NEET PhysicsUpdated 19 Oct 2026

Electrostatic Potential and Capacitance: NEET PYQs and weightage

Electrostatic Potential and Capacitance carries 4.2 of 100 Physics marks in NEET by past-paper share: 32 of the 758 tagged NEET Physics PYQs in the ViJEEta bank (NEET 2013–2026). It ranks 7 of 30 Physics chapters.

Take the 25-question Electrostatic Potential and Capacitance test Formula sheet ↓All NEET Physics chapters

4.2/100 Weight in Physics

Exam importance (3/5, by past-paper share)

32 NEET PYQs

484 Practice questions

456 With measured difficulty

How the weight is computed: this chapter's tagged NEET PYQs divided by all tagged NEET Physics PYQs in the bank (32 of 758), times 100, with repeated questions counted once. The bank does not store the year of each NEET question, so none is printed. Computed from the bank at build time. Updated 19 Oct 2026.

Electrostatic Potential and Capacitance: formula sheet for NEET

14 formulas, each with when to use it. Copy it onto one page of your notebook: that page is this chapter's handbook.

Capacitance
$C = \frac{Q}{V} (Q = charge on one plate)$
$1 F = 1 C/V$
$1 μ F = 1 0^{- 6} F$
$1 pF = 1 0^{- 12} F$
Use it when: you know any two of charge, voltage and capacitance and need the third
Parallel plate (air/vacuum)
$C = \frac{ε _{0} A}{d}$
$ε_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}$
Use it when: plate area $A$ and gap $d$ are given ( $A$ in $m^{2}$ , $d$ in m, not cm or mm)
Dielectric filling the whole gap
$C = K C_{0} = \frac{K ε _{0} A}{d} = \frac{ε A}{d}$
$ε = K ε_{0}$
Use it when: the space between the plates is completely filled with a medium of dielectric constant $K$
Series
$\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}} + \dots$
Use it when: capacitors are in one line; each carries the same $Q$ and the voltages add up
Parallel
$C = C_{1} + C_{2} + C_{3} + \dots$
Use it when: capacitors share the same two terminals; each has the same $V$ and the charges add up
Energy stored
$U = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C} = \frac{1}{2} Q V$
Use it when: asked for stored energy; pick the form whose quantity stays fixed in the problem
Field between the plates
$E = \frac{σ}{ε _{0}} = \frac{Q}{ε _{0} A}$
$V = E d$
Use it when: you need the uniform field inside a parallel plate capacitor (it is zero outside)
Energy density
$u = \frac{1}{2} ε_{0} E^{2} (inside a dielectric: u = \frac{1}{2} K ε_{0} E^{2})$
Use it when: asked for energy per unit volume of the field
Slab of thickness t
$C = \frac{ε _{0} A}{d - t + \frac{t}{K}} (t < d, same area as the plates)$
Use it when: a dielectric slab only partly fills the gap ( $t = d$ gives $K C_{0}$ , $K = 1$ gives $C_{0}$ )
Dielectric inserted after the battery is removed
$Q stays the same$
$V \to \frac{V}{K}$
$E \to \frac{E}{K}$
$U \to \frac{U}{K}$
Use it when: the capacitor is isolated (charged, then disconnected) before the slab goes in
Dielectric inserted with the battery still connected
$V stays the same$
$C \to K C$
$Q \to K Q$
$U \to K U$
Use it when: the supply stays connected while the slab goes in
Isolated sphere of radius R
$C = 4 π ε_{0} R$
Use it when: there is a single charged conducting sphere (the other "plate" is at infinity)
Spherical capacitor
$C = \frac{4 π ε _{0} ab}{b - a} (radii a < b)$
Multiply by $K$ if a dielectric fills the space between the shells.
Use it when: there are two concentric conducting shells
Charged C joined to an identical uncharged C
$V^{'} = \frac{V}{2}$
$U_{total} \to \frac{1}{2} U_{original}$
Use it when: two equal capacitors share charge (the lost half goes out as heat and radiation; the charge is conserved)

Trap

Not checking whether the battery is still connected. Connected means V is fixed. Disconnected means Q is fixed. Using ½CV² with the old V after the battery is removed gives the wrong energy.

How does NEET ask Electrostatic Potential and Capacitance?

8 real previous-year questions from this chapter, as printed in the paper, each with its worked solution. Answer the first one here with no sign-in.

