q = Q₁ + Q₂ …....(1)
But Q₁ = C₁V₁ and Q₂ = C₂V₂
∴ From eqn. (1) Q = C₁V₁ + C₂V₂ and total capacity C = C₁ + C₂
(i) Common potential: Let the conductors are connected by a wire then the charge
flows from higher potential conductor to lower potential. After sometime the system
acquires a common potential V.
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (4 marks)

Arivihan
Q.3. Prove that there is always loss of energy in redistribution of charge .
Ans: or V = Q/C
or V = (Q₁ + Q₂)/(C₁ + C₂) = (C₁V₁ + C₂V₂)/(C₁ + C₂) ...(2)
This is expression for the common potential.
(ii) Loss of energy: Let the energy of conductor A and B before joining them are
½C₁V₁² and ½C₂V₂² respectively.
∴ Total energy of system before joining them
U₁ = ½C₁V₁² + ½C₂V₂²
and total energy after connection
U₂ = ½(C₁ + C₂)V² ...(3)
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (4 marks)

Arivihan
Q.3. Prove that there is always loss of energy in redistribution of charge .
Ans: ∴ Loss of energy ΔU = U₁ - U₂
= ½C₁V₁² + ½C₂V₂² - ½(C₁ + C₂)V²
or ΔU = ½C₁V₁² + ½C₂V₂² - ½(C₁ + C₂)[(C₁V₁ + C₂V₂)²/(C₁ + C₂)²]
= ½[(C₁V₁² + C₂V₂²)(C₁ + C₂) - (C₁V₁ + C₂V₂)²]/(C₁ + C₂)
= ½[C₁²V₁² + C₁C₂V₁² + C₁C₂V₂² + C₂²V₂² - C₁²V₁² - C₂²V₂² - 2C₁C₂V₁V₂]/(C₁ + C₂)
= ½[C₁C₂/(C₁ + C₂)]( V₁² + V₂² - 2V₁V₂)
ΔU = ½[C₁C₂/(C₁ + C₂)](V₁ - V₂)² ...(4)
 PHYSICS | MP BOARD EXAM | IMPORTANT QUESTIONS

IMP QUESTIONS (4 marks)

Arivihan
Q.3. Prove that there is always loss of energy in redistribution of charge .
Ans:
∴ (V₁ - V₂)² is positive, hence (U₁ - U₂) is positive. Hence, during redistribution, there
will be always loss of energy. Thus, the loss of energy shown in eqn. (4) when two
conductors are joined by a wire then the loss of energy gets transferred into heat and
light.
Arivihan
 Physics | M.P Board Exam | Previous Year Questions

Fill in the blanks

Arivihan
Q1. Potential___________ on moving along the direction of electric field.
[2024]
Ans : decreases
 Physics | M.P Board Exam | Previous Year Questions

One word / One sentence

Arivihan
Q1. What is the effect of insulated medium on the potential?
Ans : Potential become decrease. [2022]
 Physics | M.P Board Exam | Previous Year Questions

One word / One sentence

Arivihan
Q2. In an electric field, an electron will move towards high potential
or low potential. [2023]
Ans : Towards high potential, because an electron experiences
force in the opposite direction of electric field.
 Physics | M.P Board Exam | Previous Year Questions

Very short type questions

Arivihan
Q1. Write two characteristics of equipotential surface. [2024]
Ans : (1) Electric field lines always cut the equipotential surface
normally.
(2) Two equipotential surfaces never intersect each other,
otherwise at the point of intersection of the two surfaces, there
will be two values of potential which is not possible.
 Physics | M.P Board Exam | Previous Year Questions

Very short type questions

Arivihan
Q2. Write two characteristic of electric field lines. [2024]
Ans : (1) These are open curves.
(2) They are always normal to the charged surface.
(3) They are not present inside the conductor.
 Physics | M.P Board Exam | Previous Year Questions

Short type questions

Arivihan
Q1. What is capacity of a conductor? Write the factors affecting the
capacity of a conductor [2023]
Ans : Capacity- The capacity of a conductor is the charge given to
the conductor which raises its potential by unity.

