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NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12 PDF Download

2024

Q1: 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 NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12 of magnitude, 4 × 10−6Cm, is  ±9 × 103V.
(Take 1 / 4πϵ0 = 9 × 109 SI units)
Reason R: NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12, 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:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is NOT the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.               [2024]
Ans:
(c)
The potential V at any point, at distance r from centre of dipole = NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
At axial point where  NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
At axial point where  NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Q2: In the following circuit, the equivalent capacitance between terminal A and terminal B is :

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12(a) 2 μF
(b) 1 μF
(c) 0.5 μF
(d) 4 μF        [2024]
Ans:
(a)
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12Given circuit is balanced Wheatstone bridge
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

CAB = 1 + 1
= 2 μF

Q3: 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:
(a) A, B and E only
(b) A, C and E only
(c) B, D and E only
(d) A, B and C only               [2024]
Ans:
(b)
Given V′ = V =  Constant
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
Hence, final capacitance greater than initial capacitance,
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
Hence final energy is greater than initial energy
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(iv) Product of charge and voltage
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

2023

Q1: An electric dipole is placed at an angle of 30° with an electric field of intensity 2 x 105NC-1 It experiences a torque equal to 4 N m. Calculate the magnitude of charge on the dipole, if the dipole length is 2 cm.
(a) 6 mC
(b) 4 mC
(c) 2 mC
(d) 8 mC
Ans: 
(c)
The torque τ experienced by an electric dipole in an electric field is given by the formula:

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

where p is the electric dipole moment, E is the electric field intensity, and θ is the angle between the dipole and the electric field. The electric dipole moment p can be expressed as:

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

where q is the charge on the dipole, and d is the dipole length.

We are given the following values:
Torque τ = 4 N·m

Electric field intensity E = 2 x 105 NC-1 Angle G = 30°

Dipole length d = 2 cm = 0.02 m

We need to find the charge q on the dipole. Let's first solve for the electric dipole moment p:

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Substituting the given values:

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

So, the magnitude of the charge on the dipole is 2 mC.

Q2: The equivalent capacitance of the system shown in the following circuit is:
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12 
(a) 2 μF
(b) 3 μF
(c) 6 μF
(d) 9 μF
Ans:
(a)
For parallel grouping
C1 = 3 + 3 = 6 μF
For series grouping
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Q3: According to Gauss's law of electrostatics, electric flux through a closed surface depends on:      
(a) the area of the surface
(b) the quantity of charges enclosed by the surface 

(c) the shape of the surface 

(d) the volume enclosed by the surface
Ans: (b)
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12only depends on the charge enclosed by the surface.

Q4: A charge Q /μC is placed at the center of a cube. The flux coming out from any one of its faces will be (in SI unit):    
(a) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(b) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(c) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(d) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Ans: (d)
Total flux from cube =q/ϵ0
So the flux of any one surface of the cube
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Q5: If a conducting sphere of radius R is charged. Then the electric field at a distance r(r > R) from the center of the sphere would be, (V = potential on the surface of the sphere)    
(a) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(b) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(c) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(d) V/r
Ans:
(c)
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Q6: if NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12  over a surface, then    
(a) the magnitude of the electric field on the surface is constant.
(b) all the charges must necessarily be inside the surface.
(c) the electric field inside the surface is necessarily uniform.
(d) the number of flux lines entering the surface must be equal to the number of flux lines leaving it,
Ans: 
(d)
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

The number of field lines entering is equal to the number of field lines leaving.

Q7: An electric dipole is placed as shown in the figure.
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12The electric potential (in 102 V) at point P due to the dipole is (ϵo - permittivity of free space and
 NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12   
(a) (5/8) qK
(b) (8/5) qK
(c) (8/3) qK
(d) (3/8) qK

Ans: (d)
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

2022

Q1: Six charges +q, −q, +q, −q, +q, and −q are fixed at the corners of a hexagon of side d as shown in the figure. The work done in bringing a charge q0 to the centre of the hexagon from infinity is

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

(a) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

(b) Zero

(c) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

(d) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Ans: (b)

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Q2: Two-point charges −q and +q is placed at a distance of L, as shown in the figure.

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

The magnitude of electric field intensity at a distance R(R >> L) varies as:
(a) 1/R2
(b) 1/R3
(c) 1/R4
(d) 1/R6
Ans:
(b)
For R » L, arrangement is an electric dipole

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Q3: A capacitor of capacitance C = 900 pF is charged fully by 100 V battery B as shown in Figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C = 900 pF as shown in Figure (b). The electrostatic energy stored by the system (b) is                   
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

(a) 3.25 × 10–6

(b) 2.25 × 10–6

(c) 1.5 × 10–6

(d) 4.5 × 10–6 J

Ans: (b)
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Q4: Two hollow conducting spheres of radii R1 and R2 (R1 >> R2) have equal charges. The potential would be                 
(a) More on the smaller sphere 
(b) Equal in both the spheres 
(c) Dependent on the material property of the sphere 
(d) More on the bigger sphere
Ans: (a)

The potential of conducting hollow sphere = KQ/R
Now, Q = same
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Q5: The angle between the electric lines of force and the equipotential surface is         
(a) 45°
(b) 90° 
(c) 180° 
(d) 0° 
Ans: (b)
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

2021

Q1: A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and the area of each plate is A , the energy stored in the capacitor is (ε0 = permittivity of free space).