Q1PhysicsNEET UG PYQElectrostatics of Conductors

Try right here · No sign-in

Which of the following statements are correct? A. Inside a conductor, the electrostatic field is zero. B. Electric field at the surface of a charged conductor does not depend on its surface charge density. C. The interior of a charged conductor can have no excess charge in the static situation. D. At the surface of a charged conductor, the electrostatic field must be normal to the surface at every point. E. The electrostatic potential is zero everywhere inside a charged conductor. Choose the correct answer from the options given below:

Answer key (C) & worked solution

Correct answer: Option C

A. Electrostatic field is zero inside a conductor. B. Electric field at the surface of a charged conductor

= \frac{σ}{ε _{0}} \overset{n}{^}

, depends on surface charge density (

σ

). C. The interior of a charged conductor cannot have any excess charge in the static situation. D. At the surface of a charged conductor, the electrostatic field ⟂ surface. E. The electrostatic potential is constant and can be non-zero everywhere inside a charged conductor.

Q2PhysicsNEET UG PYQCapacitors and Capacitance

A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?

AThe charge on the capacitor is not conserved.

BThe potential difference between the plates decreases K times.

CThe energy stored in the capacitor decreases K times

Dthe change in energy stored is

\frac{1}{2} C V^{2} (\frac{1}{K} - 1)

Answer key (A) & worked solution

Correct answer: Option A

A parallel plate air capacitor of capacitance

C

is connected to a cell of emf

V

and then disconnected from it. The charge on the capacitor is given by

Q = C V

The energy stored in the capacitor is E=

\frac{1}{2} C V^{2}

When a dielectric slab of dielectric constant K is inserted in it, the charge

Q

is conserved. The capacitance becomes

K

times the original capacitance.

(C^{'} = K C)

The voltage becomes

\frac{1}{K}

time the original voltage.

V^{'} = \frac{Q}{K C} = \frac{V}{K}

Energy decreases by K times :-

\frac{Q ^{2}}{2 C ^{'}} = \frac{Q ^{2}}{2 K C} = \frac{E}{K}

The change in energy stored is $\begin{gathered} \frac{{{Q^2}}}{{2C'}} - \frac{{{Q^2}}}{{2C}} = \frac{{{Q^2}}}{{2KC}} - \frac{{{Q^2}}}{{2C}} = \frac{{{Q^2}}}{{2C}}\left[ {\frac{1}{K} - 1} \right] \\ = \frac{1}{2}C{V^2}\left[ {\frac{1}{K} - 1} \right] \\ \end{gathered} $

Q3PhysicsNEET UG PYQCapacitors and Capacitance

The capacitance of a parallel plate capacitor with air as medium is 6

μ F .

With the introduction of a dielectric medium, the capacitance becomes 30

μ F .

The permittivity of the medium is :

(\in_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2})

0.44 \times 1 0^{- 10} C^{2} N^{- 1} m^{- 2}

5.00 C^{2} N^{- 1} m^{- 2}

0.44 \times 1 0^{- 13} C^{2} N^{- 1} m^{- 2}

1.77 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}

Answer key (A) & worked solution

Correct answer: Option A

C_{air} = 6 μ F, C_{medium} = 30 μ F

C_{air} = \frac{\in _{0} A}{d}, C_{medium} = \frac{\in _{0} A k}{d}

C_{med} = K C_{air}

30 μ F = K 6 μ F

K = 5

∴ \in = \in_{0} K = 8.85 \times 1 0^{- 12} \times 5

\in = 44.25 \times 1 0^{- 12}

\in = 0.4425 \times 1 0^{- 10}

\in = 0.44 \times 1 0^{- 10}

C^{2} N^{- 1} m^{- 2}

Q4PhysicsNEET UG PYQPotential Energy

An electric dipole with dipole moment

5 \times 1 0^{- 6} C m

is aligned with the direction of a uniform electric field of magnitude

4 \times 1 0^{5} N / C

. The dipole is then rotated through an angle of

6 0^{\circ}

with respect to the electric field. The change in the potential energy of the dipole is:

A1.2 J

B1.5 J

C0.8 J

D1.0 J

Answer key (D) & worked solution

Correct answer: Option D

Δ U = U_{2} - U_{1}

= (- pE cos θ_{2}) - (- pE cos θ_{1})

= - pE cos 60 + pE cos 0

= \frac{- pE}{2} + pE

= \frac{1}{2} pE = \frac{1}{2} \times 5 \times 1 0^{- 6} \times 4 \times 1 0^{5} = 1.0 J

Q5PhysicsNEET UG PYQCapacitors and Capacitance

The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is

AProportional to the square root of the distance between the plates

BLinearly proportional to the distance between the plates

CIndependent of the distance between the plates

DInversely proportional to the distance between the plates

Answer key (C) & worked solution

Correct answer: Option C

For isolated capacitor Q = Constant

F_{pl a t e} = \frac{Q ^{2}}{2 A \in _{0}}

F is Independent of the distance between plates.