Capacity C = Charge Q/ Potential V

Unit of capacity is farad (F).

 Physics | M.P Board Exam | Previous Year Questions

Short type questions

Arivihan
Q2. Describe the electric field and electric potential inside the
charged conductor. [MP 2022]
Ans : The intensity of electric field inside a hollow charged conductor
remains always zero, because the charge given to the conductor is
distributed only on the outer surface of conductor. But the potential
inside the hollow charged conductor remains same at every point
inside it.
 Physics | M.P Board Exam | Previous Year Questions

Short type questions

Arivihan
Q2. Describe the electric field and electric potential inside the
charged conductor. [MP 2022]

Ans : It is so because since the intensity of electric field inside it

remains zero, so no work is done to displace the unit positive
charge from one point to other point. Hence the potential inside the
conductor remains same.
 Physics | M.P Board Exam | Previous Year Questions

Short type questions

Arivihan
Q3. What is capacity of a conductor? Write the factors affecting the
capacity of a conductor [2023]
Ans : Factors affecting the capacity of a conductor-
The capacity of a conductor is affected by the following factors:
(i) Size of the conductor-By increasing the size of the conductor its
potential decreases, therefore its capacity increases.
 Physics | M.P Board Exam | Previous Year Questions

Short type questions

Arivihan
Q3. What is capacity of a conductor? Write the factors affecting the
capacity of a conductor [2023]
Ans : (ii) The presence of other conductors near the charged
conductor- Due to the presence of other conductors near the
charged conductor, its potential decreases and hence its capacity
increases.
 Physics | M.P Board Exam | Previous Year Questions

Short type questions

Arivihan
Q3. What is capacity of a conductor? Write the factors affecting the
capacity of a conductor [2023]
Ans : (iii) The medium surrounding the conductor-Due to the
presence of any medium of dielectric constant K (in place of air)
around the conductor, its capacity becomes K times i.e., increases.
 Physics | M.P Board Exam | Previous Year Questions

Long type questions

Arivihan
Q1. What is capacitor? Establish an expression for finding
capacitance of the parallel capacitor ? [2020 , 22]
OR
Derive the expression for capacitance of a parallel plate capacitor. If a
dielectric medium is placed in between the plates. How is the
capacitance affected. [MP 2020]
 Physics | M.P Board Exam | Previous Year Questions

Long type questions

Arivihan
Ans : Capacitor - It is a device which increases the capacity of a
conductor without increasing its size. In other words, it is a device
to store charge (electrical energy).

Let X and Y be two parallel plates of a condenser. The area of each

plate is A.
On giving a charge +Q to the X plate a charge -Q is induced on the
inner surface of Y plate and charge +Q is induced on its outer
surface. On earthing the Y plate, it becomes negatively charged.
 Physics | M.P Board Exam | Previous Year Questions

Long type questions

Arivihan
Ans : Let there be a medium of dielectric constant K
between the two plates. The surface densities
(=Charge/Area) of the X and Y plates are respectively
σ and -σ and the distance between the plates is d.
The intensity of electric field at point P due to X plate,
Ex = σ/(2Kε₀)
[∵ Intensity of electric field at a point near a flat sheet
= σ/2ε₀]
 Physics | M.P Board Exam | Previous Year Questions

Long type questions

Arivihan
Ans : Electric field intensity at point P due to Y plate,
Ey = σ/(2Kε₀)
Ex and Ey are directed in the same direction i.e.,
from X to Y, hence resultant intensity,
E = Ex + Ey = σ/(2Kε₀) + σ/(2Kε₀) = 2σ/(2Kε₀)
∴ E = σ/(Kε₀)
 Physics | M.P Board Exam | Previous Year Questions

Long type questions

Arivihan
Ans : Potential difference between the plates X and Y,
V=E×d
[∵ Potential Difference = Electric field intensity × Distance]
V = (σ/(Kε₀)) × d
On substituting the value of V in C = Q/V,
the capacity of condenser,
C = Q/(σd/(Kε₀)) = Kε₀Q/(σd)
 Physics | M.P Board Exam | Previous Year Questions

