Skip to main content
Back to ELECTRIC FIELDS AND CHARGES

Your Progress

0 / 92 Answered
0
Correct
0
Incorrect
0%
Accuracy
NEET • PHYSICS • ELECTRIC FIELDS AND CHARGES

Electric fields and charges practice · Medium

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

The practice player uses single-correct-answer MCQs. Chapter quizzes help you review a topic; they are not full-length NEET mock papers.

Take a chapter quiz →
Showing 1-30 of 92 questions

Preparing practice and saved progress…

1 Medium
1 of 92
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 92
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 Medium
3 of 92
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.
4 Medium
4 of 92
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.
5 Medium
5 of 92
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.
6 Medium
6 of 92
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.
7 Medium
7 of 92
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.
8 Medium
8 of 92
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.
9 Medium
9 of 92
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.
10 Medium
10 of 92
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.
11 Medium
11 of 92
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.
12 Medium
12 of 92
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.
13 Medium
13 of 92
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}}.
14 Medium
14 of 92
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}}.
15 Medium
15 of 92
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.

Explore Other Chapters in PHYSICS

View All →
UNITS & MEASUREMENTS 392 MCQs WORK-ENERGY-POWER 355 MCQs MECHANICAL PROPERTIES OF FLUIDS 317 MCQs MOTION IN PLANE 307 MCQs MAGNETIC EFFECT OF CURRENT 282 MCQs THERMAL PROPERTIES OF MATTER 269 MCQs
16 Medium
16 of 92
Understanding +4 | -1
The Gaussian surface for calculating the electric field due to a charge distribution is
View answer and explanation
Correct option: C
A symmetrical closed surface containing the charge distribution, at every point of which electric field has a single fixed value
Option A symmetrical closed surface containing the charge distribution, at every point of which electric field has a single fixed value is the correct answer because it accurately describes the requirements for a Gaussian surface.
17 Medium
17 of 92
Analyzing +4 | -1
Assertion (A): If two spherical conductors of different radii have the same surface charge densities, then their electric field intensities will be equal.
Reason (R): Surface charge density =Total  chargearea= \frac{{\mathrm{Total}\,\,\mathrm{charge}}}{{\mathrm{area}}}
View answer and explanation
Correct option: B
Both A and R are true but R is NOT the correct explanation of A
Option B is correct because if two spherical conductors have the same surface charge density, their electric field intensities at the surface will be equal. The reason correctly defines surface charge density but doesn't explain the equality of electric fields.
18 Medium
18 of 92
Remembering +4 | -1
Law stating that “force is directly proportional to product of charges and inversely proportional to square of separation between them” is called
View answer and explanation
Correct option: B
Coulomb’s law
Option B (Coulomb’s law) is correct because Coulomb's law describes the electrostatic force between charges as directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
19 Medium
19 of 92
Understanding +4 | -1
Two charged spheres of radii 10cm10cm and 15cm15cm are connected by a thin wire. No charge will flow, if they have
View answer and explanation
Correct option: B
The same potential
Option B is correct because charge flow stops when the spheres reach the same potential, ensuring electrostatic equilibrium.
20 Medium
20 of 92
Analyzing +4 | -1
A hemisphere is uniformly charged positively. The electric field at a point on a diameter away from the centre is directed
View answer and explanation
Correct option: A
perpendicular to the diameter
Option A (perpendicular to the diameter) is correct due to the symmetry of the charge distribution in the hemisphere.
21 Medium
21 of 92
Analyzing +4 | -1
A charge +q+ q is fixed to each of three corners of a square on the empty corner a charge QQ is placed such that there is no net electrostatic force acting is no net electrostatic force acting on the diagonally opposite charge. Then
View answer and explanation
Correct option: B
Q=−22qQ = - 2\sqrt {2q}
Option B is correct because it correctly calculates the charge Q required to balance the electrostatic forces on the diagonally opposite charge.
22 Medium
22 of 92
Remembering +4 | -1
The electrostatic force between two point charges is directly proportional to the
View answer and explanation
Correct option: D
Product of the charges
Option D (Product of the charges) is correct because Coulomb's law states that the electrostatic force is directly proportional to the product of the charges.
23 Medium
23 of 92
Remembering +4 | -1
One of the following is not a property of field lines
View answer and explanation
Correct option: D
They form closed loop
Option D is correct because electric field lines do not form closed loops; they originate from positive charges and terminate at negative charges.
24 Medium
24 of 92
Understanding +4 | -1
Identify the WRONG statement
View answer and explanation
Correct option: B
A charged particle free to move in an electric field shall always move in the direction of E⃗\vec E
Option B is the WRONG statement because a charged particle will only move in the direction of the electric field if it's a positive charge and initially at rest. A negative charge will move opposite to the field, and if the particle has an initial velocity, its path will be curved.
25 Medium
25 of 92
Analyzing +4 | -1
A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral.
View answer and explanation
Correct option: A
A potential difference appears between the two cylinders when a charge density is given to the inner cylinder
Option A (A potential difference appears between the two cylinders when a charge density is given to the inner cylinder) is correct because the electric field created by the inner cylinder will cause a potential difference between the cylinders.
26 Medium
26 of 92
Analyzing +4 | -1
Two conducting spheres of radii R1{R_1} and R2{R_2} having charges Q1{Q_1} and Q2{Q_2} respectively are connected to each other. There is
View answer and explanation
Correct option: D
A decrease in the energy of the system unless  Q1R2=Q2R1{Q_1}{R_2} = {Q_2}{R_1}
Option D (A decrease in the energy of the system unless Q1R2=Q2R1{Q_1}{R_2} = {Q_2}{R_1}) is correct because charge redistribution leads to energy loss unless the initial potentials are equal.
27 Medium
27 of 92
Analyzing +4 | -1
Assertion(A): If a bob of a simple pendulum is kept in a horizontal electric field, its period of oscillation will remain same.