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
Ans: 
(c)

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Q2: A parallel plate condenser has a uniform electric field E(V/m) in the space between the plates. If the distance between the plates is d(m) and the area of each plate is A(m2), the energy (joule) stored in the condenser is      
(a) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

(b) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(c) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(d) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
Ans: (c)
The capacitance of a parallel plate capacitor is
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
Potential difference between the plates is V = Ed      ........ (ii)
The energy stored in the capacitor is
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
Q3: The equivalent capacitance of the combination shown in the figure is:       
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12(a) C/2

(b) 3C/2

(c) 3C
(d) 2C
Ans: 
(d)

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

The voltage at node A matches the voltage at node B. Likewise, the voltage at node F is identical to the voltage at node E. Consequently, there is no voltage drop across the EF segment, leading to the absence of current (and therefore, no charge movement) within the circuit. Therefore, EF acts as if it were an open circuit. So the capacitor between the arms E and F gets short. 
⇒ Ceq = C1 + C2
= C + C
= 2C

Q4: Two charged spherical conductors of radius R1 and R2 are connected by a wire. Then the ratio of surface charge densities of the spheres (σ12) is :      
(a) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(b) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

(c) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(d) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
Ans: (d)

When two charged spherical conductors are connected by a wire, charges will flow between them until their potentials are equal. Since they are conductors, the potential at any point on each sphere will be uniform. The potential V for a charged sphere is given by the formula:
V=kQ/R
When the two spheres are connected by a wire, the charges �1Q1 and �2Q2  will redistribute between the spheres until �1=�2V1=V2. If we assume �totalQtotal  is the total charge conserved across both spheres, the equation can be written as:
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12Surface charge density (σ) is the charge per unit area. For a sphere, surface area A is 4��24πR2. Thus, the surface charge density is:
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Q5: Twenty-seven drops of the same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.    
(a) 1520 V
(b) 1980 V
(c) 660 V
(d) 1320 V
Ans: 
(b)
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12⇒ V2 = 220 × 9
= 1980 Volt

2020

Q1: The variation of electrostatic potential with radial distance r from the centre of a positively charged metallic thin shell of radius R is given by the graph

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Ans: (b)
Since, electric potential remains constant inside the metallic spherical shell and same as the surface of spherical shell.
Outside the spherical shell, V ∝ 1/r
Hence, variation of potential (V ) with distance r is given as

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Q2:  A short electric dipole has a dipole moment of 16 × 10−9 C-m. The electric potential due to the dipole at a point at a distance of 0.6 m from the centre of the dipole, situated on a line making an angle of 60° with the dipole axis is NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(a) 200 V
(b) 400 V
(c) zero
(d) 50 V
Ans: 
(a)
Given, electric dipole moment, p = 16 × 10−9 C-m
Distance, r = 0.6 m
Angle, θ = 60° ⇒ cos60° = 1/ 2
Electric potential at a point which is at a distance r at some angle θ from electric dipole is
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
Hence, correct option is (a).

Q3: In a certain region of space with a volume of 0.2 m3, the electric potential is found to be 5 V throughout. The magnitude of the electric field in this region is:      
(a) 1 N/C 
(b) 5 N/C 
(c) Zero 
(d) 0.5 N/C
Ans: (c)
given V = const. (5 volts)
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
E = 0

Q4: The capacitance of a parallel plate capacitor with air as the medium is 6 μF. With the introduction of a dielectric medium, the capacitance becomes 30 μF. The permittivity of the medium is :       
0 = 8.85×10-12 C2 N-1 m-2
(a) 0.44×10-10 C2 N-1 m-2 
(b) 5.00 C2 N-1 m-2 
(c) 0.44×10-13 C2 N-1 m-2 
(d) 1.77×10-12 C2 N-1 m-2
Ans: (a)
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
Cmed = KCair 
30μF = K6μF
K = 5
∴ ε = ε0k = 8.85 × 10-12 × 5
ε = 44.25 × 10-12
ε = 0.4425 × 10-10
ε = 0.44 × 10-10 C2N-1m-2

Q5: A 40 μF capacitor is connected to a 200 V, 50 Hz AC supply. The RMS value of the current in the circuit is, nearly:      
(a) 2.5 A
(b) 25.1 A
(c) 1.7 A
(d) 2.05 A
Ans:
(a)
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

2019

Q1: Two metal spheres, one of radius R and the other of radius 2R respectively have the same surface charge density σ. They are brought in contact and separated. What will be the new surface charge densities on them?

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Ans: (d)
The surface charge density of a closed surface area having charge Q is given by

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Thus, the charges on sphere P and Q having same charge density as shown in the figure below is given by

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

when they are brought in contact with each other, the total charge will be

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

In connection of two charged conducting bodies, the potential will become same on both, i.e.

