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JEE • PHYSICS • ELECTRIC FIELDS AND CHARGES

Electric fields and charges practice · Medium

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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 361-385 of 385 questions
361 Medium
361 of 385
Application +4 | -1
The ratio of Coulomb's electrostatic force to the gravitational force between an electron and a proton separated by some distance is 2.4×10392.4 \times 10^{39}. The ratio of the proportionality constant, K=14πε0K=\frac{1}{4 \pi \varepsilon_0} to the Gravitational constant GG is nearly

(Given that the charge of the proton and electron each =1.6×10−19C=1.6 \times 10^{-19} \mathrm{C}, the mass of the electron =9.11×10−31 kg=9.11 \times 10^{-31} \mathrm{~kg}, the mass of the proton =1.67×10−27 kg=1.67 \times 10^{-27} \mathrm{~kg})
View answer and explanation
Correct option: B
102010^{20}
Option B (102010^{20}) is the correct order of magnitude for the ratio of the proportionality constant K to the gravitational constant G, based on the given ratio of electrostatic and gravitational forces.
362 Medium
362 of 385
Remembering +4 | -1
The electric charge always resides:
View answer and explanation
Correct option: C
On the outer surface of charged conductor
Option C (On the outer surface of charged conductor) is correct because charges reside on the surface of a conductor to minimize repulsion and achieve electrostatic equilibrium.
363 Medium
363 of 385
Understanding +4 | -1
If the charge on an object is halved then electric field becomes:
View answer and explanation
Correct option: A
half
Option A is correct because the electric field is directly proportional to the charge on the object. Therefore, if the charge is halved, the electric field is also halved.
364 Medium
364 of 385
Analyzing +4 | -1
Assertion: At a point in space, the electric field points towards north. In the region, surrounding this point the rate of change of potential will be zero along the east and west.

