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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 361-390 of 420 questions

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361 Medium
361 of 420
Application +4 | -1
A particle of mass m\text{m} and charge q\text{q} travelling with a velocity v\text{v} along the X-axis enters a uniform electric field E→\overrightarrow{\mathrm{E}} directed along the Y\mathrm{Y}-axis. What will be the trajectory of the particle?
View answer and explanation
Correct option: C
Parabolic
Option C (Parabolic) is correct because the particle experiences constant acceleration in the Y-direction while maintaining constant velocity in the X-direction, resulting in a parabolic trajectory.
362 Hard
362 of 420
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 (T=2πrm2kλqT=2 \pi r \sqrt{\frac{m}{2 k \lambda q}}) is correct because it correctly equates the centripetal force with the electrostatic force and solves for the time period.
363 Medium
363 of 420
Analyzing +4 | -1
A neutron, a proton and an electron are placed in a uniform electric field.
View answer and explanation
Correct option: D
Magnitude of acceleration will be different.
Option D (Magnitude of acceleration will be different) is correct because the forces on the proton and electron are equal in magnitude but opposite in direction, and their masses are different, leading to different accelerations.
364 Medium
364 of 420
Understanding +4 | -1
A positively charged ball hangs from a silk thread. We put positive test charge q0q_0 at a point and measure F/q0,F / q_0, then it can be predicted that the electric field strength E{E}
View answer and explanation
Correct option: A
>F/q0> F / {q}_0
Option A (>F/q0> F / {q}_0) is correct because the introduction of the positive test charge causes a redistribution of charge on the ball, reducing the measured force.
365 Medium
365 of 420
Application +4 | -1
A charge QQ is divided into two parts of qq and Q−qQ - q If the coulomb repulsion between them when they are separated is to be maximum, the ratio of Qq\frac{Q}{q} should be
View answer and explanation
Correct option: A
2
Option A (2) is correct because the Coulomb repulsion is maximized when the charges are divided equally, leading to a ratio of 2.
366 Medium
366 of 420
Understanding +4 | -1
Assertion(A): In an uniform magnetic field,
speed and kinetic energy remains the same
for a moving charged particle.

Reason (R): In the case of motion of a charge
in a magnetic field, the magnetic force is
perpendicular to the velocity of the particle.
So, no work is done and no change in the
magnitude of the velocity (speed) and
kinetic energy is produced.
View answer and explanation
Correct option: A
Both A and R are true and R is the
correct explanation of A
Option A is correct because the magnetic force on a moving charged particle in a uniform magnetic field is perpendicular to its velocity, resulting in no work done and constant speed and kinetic energy.
367 Medium
367 of 420
Understanding +4 | -1
Electric charges are distributed in a small volume. The flux of the electric field through a splerical surface of radius 1 m1 \mathrm{~m} surrounding the total charge is 100 V−m100 \mathrm{~V}-\mathrm{m}. The flux over the concentric sphere of radius 2 m2 \mathrm{~m} will be:
View answer and explanation
Correct option: C
100 V-m
Option C is correct. According to Gauss's law, the electric flux through a closed surface depends only on the enclosed charge, not the radius of the surface.
368 Medium
368 of 420
Understanding +4 | -1
An electron enters in an electric field with its velocity in the direction of the electric lines of force. Then
View answer and explanation
Correct option: C
The velocity of the electron will decrease
Option C (The velocity of the electron will decrease) is correct because the electric field exerts a force on the electron opposite to its direction of motion, causing it to decelerate.
369 Medium
369 of 420
Analyzing +4 | -1
A particle of mass M and charge q is released from rest in a region of uniform electric field of magnitude E. After a time t, the distance travelled by the charge is S and the kinetic energy attained by the particle is T. Then, the ratio TS:\frac{T}{S}:
View answer and explanation
Correct option: A
remains constant with time t
Option A is correct because the ratio T/S remains constant with time t, as both kinetic energy and distance traveled are proportional to t2.