Long type questions

Arivihan
Ans : or
C = Kε₀QA/(Qd) [∵ σ = Q/A]
∴ C = Kε₀A/d
This is the required expression.
 Physics | M.P Board Exam | Previous Year Questions

Long type questions

Arivihan
Q2. Obtain an expression for the equivalent capacity of three
condensers with capacities C₁, C₂ and C₃ respectively, connected in
parallel. Draw diagram of the Combination
[2019]
 Physics | M.P Board Exam | Previous Year Questions

Long type questions

Arivihan
Ans : Parallel combination—
The combination of condensers in which the
positively charged plates of all the condensers are
connected to one point and all the negatively
charged plates are connected to another point by
the wires of zero capacity, is called parallel
combination.
 Physics | M.P Board Exam | Previous Year Questions

Long type questions

Arivihan
Ans : Parallel combination—
In Fig. , there are three condensers of capacities C₁,
C₂ and C₃ respectively. Now if a total charge +Q is
given to the point A, it
 Physics | M.P Board Exam | Previous Year Questions

Long type questions

Arivihan
Ans : divides itself in three parts Q₁, Q₂ and Q₃ in accordance with the
capacities of condensers. All the positive plates are connected together
at point A therefore they will be at same potential V₁ whereas all the
negative plates are connected at point B and their potential will be V₂.
Therefore in parallel combination, the potential difference between the
plates of each condenser is the same, but charge is different on the
plates of different condensers.
 Physics | M.P Board Exam | Previous Year Questions

Long type questions

Arivihan
Ans : The charge gets divided in the ratio of their capacities.
Charge on first condenser
Q₁ = C₁(V₁ - V₂) ...(1)
Charge on second condenser
Q₂ = C₂(V₁ - V₂) ...(2)
Charge on third condenser
Q₃ = C₃(V₁ - V₂) ...(3)
 Physics | M.P Board Exam | Previous Year Questions

Long type questions

Arivihan
Ans : On adding eqns. (1), (2) and (3), we get
Q = Q₁ + Q₂ + Q₃
i.e., Q = (V₁ - V₂)(C₁ + C₂ + C₃) ...(4)
If V₁ - V₂ = V and the equivalent capacitance
of the condenser is C,then
substituting Q = CV in eqn. (4),
C V = V(C₁ + C₂ + C₃)
⇒ C = C₁ + C₂ + C₃
This is the required expression.
 Physics | M.P Board Exam | Previous Year Questions

Long type questions

Arivihan
Q3. Obtain an expression for the equivalent capacitance of the
condensers connected in series combination. [2019]
Ans : As shown in Figure , the condensers of capacities C₁,
C₂ and C₃ respectively are connected in series by
conducting wires.
 Physics | M.P Board Exam | Previous Year Questions

Long type questions

Arivihan
Q3. Obtain an expression for the equivalent capacitance of the
condensers connected in series combination. [2019]
Ans : The potential of plate a of first
condenser is V₁, the potential of its second
plate b is V₂.
The potential of first plate c of second
condenser will also be V₂ because b and c are
connected together by a conducting wire of
zero capacitance.
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Arivihan
Ans : Similarly, the potential of second plate d of second condenser and
first plate e of the third condenser both, will be V₃. Let the potential of f
plate be V₄. The charge (Q) on all the condensers is same.
Therefore by V = Q/C, the potential difference for the first condenser,
V₁ - V₂ = Q/C₁ ...(1)
The potential difference for second condenser
V₂ - V₃ = Q/C₂ ...(2)
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Arivihan
The potential difference for third condenser
V₃ - V₄ = Q/C₃ ...(3)
Adding eqns. (1), (2) and (3), we get
V₁ - V₄ = Q/C₁ + Q/C₂ + Q/C₃ ...(4)
If the equivalent potential difference V = V₁ - V₄
and the equivalent capacity of the combination is
C then substituting V =Q/C in eqn. (4)
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Arivihan
Q/C = Q/C₁ + Q/C₂ + Q/C₃
⇒ 1/C = 1/C₁ + 1/C₂ + 1/C₃
This is the required expression.
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Arivihan
Q4. Obtain the potential at a point equidistant from both the charges
of an electric dipole.
[MP 2014, 16]
Ans: Let AB be an electric dipole consists of +q
and -q charges separated by a distance 2l. A
point P lies on the neutral axis at a distance r
from its centre. If a point P is taken on the
neutral axis then it will be equidistant from
both the charges of dipole as shown in Fig.
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Arivihan
Now, AP = BP = √(r² + l²)
∴ Potential at P due to +q charge will be
V₁ = 1/(4πε₀) q/AP
or V₁ = 1/(4πε₀) q/√(r² + l²)
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Arivihan
Similarly potential at P due to -q charge will be
V₂ = 1/(4πε₀) [-q/BP]
or V₂ = 1/(4πε₀) (-q)/√(r² + l²)
∴ Total potential due to dipole at point P
V = V₁ + V₂
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Arivihan
= 1/(4πε₀) q/√(r² + l²) + 1/(4πε₀) (-q)/√(r² + l²)
= 1/(4πε₀) q/√(r² + l²) - 1/(4πε₀) q/√(r² + l²)
or V = 0.
Therefore, the potential due to a dipole at a point in the equatorial
position is zero.
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Arivihan
Q1. Three capacitance 3 pF, 4 pF and 5 pF are connected in parallel to
a 120 V battery. Find the total capacitance of the combination and
charge on each capacitor. (2024)

