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NEET • PHYSICS • ELECTRIC FIELDS AND CHARGES

Electric fields and charges practice · Mixed difficulty

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Showing 1-30 of 156 questions

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1 Medium
1 of 156
Remembering +4 | -1
The ratio of magnitude of electric displacement (D) and electric field (E) is
View answer and explanation
Correct option: B
Permittivity
Option B (Permittivity) is correct because the ratio of electric displacement (D) to electric field (E) is defined as the permittivity of the medium.
2 Medium
2 of 156
Remembering +4 | -1
Gauss’s Law is valid for
View answer and explanation
Correct option: A
Any closed surface
Gauss’s Law is valid for any closed surface, regardless of its shape or size. The surface is called the Gaussian surface.
3 Easy
3 of 156
Application +4 | -1
In hydrogen like system the ratio of coulombian force and gravitational force between an electron and a proton is in the order of:
View answer and explanation
Correct option: A
103910^{39}
Option A (103910^{39}) is correct because the ratio of Coulomb force to gravitational force between an electron and a proton is of this order.
4 Hard
4 of 156
Understanding +4 | -1
Statement-1: E→\overrightarrow{{E}} in outside velocity of a conductor depends only on the local charge density α\alpha.
Statement-2: E→\overrightarrow{{E}} in outside vicinity of a conductor is given by σε0\frac{\sigma}{\varepsilon_0}.
View answer and explanation
Correct option: D
Statement-1 is false, Statement-2 is true
Option D is correct because Statement-1 is false as the electric field outside a conductor depends on the local charge density, and Statement-2 is true, providing the correct formula.
5 Easy
5 of 156
Application +4 | -1
Two charges each of 1 coulomb are at a distance 1 kmkm apart the force between them is P×103P \times 10^3 then value of P{P} is
View answer and explanation
Correct option: A
9
Option A (9) is correct because the force between two 1 Coulomb charges 1 km apart is approximately 9 × 103 N.
6 Medium
6 of 156
Remembering +4 | -1
Electric field due to point charge depends on distance rr as
View answer and explanation
Correct option: D
1r2\frac{1}{r^2}
Option D is correct because the electric field due to a point charge is inversely proportional to the square of the distance, following Coulomb's Law.
7 Easy
7 of 156
Analyzing +4 | -1
A tennis ball which has been covered with charges is suspended by a thread so that it hangs between two metal plates. One plate is earthed, while other is attracted to a high voltage generator. The ball
View answer and explanation
Correct option: A
swings backward & forward hitting each plate in turn
The ball swings back and forth because it gets charged by induction when it touches a plate, then is repelled and attracted to the other plate, repeating the process.
8 Easy
8 of 156
Understanding +4 | -1
A large aluminium sheet duly insulated from ground is fixed in the open. Next day, when a person touches the sheet, then he
View answer and explanation
Correct option: A
Experiences electric shock.
The correct answer is A because the person will experience an electric shock due to the accumulation of static charge on the insulated aluminum sheet.
9 Medium
9 of 156
Application +4 | -1
A solid sphere of radius R1{R_1} and volume charge density ρ=ρ0r\rho = \frac{{{\rho _0}}}{r} is enclosed by a hollow sphere of radius R2{R_2} with negative surface charge density σ\sigma, such that the total charge in the system is zero, ρ0{\rho _0} is a positive constant and rr is the distance from the centre of sphere. The ratio R2R1\frac{{{R_2}}}{{{R_1}}} is
View answer and explanation
Correct option: C
ρ0/2σ\sqrt {{\rho _0} /2\sigma}
Option C (ρ0/2σ\sqrt {{\rho _0} /2\sigma}) is correct because it is derived from equating the total positive charge of the sphere to the total negative charge on the hollow sphere to ensure the system's net charge is zero.
10 Easy
10 of 156
Analyzing +4 | -1
Given below are two statements:
Statement I: The electric force changes the speed of the charged particle and hence changes its kinetic energy; whereas the magnetic force does not change the kinetic energy of the charged particle.
Statement II: The electric force accelerates the positively charged particle perpendicular to the direction of electric field. The magnetic force accelerates the moving charged particle along the direction of magnetic field.
In the light of the above statements, choose the most appropriate answer from the options given below:
View answer and explanation
Correct option: C
Statement I is correct but statement II is incorrect
Statement I is correct, as electric force can change kinetic energy, while magnetic force cannot. Statement II is incorrect because electric force accelerates a positive charge *along* the field, and magnetic force accelerates it *perpendicular* to both velocity and field.
11 Medium
11 of 156
Application +4 | -1
A conductor has been given a charge −3×10−7C- 3 \times {10^{ - 7}}C by transferring electrons. Mass increase (in kg) of the conductor and the number of electrons added to the conductor are respectively
View answer and explanation
Correct option: D
2×10−18  and  2×10122 \times {10^{ - 18}}\;\mathrm{and\;}2 \times {10^{12}}
Option D (2×10−18  and  2×10122 \times {10^{ - 18}}\;\mathrm{and\;}2 \times {10^{12}}) is correct. It involves calculating the number of electrons transferred and the corresponding mass increase using the charge and mass of a single electron.
12 Easy
12 of 156
Remembering +4 | -1
Choose the correct relation regarding units of charge (e.s.u. →\rightarrow electrostatic unit)
View answer and explanation
Correct option: B
1 coulomb =3×109=3 \times 10^9 e.s.u.
Option B (1 coulomb =3×109=3 \times 10^9 e.s.u.) is correct because it represents the correct conversion factor between coulombs and electrostatic units.
13 Easy
13 of 156
Understanding +4 | -1
A neutral body is charged positively by rubbing it, its weight
View answer and explanation
Correct option: B
Decreases slightly
Option B (Decreases slightly) is correct because when a neutral body is charged positively, it loses electrons, hence its mass decreases slightly.
14 Medium
14 of 156
Remembering +4 | -1
The magnitude of electric force experienced by
a charged particle in an electric field depends
on:
View answer and explanation
Correct option: A
Charge of the particle
The correct answer is A because the magnitude of the electric force is directly proportional to the charge of the particle, as given by the formula F = qE.
15 Medium
15 of 156
Understanding +4 | -1
Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because the net charge on a neutral conducting sphere remains zero, even when a charge is placed outside. The charges redistribute, but the total charge remains the same.