370 Medium
370 of 420
Remembering +4 | -1
Flux coming out from a unit positive charge enclosed in air is
View answer and explanation
Correct option: B
(ε0)−1{\left( {{\varepsilon _0}} \right)^{ - 1}}
Option B ((ε0)−1{\left( {{\varepsilon _0}} \right)^{ - 1}}) is correct because Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space.
371 Medium
371 of 420
Analyzing +4 | -1
For a uniformly charged ring of radius R,R, the electric field on its axis has the largest magnitude at a distance hh from its centre. Then, value of hh is
View answer and explanation
Correct option: A
R2\frac{R}{{\sqrt 2 }}
Option A (R2\frac{R}{{\sqrt 2 }}) is correct because the electric field is maximum at this distance, which is found by differentiating the electric field equation and setting it to zero.
372 Hard
372 of 420
Analyzing +4 | -1
Given below are two statements:
Statement I An electric diploe is placed at the centre of a hollow sphere. The flux of electric field through the sphere is zero but the electric field is not zero anywhere in the sphere.
Statement II If RR is the radius of a solid metallic sphere and QQ be the total charge on it. The electric field at any point on the spherical surface of radius r(<R)r( < R) is zero but the electric flux passing through this closed spherical surface of radius rr is not zero.
In the light of the above statements, choose the correct answer from the options given below.
View answer and explanation
Correct option: B
Statement I is true but Statement II is false.
Option B is correct because Statement I is true (flux through the sphere is zero due to the dipole's net charge being zero), but Statement II is false (electric field inside a charged solid sphere is non-zero).
373 Medium
373 of 420
Analyzing +4 | -1
Assertion: If an electron and proton possessing same kinetic energy an electric field in a perpendicular direction, the path of the electron is more curved than that of the proton

Reason: Electron forms a large curve due to its small mass
View answer and explanation
Correct option: D
Assertion is false and Reason is false
Option D is correct because both the assertion and reason are false; the radius of curvature depends on the charge and electric field, not just the mass.
374 Medium
374 of 420
Application +4 | -1
An infinite number of point charges of same magnitude are kept along XX-axis at x=1 cm,xx=1 \mathrm{~cm}, x =2 cm,x=4 cm,x=8 cm,…=2 \mathrm{~cm}, x=4 \mathrm{~cm}, x=8 \mathrm{~cm}, \ldots and so on. If the magnitude of charge is 5 nC5 \mathrm{~nC} and consecutive charges are of opposite sign, the magnitude of electric field at x=0x=0 is
View answer and explanation
Correct option: D
36×104 N/C36 \times 10^4 \mathrm{~N} / \mathrm{C}
Option D (36×104 N/C36 \times 10^4 \mathrm{~N} / \mathrm{C}) is correct because it correctly sums the electric field contributions from each charge, considering alternating signs.
375 Medium
375 of 420
Understanding +4 | -1
Van de Graaff electrostatic generator is based on:
View answer and explanation
Correct option: D
both (1) and (2)
Option D (both (1) and (2)) is correct because the Van de Graaff generator works based on corona discharge and the principle that charge resides on the outer surface of a conductor.
376 Medium
376 of 420
Application +4 | -1
A point charge q produces an electric field of magnitude 2NC−12\mathrm{ }N{C^{ - 1}} at a point distance 0.25 m from it. What is the value of charge?
View answer and explanation
Correct option: A
1.39×10−11  \times {10^{ - 11\;}}C
Option A (1.39 ×10−11  \times {10^{ - 11\;}}C) is correct because it is calculated using the formula for electric field due to a point charge.
377 Medium
377 of 420
Understanding +4 | -1
Assertion: Electric lines of force never cross each other.
Reason: Electric field at a point superimpose to give one resultant electric field
In above question a statement of assertion is followed by statement of reason mark the correct choice.
View answer and explanation
Correct option: B
If both assertion and reason are true and reason is not the correct explanation of assertion
Option B is correct because both the assertion and the reason are true, but the reason doesn't directly explain why electric field lines don't cross. The non-crossing is due to the uniqueness of the electric field direction at any point.