Reason(R): If bob is charged and kept in horizontal electric field, then the time period will be decreased.
View answer and explanation
Correct option: B
If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
Both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion. The period remains the same if only gravity acts, but if the bob is charged, the effective 'g' changes, thus changing the period.
28 Medium
28 of 92
Application +4 | -1
LetP(r)=QπR4rP\left( r \right) = \frac{Q}{{\pi {R^4}}}r be the charge density distribution for a solid sphere of radius RR and total charge Q.Q.. For a point ′p′'p' inside the sphere at distance r1{r_1} from the centre of the sphere, the magnitude of electric field is
View answer and explanation
Correct option: C
Qr124π∈0R4\frac{{Qr_1^2}}{{4\pi { \in _0}{R^4}}}
Option C is correct because the electric field inside the sphere is calculated using Gauss's law, considering the charge enclosed within the radius r1.
29 Medium
29 of 92
Application +4 | -1
The distance between a proton and electron both having a charge 1.6×10−19coulomb,1.6 \times {10^{ - 19}}coulomb, , of a hydrogen atom is 10−10metre.{10^{ - 10}}metre.. The value of intensity of electric field produced on electron due to proton will be
View answer and explanation
Correct option: D
1.44×1011N/C1.44 \times {10^{11}}N/C
Option D is correct because the electric field is calculated using Coulomb's law for the electric field due to a point charge.
30 Medium
30 of 92
Understanding +4 | -1
Identify the wrong statement in the following. Coulomb’s law correctly describes the electric force that
View answer and explanation
Correct option: B
Binds the protons and neutrons in the nucleus of an atom
Option B (Binds the protons and neutrons in the nucleus of an atom) is correct because Coulomb's law describes the electric force between charged particles, but the force binding protons and neutrons in the nucleus is the strong nuclear force.
Revision notes and key takeaways

Review ELECTRIC FIELDS AND CHARGES in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

  • ✓ This set contains 92 single-correct-answer MCQs on ELECTRIC FIELDS AND CHARGES.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.
Showing 1-30 of 92 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ NEET PHYSICS - ELECTRIC FIELDS AND CHARGES MCQs (All) 156 MCQs → NEET PHYSICS - ELECTRIC FIELDS AND CHARGES MCQs (Easy) 51 MCQs → NEET PHYSICS - ELECTRIC FIELDS AND CHARGES MCQs (Hard) 13 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ UNITS & MEASUREMENTS 392 MCQs → WORK-ENERGY-POWER 355 MCQs → MECHANICAL PROPERTIES OF FLUIDS 317 MCQs → MOTION IN PLANE 307 MCQs

Your practice

0 / 92 answered · 92 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →