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JEE • PHYSICS • ELECTRIC FIELDS AND CHARGES

Electric fields and charges practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 331-360 of 420 questions

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331 Medium
331 of 420
Understanding +4 | -1
Assertion:- Electric force is the reason for existence of molecules.
Reason:- At atomic level electric force is the most effective force or strongest force.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
The correct answer is A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) because electric forces between atoms are responsible for the formation of molecules, and at the atomic level, electric force is the most effective.
332 Medium
332 of 420
Application +4 | -1
Electric field strength due to a point charge of 10  μC10\,\,\mu C at a distance of 80 cm from the charge is [N/C][N/C]
View answer and explanation
Correct option: D
1.4×1051.4 \times {10^5}
Option D (1.4×1051.4 \times {10^5}) is correct because it is the electric field strength calculated using the formula E=14π∈0Qd2E = \frac{1}{{4\pi { \in _0}}}\frac{Q}{{{d^2}}} with the given charge and distance.
333 Medium
333 of 420
Understanding +4 | -1
Assertion: A line of force has sudden breaks.

Reason: An electrostatic line of force is a continuous curve.
View answer and explanation
Correct option: A
Both Assertion & Reason are true. Reason is the correct explanation for assertion
Option D is incorrect. Assertion is false and Reason is true. Electrostatic lines of force are continuous curves without breaks, which contradicts the assertion.
334 Medium
334 of 420
Application +4 | -1
Two charges each ‘Q’ are released when the distance between is ‘d’. Then velocity of each charge of mass ‘m’ each when the distance between them is ‘2d’ is
View answer and explanation
Correct option: A
Q8πε0dm\frac{Q}{{\sqrt {8\pi {\varepsilon _0}dm} }}
Option A (Q8πε0dm\frac{Q}{{\sqrt {8\pi {\varepsilon _0}dm} }}) is correct because it is derived from the conservation of energy, equating initial potential energy to final kinetic and potential energies.
335 Medium
335 of 420
Application +4 | -1
A point charge qq is placed at origin. Let E→A,E→B\overrightarrow{\mathrm{E}}_{\mathrm{A}}, \overrightarrow{\mathrm{E}}_{\mathrm{B}} and E→C\overrightarrow{\mathrm{E}}_{\mathrm{C}} be the electric field at three points A(1,2,3),B(1,1,−1)\mathrm{A}(1,2,3), \mathrm{B}(1,1,-1) and C(2,2,2)\mathrm{C}(2,2,2) due to charge q. Then
View answer and explanation
Correct option: C
∣E→B∣=4∣E→C∣\left|\overrightarrow{\mathrm{E}}_{\mathrm{B}}\right|=4\left|\overrightarrow{\mathrm{E}}_{\mathrm{C}}\right|
Option D (∣E→B∣=4∣E→C∣\left|\overrightarrow{\mathrm{E}}_{\mathrm{B}}\right|=4\left|\overrightarrow{\mathrm{E}}_{\mathrm{C}}\right|) is correct because the electric field magnitude is inversely proportional to the square of the distance from the charge.
336 Medium
336 of 420
Analyzing +4 | -1
Statement-1: Two point charges +Q+Q and −Q-Q are fixed at point A(+a,0,0)A(+a, 0,0) and point B(−a,0,0)B(-a, 0,0) respectively. Then the magnitude of electric flux due to electric field of either point charge through infinite y−zy-z plane (that is x=0x=0 plane) is less than magnitude of net electric flux due to electric field of both charges through that plane (x=0x=0 plane).
Statement-2: The magnitude of net electric flux through a surface due to a system of point charges is equal to sum of magnitude of electric flux through that surface due to each of the point charge of the system.
View answer and explanation
Correct option: C
Statement-1 is true, Statement-2 is false
Option C is correct because Statement-1 is true (flux due to individual charges is less than the net flux), but Statement-2 is false (flux is a vector quantity, so magnitudes don't simply add).
337 Medium
337 of 420
Understanding +4 | -1
Eight dipoles of charges of magnitude ee are placed inside a cube. The total electric flux coming out of the cube will be
View answer and explanation
Correct option: D
ZeroZero
Option D (Zero) is correct because the net charge of a dipole is zero, and the total electric flux is proportional to the enclosed charge.