Q6PhysicsNEET UG PYQElectrostatic Potential

In a region, the potential is represented by V(x, y, z) = 6x - 8xy - 8y + 6yz, where V is in volts and x, y, z are in meters. The electric force experienced by a charge of 2 coulomb situated at point (1,1,1) is

A24N

435

65

D30N

Answer key (B) & worked solution

Correct answer: Option B

E_{x} = - \frac{\partial V}{\partial x} = - (6 - 8 y) = 2 E_{y} = - \frac{\partial V}{\partial y} = - (- 8 x - 8 + 6 z) = 10 E_{z} = - \frac{\partial V}{\partial z} = - (- 6 y) = - 6 E = 2^{2} + 10^{2} + 6^{2} = 140 = 235 F = 435

Q7PhysicsNEET UG PYQCapacitors and Capacitance

If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then A. The charge stored in it, increases. B. The energy stored in it decreases. C. Its capacitance increases. D. The ratio of charge to its potential remains the same. E. The product of charge and voltage increases. Choose the most appropriate answer from the options given below:

AA, C and E only

BB, D and E only

CA, B and C only

DA, B and E only

Answer key (A) & worked solution

Correct answer: Option A

c = \frac{ε _{0} A}{d} d ↓ c ↑

∵

V = constant

q = c v (q ↑)

U = \frac{1}{2} C V^{2} (U ↑)

\frac{q}{V} = c \times

not constant

q v = c v . v = c v^{2}

Q8PhysicsNEET UG PYQElectrostatic Potential

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The potential (V) at any axial point, at 2m distance (r) from the centre of the dipole of dipole moment vector

P

of magnitude,

4 \times 1 0^{- 6} C m

, is

\pm 9 \times 1 0^{3} V

(Take

\frac{1}{4 π \in _{0}} = 9 \times 1 0^{9}

SI units) Reason R:

V = \pm \frac{2 P}{4 π \in _{0} r ^{2}},

where r is the distance of any axial point, situated at 2m from the centre of the dipole. In the light of the above statements, choose the correct answer from the options given below:

ABoth A and R are true and R is NOT the correct explanation of A.

BA is true but R is false

CA is false but R is true

DBoth A and R are true and R is the correct explanation of A.

Answer key (B) & worked solution

Correct answer: Option B

Potential due to dipole at axial position is

= \frac{k P c o s θ}{r ^{2}}

(where

θ = 0^{\circ}

18 0^{\circ}

)

V_{a x ia l} = \pm \frac{9 \times 10 ^{9} \times 4 \times 10 ^{- 6}}{( 2 ) ^{2}} = \pm 9 \times 1 0^{3} V

Which topics in Electrostatic Potential and Capacitance carry the most questions?

All 8 topics, by tagged PYQs, then practice questions. Counted from the bank. Updated 19 Oct 2026.

Topic	PYQs	Practice questions
Electrostatic Potential	9	102
Energy Stored in a Capacitor	6	47
Capacitors and Capacitance	5	82
Combination of Capacitors	4	123
Potential Energy	3	41
Electrostatics of Conductors	2	31
Dielectrics and Polarization	2	30
Equipotential Surfaces	1	28

Why does Electrostatic Potential and Capacitance show "–" until you answer 3 questions?

Unknown is not zero. Until you answer at least 3 graded questions in Electrostatic Potential and Capacitance, ViJEEta shows – instead of a zero or a 50% default. Book exercises can raise chapter mastery up to 0.60, adaptive practice up to 0.75, and only a timed chapter test or mock can certify 0.80 and above.

How ViJEEta scores you

What happens when you test Electrostatic Potential and Capacitance?

Every chapter test is 25 questions. It starts at moderate difficulty and adapts after each saved answer. If you open a solution before answering, that attempt earns no credit, so your mastery stays honest.

Start the Electrostatic Potential and Capacitance test

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Electrostatic Potential and Capacitance: NEET PYQs and weightage

Electrostatic Potential and Capacitance: formula sheet for NEET

Capacitance

Parallel plate (air/vacuum)

Dielectric filling the whole gap

Series

Parallel

Energy stored

Field between the plates

Energy density

Slab of thickness t

Dielectric inserted after the battery is removed

Dielectric inserted with the battery still connected

Isolated sphere of radius R

Spherical capacitor

Charged C joined to an identical uncharged C

How does NEET ask Electrostatic Potential and Capacitance?

Which topics in Electrostatic Potential and Capacitance carry the most questions?

Why does Electrostatic Potential and Capacitance show "–" until you answer 3 questions?

What happens when you test Electrostatic Potential and Capacitance?

Chapters either side of Electrostatic Potential and Capacitance by weight