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

So, the charges on the sphere P and Q after separation will be distributed as

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

After separation, the new surface charge densities on P and Q will be

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Q2: Two identical capacitors C1 and C2 of equal capacitance are connected as shown in the circuit.
Terminals a and b of the key k are connected to charge capacitor C1 using battery of emf V volt. Now, disconnecting a and b the terminals b and c are connected.
Due to this, what will be the percentage loss of energy?

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

(a) 75%
(b) 0%
(c) 50%
(d) 25%
Ans: 
(c)
When C1 is connected to voltage source, it is charged to a potential V and this will be stored as a potential energy in the capacitor given by

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

When key is disconnected from battery and b and c are connected, the charge will be transformed from the capacitor C1 to capacitor C2 , then

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

The loss of energy due to redistribution of charge is given by

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

2018

Q1: The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is:    
(a) independent of the distance between the plates.
(b) linearly proportional to the distance between the plates
(c) proportional to the square root of the distance between the plates.
(d) inversely proportional to the distance between the plates.
Ans: (a)

Solution:

F =  QE
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
∴ The electrostatic force is independent of the distance between plates.

2017

Q1: A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system:  
(a) Decreases by a factor of 2
(b) Remains the same
(c) Increases by a factor of 2
(d) Increases by a factor of 4
Ans: (a)
Solution:
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
Charge on capacitor
q = CV
when it is connected to another uncharged capacitor.
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
Q2: The diagrams below show regions of equipotentials:     
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12A positive charge is moved from A to B in each diagram.
(a) Maximum work is required to move q in Figure (c)
(b) Minimum work is required to move q in Figure (a)
(c) Maximum work is required to move q in Figure (b)
(d) In all four cases the work done is the same
Ans:
(d)
Solution:
W = qΔV
as ΔV is the same in all conditions, work will be the same.

2016

Q1: A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials having dielectric constants k1, k2, k3 and k4 as shown in the figure below. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Ans: (*)
Given capacitor is equivalent to capacitors K1 , K2 and K3 in parallel and part of K4 in series with them

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

No option is matching.

Q2: A capacitor of 2µF is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is :   
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(a) 80%
(b) 0%
(c) 20%
(d) 75%
Ans: (a)

Solution:

Consider the figure given above.
When switch S is connected to point 1, the initial energy stored in the capacitor is given as,
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
When the switch S is connected to point 2, energy dissipated on connection across 8μF will be,
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12Therefore, the percentage loss of energyNEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

2015

Q1: If potential (in volts) in a region is expressed as V (x, y, z) = 6xy − y + 2yz, the electric field (in N/C) at point (1, 1, 0) is [CBSE AIPMT 2015] (a) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(b) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(c) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(d) NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
Ans
: (b)
Given, potential in a region,
V = 6xy − y + 2yz.
Electric field in a region,

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

At, (1, 1, 0), electric field can be expressed,

NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

Q2: 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?    
(a) The charge on the capacitor is not conserved.
(b) The potential difference between the plates decreases K times
(c) The energy stored in the capacitor decreases K times

(d) the change in energy stored is  NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12Ans: (a)
Solution:
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 = CV
The energy stored in the capacitor is
E = 1/2 CV2
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' = KC)
The voltage becomes 1/K time the original voltage
V' = V/K
The change in energy stored is
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

2014

Q1: In a region, the potential is represented by V(x, y, z) = 6x − 8xy − 8y + 6yz, where V is in volts and x, y, and z are in meters. The electric force experienced by a charge of 2 coulombs situated at point (1,1,1) is:    
(a) 24N
(b)NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(c)NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(d) 30N
Ans:
(b)
Solution:
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12Q2: A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the center of the sphere respectively are :    
(a)NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(b) Both are zero
(c)NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
(d)NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
Ans: 
(d)
Solution:
NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12
The electric field inside, Einside = 0
Potential, Vinside = VsurfaceNEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12

The document NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance | Physics Class 12 is a part of the NEET Course Physics Class 12.
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FAQs on NEET Previous Year Questions (2014-2024): Electrostatics Potential & Capacitance - Physics Class 12

1. What is the formula for calculating the electric potential due to a point charge?
Ans. The formula for calculating the electric potential due to a point charge is V = kQ/r, where V is the electric potential, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q is the charge, and r is the distance from the charge.
2. How is the electric potential energy related to the electric potential in a system of charges?
Ans. The electric potential energy (U) of a system of charges is related to the electric potential (V) by the equation U = qV, where q is the charge. This means that the electric potential energy is the charge multiplied by the electric potential at that point.
3. What is the difference between electric potential and electric potential energy?
Ans. Electric potential (V) is the amount of electric potential energy (U) per unit charge at a point in an electric field. Electric potential energy, on the other hand, is the energy a charge possesses due to its position in an electric field.
4. How does the capacitance of a capacitor affect its ability to store electric charge?
Ans. The capacitance of a capacitor is a measure of its ability to store electric charge. A higher capacitance means that the capacitor can store more charge for a given potential difference across its plates.
5. How can the capacitance of a parallel plate capacitor be increased?
Ans. The capacitance of a parallel plate capacitor can be increased by increasing the area of the plates, decreasing the distance between the plates, or using a material with a higher dielectric constant between the plates.
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