Ans : Since 3 pF, 4 pF and 5 pF are connect in parallel order, hence

equivalent capacitance, from,
C = C₁ + C₂ + C₃
C=3+4+5
C = 12 pF.
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Arivihan
Charge on 3 pF capacitor, (let) q₁ = C₁V
= 3 × 120
= 360 pC.
Charge on 4 pF capacitor, (let) q₂ = C₂V
= 4 × 120
= 480 pC.
Charge on 5 pF capacitor, (let) q₃ = C₃V
= 5 × 120
q₃ = 600 pC.
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Electrostatic Potential and Capacitance

Arivihan
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Q 1. Three capacitance 3 pF, 4 pF and 5 pF are connected in parallel to a 120

V battery. Find the total capacitance of the combination and charge on each

Arivihan
capacitor. [2024]
Solution: Three capacitors of 3 pF, 4 pF and 5 pF are connect in parallel order, hence
equivalent capacitance, from,
C = C₁ + C₂ + C₃
C=3+4+5
C = 12 pF.
Charge on 3 pF capacitor, (let) q₁ = C₁V
= 3 × 120
= 360 pC.
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Q 1. Three capacitance 3 pF, 4 pF and 5 pF are connected in parallel to a 120

V battery. Find the total capacitance of the combination and charge on each

Arivihan
capacitor. [2024]
Solution:
Charge on 4 pF capacitor, (let) q₂ = C₂V
= 4 × 120
= 480 pC.
Charge on 5 pF capacitor, (let) q₃ = C₃V
= 5 × 120
q₃ = 600 pC.
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Q 2. 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

Arivihan
zero? Take the potential at infinity to be zero
Solution:
There are two charges,
q₁ = 5 × 10⁻⁸ C
q₂ = -3 × 10⁻⁸ C
Distance between the two charges,
d = 16 cm = 0.16 m
Consider a point P on the line joining the two charges,
r = Distance of point P from charge q₁
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Solution: Let the electric potential (V) at point P be zero.

Potential at point P is the sum of potentials caused by charges q₁ and q₂ respectively.

∴
Where,
Arivihan
V = 1/(4πε₀) × q₁/r + 1/(4πε₀) × q₂/(d-r) ...(1)

ε₀ = Permittivity of free space

For V = 0, equation (1) reduces to

0 = 1/(4πε₀) × q₁/r + 1/(4πε₀) × q₂/(d-r) ⇒ 1/(4πε₀) × q₁/r

= -1/(4πε₀) × q₂/(d-r)
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Solution: ⇒ q₁/r = -q₂/(d-r) ⇒ 5×10⁻⁸/r = -(-3×10⁻⁸)/(0.16-r)

⇒ 5(0.16-r) = 3r

Arivihan
⇒ 0.8 = 8r ⇒ r = 0.1 m = 10 cm
Therefore, the potential is zero at a distance of 10 cm from the positive charge
between the charges.