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16 Medium
16 of 156
Understanding +4 | -1
For a closed surface through which the net flux is zero, each of the following four statements could be true. Which of the statements must be true ? a) There are no charges inside the surface b) The net charge inside the surface is zero c) The electric field is zero everywhere on the surface d) The number of electric field lines entering the surface equals the number leaving the surface
View answer and explanation
Correct option: B
b, d
The correct answer is B because, according to Gauss's law, zero net flux implies zero net charge inside the surface, and the number of electric field lines entering must equal the number leaving.
17 Easy
17 of 156
Understanding +4 | -1
Statement-1: Any net charge of a conductor resides on its outer surface.
Statement-2: The electrostatic energy of a conductor is minimum when charge is spread over the surface.
View answer and explanation
Correct option: B
Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation for Statement-1
Option B is correct because Statement-1 and Statement-2 are both true, but Statement-2 does not directly explain Statement-1. The charge resides on the surface due to minimization of potential energy, but the direct reason is related to the electric field inside the conductor being zero.
18 Easy
18 of 156
Application +4 | -1
Force between two point charges q1q_1 and q2q_2 placed in vacuum at ' rr ' cm\mathrm{cm} apart is FF. Force between them when placed in a medium having dielectric constant K=5K=5 at r/5r / 5  cm\mathrm{~cm} apart will be
View answer and explanation
Correct option: C
5F5 F
Option C (5F5F) is correct because the force is inversely proportional to the square of the distance and directly proportional to the dielectric constant.
19 Easy
19 of 156
Application +4 | -1
In a parallel-plate capacitor with plate area AA and charge Q,Q, the force on the plate because of the charge on the other is equal to
View answer and explanation
Correct option: D
Q22ε0A\frac{Q^2}{2 \varepsilon_0 A}
Option D (Q22ε0A\frac{Q^2}{2 \varepsilon_0 A}) is correct because the force is calculated using the electric field due to one plate acting on the charge of the other.
20 Medium
20 of 156
Application +4 | -1
Three equal charges Q each are placed at the vertices of an equilateral triangle of side ll. How much work has to be done in bringing them closer to an equilateral triangle of side l2\frac{l}{2}.
View answer and explanation
Correct option: B
3Q24πε0l\frac{{3{Q^2}}}{{4\pi {\varepsilon _0}l}}
Option B (3Q24πε0l\frac{{3{Q^2}}}{{4\pi {\varepsilon _0}l}}) is correct because the work done is the change in potential energy of the system, which is calculated by considering the interactions between all pairs of charges.
21 Easy
21 of 156
Remembering +4 | -1
There are two kinds of charges-positive charge and negative charge. The property which differentiates the two kinds of charges is called
View answer and explanation
Correct option: D
polarity of charge
Option D (polarity of charge) is correct because polarity refers to the positive or negative nature of the charge, which distinguishes the two types.
22 Easy
22 of 156
Application +4 | -1
Two charges of 5Q5 Q and −2Q-2 Q are situated at the points (3a,0)(3 a, 0) and (−5a,0)(-5 a, 0) respectively. The electric flux through a sphere of radius '4a4 a' having center at origin is
View answer and explanation
Correct option: D
5Qε0\frac{5 Q}{\varepsilon_0}
Option D (5Qε0\frac{5 Q}{\varepsilon_0}) is correct because only the charge at (3a, 0) is enclosed within the sphere of radius 4a centered at the origin.