378 Easy
378 of 420
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 it applies Gauss's law to find the electric flux through the sphere, considering only the charge enclosed within the sphere.
379 Medium
379 of 420
Understanding +4 | -1
The electrostatic field lines
View answer and explanation
Correct option: D
Do not form closed loops
Option D is correct because electrostatic field lines originate from positive charges and terminate at negative charges, never forming closed loops.
380 Medium
380 of 420
Understanding +4 | -1
A ring of radius rr carries a charge QQ uniformly distributed over its length. A charge qq is placed at its centre will experience a force equal to
View answer and explanation
Correct option: C
Zero
Option C (Zero) is correct because the electric field at the center of a uniformly charged ring is zero, resulting in no force on the charge placed there.
381 Easy
381 of 420
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 the property that distinguishes positive and negative charges is their polarity.
382 Medium
382 of 420
Understanding +4 | -1
There is 10 units of charge at the centre of a circle of radius 10m.10m. The work done in moving 1 unit of charge around the circle once is
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because the work done in moving a charge around a closed path on an equipotential surface is zero.
383 Medium
383 of 420
Understanding +4 | -1
Which is the incorrect statement regarding
electric lines of force?
View answer and explanation
Correct option: D
Electric lines of force are always parallel to
equipotential surface
Option D (Electric lines of force are always parallel to equipotential surface) is correct because electric field lines are always perpendicular to equipotential surfaces.
384 Medium
384 of 420
Application +4 | -1
Two isolated metallic solid spheres of radii RR and 2R2 R are charged such that both have same charge density σ\sigma. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is σ′\sigma^{\prime}. The ratio σ′σ\frac{\sigma^{\prime}}{\sigma} is
View answer and explanation
Correct option: A
56\frac{5}{6}
Option A (56\frac{5}{6}) is correct because the new charge density is calculated by considering the total charge and the new surface area of the larger sphere.
385 Medium
385 of 420
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 from the charge.
386 Medium
386 of 420
Understanding +4 | -1
What happens when some charge is placed on a soap bubble?
View answer and explanation
Correct option: B
Its radius increases
Option B (Its radius increases) is correct because the charge distributes evenly on the surface, creating an outward electric force that expands the bubble.
387 Medium
387 of 420
Understanding +4 | -1
Assertion: Work done by electric force is path independent.

Reason: Electric force is conservative.
View answer and explanation
Correct option: A
Both Assertion & Reason are true. Reason is the correct explanation for assertion
Option A is correct because the assertion that work done by an electric force is path independent is true, and the reason that electric force is conservative is the correct explanation for it.
388 Medium
388 of 420
Understanding +4 | -1
Which of the following statement is correct?
View answer and explanation
Correct option: B
Electric field lines may form closed loop
Option B (Electric field lines may form closed loop) is correct because induced electric fields can form closed loops.
389 Medium
389 of 420
Understanding +4 | -1
A charge QQ is uniformly distributed over a large square plate of copper. The electric field at a point very close to the centre of the plate is 10 V/m10 \mathrm{~V} / \mathrm{m}. If the copper plate is replaced by a plastic plate of the same geometrical dimensions and carrying the same charge Q uniformly distributed, then the electric field at the point PP will be :-
View answer and explanation
Correct option: C
10 V/m10 \mathrm{~V} / \mathrm{m}
Option C (10 V/m10 \mathrm{~V} / \mathrm{m}) is correct because the electric field due to a uniformly charged plate depends only on the surface charge density and is independent of the material of the plate.
390 Medium
390 of 420
Application +4 | -1
Find the ratio of the magnitude of the electric force to the gravitational force acting between two protons.
View answer and explanation
Correct option: A
1.24×10361.24 \times 10^{36}
Option A (1.24×10361.24 \times 10^{36}) is correct because it is the ratio of electric force to gravitational force calculated using the respective formulas.
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