For this arrangement, potential is given by,

V = 1/(4πε₀) × q₁/s + 1/(4πε₀) × q₂/(s-d) ...(2)

Where,
ε₀ = Permittivity of free space
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Solution: Suppose point P is outside the system of two charges at a distance s from
the negative charge, where potential is zero as shown in the following figure.

Arivihan
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Solution: For V = 0, equation (2) reduces to

Arivihan
0 = 1/(4πε₀) × q₁/s + 1/(4πε₀) × q₂/(s-d) ⇒ 1/(4πε₀) × q₁/s = -1/(4πε₀) × q₂/(s-
d)
⇒ q₁/s = -q₂/(s-d)
⇒ 5×10⁻⁸/s = -(-3×10⁻⁸)/(s-0.16)
⇒ 5(s-0.16) = 3s
⇒ 0.8 = 2s ⇒ s = 0.4 m = 40 cm
Therefore, the potential is zero at a distance of 40 cm from the positive charge
outside the system of charges.
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Q 3. 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.

Arivihan
Solution: The given figure shows six equal amount of charges, 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 centre O, d = 10 cm
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Q 3. 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.

Arivihan
Solution: Electric potential at point O,
V = (1/4πε₀) × (6 × q)/d
Where,
Where, ε₀ = Permittivity of free space and 1/4πε₀ = 9 × 10⁹ Nm²C⁻²
∴ V = (9 × 10⁹ × 6 × 5 × 10⁻⁶)/0.1 = 2.7 × 10⁶ V
Therefore, the potential at the centre of the hexagon is 2.7 × 10⁶ V.
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Q 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

Arivihan
(a) inside the sphere
(b) just outside the sphere
(c) at a point 18 cm from the centre of the sphere?
Solution: (a) Radius of the spherical conductor, r = 12 cm = 0.12 m
Charge is uniformly distributed over the conductor, q = 1.6 × 10⁻⁷ C
Electric field inside a spherical conductor is zero. This is because if there is field inside
the conductor, then charges will move to neutralize it.

(b) Electric field E just outside the conductor is given by the relation,
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Solution: E = (1/4πε₀) × q/r²

Arivihan
Where, ε₀ = Permittivity of free space and 1/4πε₀ = 9 × 10⁹ Nm²C⁻²
Therefore, E = (9×10⁹×1.6×10⁻⁷)/(0.12)² = 10⁵ NC⁻¹
Therefore, the electric field just outside the sphere is 10⁵ NC⁻¹.
c) Electric field at a point 18 cm from the centre of the sphere
Using E₁ = (1/4πε₀) × (q/d²)
d = 18 cm = 0.18 m
E₁ = (9×10⁹ × 1.6×10⁻⁷)/(0.18)² = 4.4×10⁴ NC⁻¹
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Q 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

Arivihan
plates is reduced by half, and the space between them is filled with a
substance of dielectric constant 6?
Solution : Capacitance between the parallel plates of the capacitor, C = 8 pF
Initially, distance between the parallel plates was d and it was filled with air.
Dielectric constant of air, k = 1
Capacitance, C, is given by the formula,
C = kε₀A/d = ε₀A/d ...(1)
Where,
A = Area of each plate
ε₀ = Permittivity of free space
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Solution : If distance between the plates is reduced to half, then new distance,

Arivihan
d₁ = d/2
Dielectric constant of the substance filled in between the plates, k₁ = 6
Hence, capacitance of the capacitor becomes
C₁ = k₁ε₀A/d₁ = 6ε₀A/d = 12ε₀A/d ...(2)
Taking ratios of equations (1) and (2), we obtain
C₁ = 2 × 6 C = 12 C = 12 × 8 pF = 96 pF
Therefore, the capacitance between the plates is 96 pF.
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Q 6. Three capacitors each of capacitance 9 pF are connected in series.

(a) What is the total capacitance of the combination?