Reason: Electric field due to a charge is the space around the charge.
View answer and explanation
Correct option: B
Both Assertion & Reason are true. Reason is not the correct explanation for assertion
Option B is correct because both the assertion and reason are true, but the reason doesn't directly explain the assertion.
365 Medium
365 of 385
Understanding +4 | -1
Consider two concentric spherical surface S1S_1 with radius aa and S2S_2 within radius 2a2a, both centeredon the origin. There is a charger +q+ q at the origin, and no other charges. Compare the flux ϕ1\phi 1 through S1S_1 with the flux ϕ2\phi 2 through S2S_2
View answer and explanation
Correct option: C
ϕ1=ϕ2{\phi _1} = {\phi _2}
Option C (ϕ1=ϕ2{\phi _1} = {\phi _2}) is correct because Gauss's law states that the electric flux through a closed surface depends only on the enclosed charge.
366 Medium
366 of 385
Analyzing +4 | -1
Given below are two statements
Statement-I: A point charge is brought in an electric field. The value of electric field at a point near to the charge may increase if the charge is positive.
Statement-II: An electric dipole is placed in a non-uniform electric field. The net electric force on the dipole will not be zero.
Choose the correct answer from the options given below
View answer and explanation
Correct option: A
Both statement-I and statement-II are true
Option A is correct because both statements are true. A positive charge increases the electric field strength nearby, and a non-uniform field exerts a net force on a dipole.
367 Medium
367 of 385
Analyzing +4 | -1
A body of mass M and charge q is connected to a spring of spring constant k. It is oscillating along x-direction about its equilibrium position, taken to be at x = 0, with an amplitude A. An electric field E is applied along the x-direction. Which of the following statements is correct?
View answer and explanation
Correct option: A
The total energy of the system is 12mω2A2+12q2E2k\frac{1}{2} m \omega^2 A^2+\frac{1}{2} \frac{q^2 E^2}{k}.
Option A is correct because the total energy includes the potential energy due to the spring and the additional potential energy due to the electric field's effect on the charge.
368 Medium
368 of 385
Understanding +4 | -1
A sphere encloses an electric dipole with charge ±3×10−6C.\pm 3 \times 10^{-6} \mathrm{C}. What is the total electric flux across the sphere?
View answer and explanation
Correct option: B
Zero
Option B (Zero) is correct because the net charge enclosed by the sphere is zero due to the dipole.
369 Medium
369 of 385
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 (Qr124π∈0R4\frac{{Qr_1^2}}{{4\pi { \in _0}{R^4}}}) is correct because it is derived from Gauss's law, considering the charge enclosed within the sphere of radius r1r_1.
370 Medium
370 of 385
Remembering +4 | -1
The force on qq charge is placed in an uniform electric field EE will be
View answer and explanation
Correct option: B
qE{qE}
Option B (qE{qE}) is correct because the force on a charge in an electric field is given by the product of the charge and the electric field.
371 Medium
371 of 385
Understanding +4 | -1
If same charge qq is placed inside a sphere and cube hacing
radius 1 m and side 2 m respectively. What will be the ratio of
flux passing through them?
View answer and explanation
Correct option: A
1:11:1
Option A (1:11:1) is correct because Gauss's law states that the electric flux depends only on the enclosed charge, not the shape of the surface.
372 Medium
372 of 385
Understanding +4 | -1
An electron is placed at the centre of a conducting sphere of radius 0.2  0.2\,\, metre having a charge 5×10−25 \times {10^{ - 2}} coulomb. The force on the electron
View answer and explanation
Correct option: C
0
Option C (0) is correct because the electric field inside a conductor is zero, hence the force on the electron is zero.
373 Medium
373 of 385
Application +4 | -1
Point charge 5q5 q is placed inside a cube at centre, then electrostatic flux passing through one face of cube is
View answer and explanation
Correct option: C
5q6ε0\frac{5 q}{6 \varepsilon_0}
Option C (5q6ε0\frac{5 q}{6 \varepsilon_0}) is correct because it represents the electric flux passing through one face of the cube when a charge is placed at the center.
374 Medium
374 of 385
Understanding +4 | -1
The force of repulsion between two electrons at a certain distance is FF. The force between two protons separated by the same distance is: (mp=1836me)\left(m_p=1836 {m}_e\right)
View answer and explanation
Correct option: B
FF
Option B (FF) is correct because the electrostatic force depends on the magnitude of the charges and the distance between them, and both electrons and protons have the same magnitude of charge.
375 Medium
375 of 385
Understanding +4 | -1
Which of the following items is used in the
household wirings to prevent accidental fire in
case of short circuit?
View answer and explanation
Correct option: D
Electric fuse
Option D (Electric fuse) is correct because an electric fuse is designed to break the circuit in case of a short circuit or overload, preventing fire.
376 Medium
376 of 385
Understanding +4 | -1
The magnitude of electric field EE in the annular region of a charged cylindrical capacitor
View answer and explanation
Correct option: C
Varies as 1/r,1/r, where r is the distance from the axis
Option C (Varies as 1/r,1/r, where r is the distance from the axis) is correct because the electric field in the annular region of a cylindrical capacitor is inversely proportional to the distance from the axis.
377 Medium
377 of 385
Application +4 | -1
A charge having e/me/m equal to 108   c/kg{10^8}\,\,\,c/kg and with velocity 3×105   m/s3 \times {10^5}\,\,\,m/s enters into a uniform magnetic field B=0.4   teslaB = 0.4\,\,\,{\text{tesla}} at an angle 30∘30^\circ with direction of field. The radius of curvature will be
View answer and explanation
Correct option: A
3.75×10−3   m3.75 \times {10^{ - 3}}\,\,\,m
Option A (15.75 cm) is correct because it calculates the new null point distance when the magnet is rotated, using the relationship between axial and equatorial magnetic fields.
378 Medium
378 of 385
Understanding +4 | -1
The electric charge in uniform motion
produces
View answer and explanation
Correct option: C
Both electric and magnetic fields
Option C is correct because a charge in uniform motion creates both an electric field (due to the charge itself) and a magnetic field (due to the moving charge, which constitutes a current).
379 Medium
379 of 385
Understanding +4 | -1
When a proton is accelerated through 1V,1V, then its kinetic energy will be
View answer and explanation
Correct option: C
1  eV1\;eV
Option C is correct because the kinetic energy gained by a proton accelerated through a potential difference of 1V is equal to the charge of the proton (e) times the potential difference (1V), which is 1 eV.
380 Medium
380 of 385
Analyzing +4 | -1
Point charges +4q,  −q  and+4q+ 4q,\,\, - q\;\mathrm{ and } + 4q are kept on the x−axis\mathrm{x - axis} at points x=0,  x=a  and   x=2ax = 0,\,\,x = a\mathrm{ \,\,and}\,\,\,\mathrm{ }x = 2a respectively, for displacement along x−axis\mathrm{x - axis} :-
View answer and explanation
Correct option: C
all the charges are in unstable equilibrium
Option C (all the charges are in unstable equilibrium) is correct because any slight displacement from their positions will cause them to move further away, indicating unstable equilibrium.
381 Medium
381 of 385
Understanding +4 | -1
If there is only one type of charge in the universe, then
View answer and explanation
Correct option: D
∮E.ds=q/ε0\oint E . d s=q / \varepsilon_0 if charges of magnitude qq is inside the surface.
Option D (∮E.ds=q/ε0\oint E . ds=q / \varepsilon_0 if charges of magnitude qq is inside the surface.) is correct because Gauss's law states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space.
382 Medium
382 of 385
Understanding +4 | -1
Which of the following is incorrect regarding electrostatic field?
View answer and explanation
Correct option: C
It can form closed loops
Option C is correct because electrostatic field lines originate from positive charges and terminate on negative charges, and they cannot form closed loops.
383 Medium
383 of 385
Remembering +4 | -1
The law, governing the force between electric charge is known as
View answer and explanation
Correct option: D
Coulomb’s law
Option D (Coulomb’s law) is correct because Coulomb's law governs the force between electric charges.
384 Medium
384 of 385
Understanding +4 | -1
Assertion: If a conductor is given charge then no excess inner charge appears

Reason: Electric field inside conductor is zero
View answer and explanation
Correct option: B
Both Assertion & Reason are true. Reason is not the correct explanation for assertion
Option B is correct because both the assertion and reason are true, but the reason does not directly explain why the inner charge is zero.
385 Medium
385 of 385
Understanding +4 | -1
Assertion:- If more electric field lines leave a Gaussian surface than those which enter into it, then the net charge enclosed by that surface will be positive.
Reason:- If the net flux passing through a Gaussian surface is zero, then the net charge enclosed by that surface will be zero.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because both the assertion and the reason are true statements related to Gauss's law, but the reason doesn't explain the assertion.
Revision notes and key takeaways

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

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