338 Medium
338 of 420
Application +4 | -1
A free alpha particle and a free proton are separated by distance of 10−10 m10^{-10} \mathrm{~m} and are released. The kinetic energy of alpha particle when at infinite separation is
View answer and explanation
Correct option: B
9.2×10−19 J9.2 \times 10^{-19} \mathrm{~J}
Option B (9.2×10−19 J9.2 \times 10^{-19} \mathrm{~J}) is correct because it correctly calculates the kinetic energy of the alpha particle at infinite separation using conservation of energy and momentum.
339 Medium
339 of 420
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 the wrong statement because nuclear force, not Coulomb's law, binds protons and neutrons in the nucleus.
340 Medium
340 of 420
Analyzing +4 | -1
Assertion: With the help of Gauss's theorem we can find electric field at any point.

Reason: Gauss's theorem can be applied for any type of charge distribution.
View answer and explanation
Correct option: C
If Assertion is true but Reason is false.
Option C is correct. Gauss's theorem can be used to find the electric field at points with sufficient symmetry, but it is not applicable for all charge distributions.
341 Medium
341 of 420
Analyzing +4 | -1
A positively charged rod is brought near to a nonconducting object. The rod is attracted to the object. From this observation we can't predict whether the object is charged or uncharged. Some additional experiments have been performed to identify whether the object is charged or uncharged, these experiments with their observations and conclusions drawn are given. Select the option in which conclusion drawn is correct.
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because each scenario provides a valid conclusion based on the principles of electrostatic attraction and repulsion.
342 Medium
342 of 420
Analyzing +4 | -1
Assertion: If a charged particle is released from rest in a region of uniform electric and magnetic fields parallel to each other, it will move in a straight line.
Reason: The electric field exerts no force on the particle but the magnetic field does.
View answer and explanation
Correct option: C
If assertion is true but reason is false
Option D (If both assertion and reason are false) is correct because the electric field exerts a force, and the magnetic field exerts no force since they are parallel.
343 Medium
343 of 420
Understanding +4 | -1
A cube of metal is given a positive charge Q.Q. For the above system, which of the following statements is true
View answer and explanation
Correct option: C
Electric field is normal to the surface of the cube
Option C (Electric field is normal to the surface of the cube) is correct because electric fields are always perpendicular to the surface of a conductor in electrostatic equilibrium.
344 Medium
344 of 420
Understanding +4 | -1
Assertion:- Charge is invariant.
Reason:- Charge does not depend on speed or frame of reference.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A is correct because both the Assertion and Reason are true, and the Reason correctly explains why charge is invariant.
345 Easy
345 of 420
Understanding +4 | -1
Two conducting spheres of radii r1r_1 and r2r_2 are charged to equal surface charge density. The ratio of electric fields just outside the surface is:
View answer and explanation
Correct option: D
1:11: 1
Option D (1:11: 1) is correct because the electric field just outside a charged conductor depends only on the surface charge density and the permittivity of free space, and since the surface charge densities are equal, the electric fields are equal.
346 Medium
346 of 420
Understanding +4 | -1
When a body is charged by conduction, its mass
View answer and explanation
Correct option: D
increase or decrease
Option D is correct. When a body is charged by conduction, it can either gain electrons (increasing mass) or lose electrons (decreasing mass), so the mass can increase or decrease.
347 Medium
347 of 420
Understanding +4 | -1
The angle between the electric lines of force and the equipotential surface is:
View answer and explanation
Correct option: D
90∘\mathrm{90}^\circ
Option D (90∘\mathrm{90}^\circ) is correct because electric field lines are always perpendicular to equipotential surfaces.
348 Medium
348 of 420
Remembering +4 | -1
The charge on an electron was calculated by:
View answer and explanation
Correct option: C
Millikan
Option C (Millikan) is correct because Millikan's oil drop experiment determined the charge of an electron.
349 Medium
349 of 420
Understanding +4 | -1
A hollow cylinder of length LL and radius RR having surface area AA is placed horizontally with its axis parallel to an external field E.{E}. The electric flux through the surface of the cylinder is
View answer and explanation
Correct option: D
E.AE.A
Option D (E.AE.A) is correct because the electric flux is the dot product of the electric field and the area vector.
350 Medium
350 of 420
Application +4 | -1
A charged spherical conductor has a surface density of 0.7 Cm−20.7 \mathrm{~Cm}^{-2}. When its charge is increased by 0.44 C0.44 \mathrm{~C}, the charge density changes by 0.14Cm−20.14 \mathrm{Cm}^{-2}. Find the radius of the sphere and initial charge on it.