Arivihan
(b) What is the potential difference across each capacitor if the combination
is connected to a 120 V supply?
Solution : (a) Capacitance of each of the three capacitors C=9 pF
Equivalent capacitance (Ceq) of the combination of the capacitors is given by the
relation,
1/Cₑq = 1/C1+ 1/C2 +1/C3 = 1/C+1/C+1/C = 3/C = 3/9 = 1/3
Cₑq = 3 pF
Therefore , total capacitance of the combination is 3 pF
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Solution : b) Potential difference across each capacitor at 120 V supply:

Potential difference ( V1) across each capacitor is equal to one third of the supply

Arivihan
voltage.

∴ V₁ = V/3 = 120/3 = 40 V
Therefore, the potential difference across each capacitor is 40 V.
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Q 7. Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in

parallel.

Arivihan
(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.
Solution : a) Capacitance of the given capacitors C1=2pF, C2=3pF, C3=4pF
For the parallel combination of the capacitors, equivalent capacitors is given by Ceq
the algebraic sum ,Therefore
Total capacitance:
Cₑq = C₁ + C₂ + C₃ = 2 + 3 + 4 = 9 pF
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b) Supply voltage, V = 100 V

Arivihan
The voltage through all the three capacitors is same = V = 100 V
Charge on a capacitors of capacitance C and Potential difference V is given by the
equation. q = V C
Charge on each capacitor at 100 V supply:
For C = 2 pF, charge = VC = 100 × 2 = 200 pC = 2 × 10⁻¹⁰ C
For C = 3 pF, charge = VC = 100 × 3 = 300 pC = 3 × 10⁻¹⁰ C
For C = 4 pF, charge = VC = 100 × 4 = 400 pC = 4 × 10⁻¹⁰ C
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Q 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

Arivihan
capacitance of the capacitor. If this capacitor is connected to a 100 V supply,
what is the charge on each plate of the capacitor?
Solution : Area of each plate of the parallel plate capacitor, 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 plate capacitor is given by, C = ε₀A/d
Where,
ε₀ = Permittivity of free space = 8.854 × 10⁻¹² N⁻¹ m⁻² C²
∴ C = (8.854 × 10⁻¹² × 6 × 10⁻³)/(3 × 10⁻³)
= 17.71 × 10⁻¹² F = 17.71 pF
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So, charge on each plate

Arivihan
q = VC = 100 × 17.71 × 10⁻¹² C = 1.771 × 10⁻⁹ C
Therefore, capacitance of the capacitor is 17.71 pF and charge on each plate is 1.771
× 10⁻⁹ C.
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Q 9. A 12pF capacitor is connected to a 50V battery. How much electrostatic

energy is stored in the capacitor?

Arivihan
Solution : Capacitor of the capacitance, C = 12 pF = 12 × 10⁻¹² F

Potential difference, V = 50 V

Electrostatic energy stored in the capacitor is given by the relation,

E = ½CV² = ½ × 12 × 10⁻¹² × (50)² J = 1.5 × 10⁻⁸ J

Therefore, the electrostatic energy stored in the capacitor is 1.5 × 10⁻⁸ J.

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Q 10. 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.

Arivihan
How much electrostatic energy is lost in the process?
Solution : Capacitance of the capacitor, C = 600 pF
Potential difference, V = 200 V
Electrostatic energy stored in the capacitor is given by,
E₁ = ½CV²
= ½ × (600 × 10⁻¹²) × (200)² J
= 1.2 × 10⁻⁵ J
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Solution : If supply is disconnected from the capacitor and another capacitor of

Arivihan
capacitance C = 600 pF is connected to it, then equivalent capacitance (Ceq) of the
combination is given by,
1/Ceq = 1/C + 1/C
⇒ 1/Ceq = 1/600 + 1/600 = 2/600 = 1/300
⇒ Ceq = 300 pF
New electrostatic energy can be calculated as
E₂ = ½CeqV²
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Solution := ½ × 300 × (200)² J

Arivihan
= 0.6 × 10⁻⁵ J
Loss in electrostatic energy = E₁ - E₂
= 1.2 × 10⁻⁵ - 0.6 × 10⁻⁵ J
= 0.6 × 10⁻⁵ J
= 6 × 10⁻⁶ J
Therefore, the electrostatic energy lost in the process is 6 × 10⁻⁶ J.