23 Medium
23 of 156
Understanding +4 | -1
When a glass rod is rubbed with silk, it
View answer and explanation
Correct option: B
Gives electrons to silk
Option B (Gives electrons to silk) is correct because when glass is rubbed with silk, electrons are transferred from the glass to the silk, making the glass positively charged and the silk negatively charged.
24 Medium
24 of 156
Understanding +4 | -1
One metallic sphere AA is given positive charge whereas another identical sphere BB of exactly same mass as of AA is given equal amount of negative charge. Then
View answer and explanation
Correct option: D
Mass of BB increases
Option D (Mass of BB increases) is correct because when sphere B is given a negative charge, it gains electrons, thus increasing its mass.
25 Hard
25 of 156
Application +4 | -1
A charged particle q1q_1 is at position (2,−1,3)(2,-1,3). The electrostatic force on another charge particle q2q_2 at (0,0,0)(0,0,0) is
View answer and explanation
Correct option: D
q1q25614πε0(−2i^+j^−3k^)\frac{q_1 q_2}{56 \sqrt{14} \pi \varepsilon_0}(-2 \hat{i}+\hat{j}-3 \hat{k})
Option D is correct because it accurately calculates the electrostatic force vector using Coulomb's law in vector form, considering the position vectors of the charges.
26 Medium
26 of 156
Application +4 | -1
An infinite line charge produce a field of 7.182×108  N/C7.182 \times {10^8}\;N/C at a distance of 2  cm.2\;cm. The linear charge density is
View answer and explanation
Correct option: B
7.98×10−4  C/m7.98 \times {10^{ - 4}}\;C/m
Option B (7.98×10−4  C/m7.98 \times {10^{ - 4}}\;C/m) is correct because it is calculated directly from the formula for the electric field due to an infinite line charge.
27 Medium
27 of 156
Remembering +4 | -1
Which of the following is the correct statement of Gauss law for electrostatics in a region of charge distribution in free space
View answer and explanation
Correct option: B
∮E.ds=ρε0\oint E.ds = \frac{\rho }{{{\varepsilon _0}}}
Gauss's law in electrostatics relates the electric flux through a closed surface to the charge density within that surface: ∮E.ds=ρε0\oint E.ds = \frac{\rho }{{\varepsilon _0}}.
28 Medium
28 of 156
Remembering +4 | -1
Which of the following is the correct statement of Gauss law for electrostatics in a region of charge distribution in free space?
View answer and explanation
Correct option: B
∮E.ds=ρε0\oint E.ds = \frac{\rho }{{{\varepsilon _0}}}
Gauss's law in electrostatics relates the electric flux through a closed surface to the charge density within that surface: ∮E.ds=ρε0\oint E.ds = \frac{\rho }{{\varepsilon _0}}.
29 Medium
29 of 156
Application +4 | -1
Two spherical conductors A and B of radii 1mm and 2mm are separated by a distance of 5cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is
View answer and explanation
Correct option: A
4:14:1
Option A (4:14:1) is correct because after connecting the spheres, they have the same potential, and the electric field is inversely proportional to the radius.
30 Hard
30 of 156
Application +4 | -1
A particle of charge −q-q and mass mm moves in a circle of radius rr around an infinitely long line charge of linear charge density +λ.+\lambda. Then time period will be given as:

(Consider kk as Coulomb's constant)
View answer and explanation
Correct option: C
T=2πrm2kλqT=2 \pi r \sqrt{\frac{m}{2 k \lambda q}}
Option C is correct because it accurately applies the concept of centripetal force provided by the electric field of the line charge to derive the time period of the particle's circular motion.
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