View answer and explanation
Correct option: A
2.2 C2.2 \mathrm{~C} , 0.5 m0.5 \mathrm{~m}
Option A (2.2 C2.2 \mathrm{~C} , 0.5 m0.5 \mathrm{~m}) is correct because the radius and initial charge are calculated using the given surface charge densities and the change in charge.
351 Medium
351 of 420
Understanding +4 | -1
A particle of ′m′'m' and charge ′q′'q' is accelerated through a potential difference of V  volt,V\;volt, its energy will be
View answer and explanation
Correct option: A
qVqV
Option A (qVqV) is correct because the energy gained by a charge accelerated through a potential difference is equal to the charge times the potential difference.
352 Medium
352 of 420
Understanding +4 | -1
A cylinder of radius R and length L is placed in the uniform electric field E to the cylinder axis. The total flux for the surface of the cylinder is given by
View answer and explanation
Correct option: D
zero
Option D is correct because the net electric flux through any closed surface in a uniform electric field is always zero.
353 Medium
353 of 420
Application +4 | -1
Two identical charged spheres are suspended by strings of equal lengths. The stings make an angle of 30o{30^\mathrm{o}} with each other. When suspended in a liquid of density 0.8g  cm−3,0.8g\;c{m^{ - 3}}, the angle remains the same. If density of the material of the sphere is 1.6g  cm−3,1.6g\;c{m^{ - 3}}, the dielectric constant of the liquid is
View answer and explanation
Correct option: D
2
Option D (2) is correct because the dielectric constant is calculated using the ratio of the density of the sphere to the difference between the density of the sphere and the density of the liquid.
354 Medium
354 of 420
Understanding +4 | -1
Assertion:- Under electrostatic conditions net electric field inside a solid conductor will be zero.
Reason:- Under electrostatic conditions there will be no free electrons inside a conductor.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C is correct because the assertion is true (electric field inside a conductor is zero under electrostatic conditions), but the reason is false (there are free electrons inside a conductor).
355 Medium
355 of 420
Application +4 | -1
A non-uniformly charged sphere has charge density given by ρ=a2R\rho=\frac{a^2}{R} where 'aa' is a constant. 'rr' is the distance from the centre and 'RR' is the radius of the sphere. Then the total charge inside the sphere is proportional to
View answer and explanation
Correct option: D
R4{R}^4
Option D (R4{R}^4) is correct because the total charge inside the sphere is proportional to R4{R}^4 after integrating the charge density over the volume of the sphere.
356 Medium
356 of 420
Application +4 | -1
Assuming that the charge on an atom is distributed uniformly in a sphere of radius 10−10 m10^{-10} \mathrm{~m}, what will be the electric field at the surface of the gold atom? For gold, Z=79\mathrm{Z}=79.
View answer and explanation
Correct option: B
1.138×1013 NC−11.138 \times 10^{13} \mathrm{~NC}^{-1}
Option B (1.138×1013 NC−11.138 \times 10^{13} \mathrm{~NC}^{-1}) is correct because it is the electric field calculated using Coulomb's law with the given charge and radius.
357 Medium
357 of 420
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}}}
Option B is correct. Gauss's law states that the electric flux through a closed surface is equal to the enclosed charge density divided by the permittivity of free space: ∮E.ds=ρ{ε0\oint E.ds = \frac{\rho}{{\{\varepsilon _0}}.
358 Medium
358 of 420
Application +4 | -1
An electron of mass 'mm' charge 'ee' falls through a distance 'hh' metre in a uniform electric field 'EE'. Then, time of fall:
View answer and explanation
Correct option: A
t=2hmeE{t}=\sqrt{\frac{2 {hm}}{{eE}}}
Option A (t=2hmeE{t}=\sqrt{\frac{2 {hm}}{{eE}}}) is correct because the time of fall is derived from kinematics and the electrostatic force acting on the electron.
359 Medium
359 of 420
Application +4 | -1
The electric flux from a cube of edge ll is ϕ\phi in an enclosed charge. If the edge of the cube is made 23l\frac{2}{3} l and the charge enclosed in the cube is doubled, then the electric flux value will be
View answer and explanation
Correct option: B
2ϕ2 \phi
Option B (2ϕ2 \phi) is correct because the electric flux is directly proportional to the enclosed charge, and doubling the charge doubles the flux.
360 Medium
360 of 420
Understanding +4 | -1
Units of electric flux are
View answer and explanation
Correct option: C
volt-m
Option C (volt-m) is correct because electric flux is the product of electric field (V/m) and area (m²), resulting in units of volt-m.
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