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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 91-120 of 420 questions

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91 Medium
91 of 420
Understanding +4 | -1
Assertion: All the charges in a conductor get
distributed on whole of its outer surface.

Reason: In a dynamic system, charges try to
keep their potential energy minimum.
View answer and explanation
Correct option: A
If both Assertion and Reason are correct and
reason is the correct explanation of Assertion
Option A is correct because charges distribute on the outer surface to minimize potential energy, leading to a stable dynamic system.
92 Medium
92 of 420
Remembering +4 | -1
The surface charge density (in  C/m2)\left( {\mathrm{in\;}C/{m^2}} \right) of the earth is about
View answer and explanation
Correct option: D
−10−9- {10^{ - 9}}
Option D (−10−9-{10^{ - 9}}) is correct because the surface charge density is calculated by dividing the total charge by the surface area of the Earth.
93 Medium
93 of 420
Application +4 | -1
If the electric field near the earth's surface be 300 V/m300 \mathrm{~V} / \mathrm{m} directed downwards, then the surface density of charge on earth's surface is
View answer and explanation
Correct option: C
2.6×10−9C/m22.6 \times 10^{-9} \mathrm{C} / \mathrm{m}^2
Option C (2.6×10−9C/m22.6 \times 10^{-9} \mathrm{C} / \mathrm{m}^2) is correct because it's calculated using the formula relating electric field and surface charge density: E=σ/ε0E = \sigma / \varepsilon_0.
94 Medium
94 of 420
Application +4 | -1
A charge QQ is to be distributed on two conducting spheres. Find out values of charges on spheres so, that the repulsive force between them is maximum at fixed distance.
View answer and explanation
Correct option: A
Q2\frac{Q}{2}
Option A (Q2\frac{Q}{2}) is correct because the repulsive force is maximized when the charge is equally distributed between the two spheres.
95 Medium
95 of 420
Remembering +4 | -1
The force per unit charge is known as_____.
View answer and explanation
Correct option: B
Electric field
Option B (Electric field) is correct because the electric field is defined as the force per unit charge.
96 Medium
96 of 420
Application +4 | -1
What is the magnitude of a point charge due to which the electric field 30cm away has the magnitude 2  newton/coulomb2\;newton/coulomb [1/4πε0=9×109Nm2/C2][1/4\pi {\varepsilon _0} = 9 \times {10^9}N{m^2}/{C^2}]
View answer and explanation
Correct option: A
2×10−11  coulomb2 \times {10^{ - 11}}\;coulomb
Option A (2×10−11  coulomb2 \times {10^{ - 11}}\;coulomb) is correct because it is calculated directly from the formula for the electric field due to a point charge.
97 Medium
97 of 420
Understanding +4 | -1
Which of the following is false for electric lines
of force?
View answer and explanation
Correct option: C
They always form closed loops
Option C is correct because total internal reflection is not directly involved in the formation of a rainbow; the process involves refraction, dispersion, and internal reflection.
98 Medium
98 of 420
Understanding +4 | -1
Assertion: The coulomb force is the dominating force in the universe.
Reason: The coulomb force is weaker than the gravitational force.
View answer and explanation
Correct option: D
If the assertion and reason both are false.
Option D is correct because the Coulomb force is actually much stronger than the gravitational force, and it is not the dominating force in the universe.
99 Medium
99 of 420
Analyzing +4 | -1
A charged bead is capable of sliding freely
through a string held vertically in tension. An
electric field is applied parallel to the string, so
that the bead stays at rest at middle of the
string. If the electric field is switched off
momentarily and switched on
View answer and explanation
Correct option: D
The bead moves downwards with constant
velocity till it reaches the bottom of the string
Option D is correct. After the field is switched off and then on again, the forces balance, resulting in constant velocity.
100 Medium
100 of 420
Understanding +4 | -1
A proton and an electron are placed in a uniform electric field then
View answer and explanation
Correct option: D
The magnitude of the forces acting on them will be equal
Option D (The magnitude of the forces acting on them will be equal) is correct because both proton and electron have the same magnitude of charge, and the electric force depends on the charge and electric field.
101 Medium
101 of 420
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 square of the radius.
102 Medium
102 of 420
Application +4 | -1
The electric field in a region of space is given by E→=5i^+2j^  NC−1\overrightarrow E = 5\widehat i + 2\widehat j\,\,N{C^{ - 1}}. The electric flux due to this field through an area 2m22{m^2} lying in the YZ plane in S.I. Units is
View answer and explanation
Correct option: A
10
Option A (10) is correct because the electric flux is the dot product of the electric field and the area vector, which simplifies to E*A when the area is in the YZ plane.
103 Medium
103 of 420
Understanding +4 | -1
Assertion:- Electric field lines are always perpendicular to the surface of conductor.
Reason:-Electric field is always perpendicular to equipotential surface.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because electric field lines are always perpendicular to the surface of a conductor, and this is because the electric field is always perpendicular to an equipotential surface, and the surface of a conductor is an equipotential surface.
104 Easy
104 of 420
Analyzing +4 | -1
Assertion: In a region, where uniform electric field exists, the net charge within volume of any size is zero.

Reason: The electric flux within any closed surface in region of uniform electric field is zero.
View answer and explanation
Correct option: A
If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
Option A is correct. In a uniform electric field, the net charge within any volume is zero because the electric flux through any closed surface is zero, implying no net enclosed charge according to Gauss's law.
105 Medium
105 of 420
Application +4 | -1
If point charges Q1=2×10−7CQ_1=2 \times 10^{-7} \mathrm{C} and Q2=3×10−7CQ_2=3 \times 10^{-7} \mathrm{C} are at 30 cm30 \mathrm{~cm} separation, then find electrostatic force between them
View answer and explanation
Correct option: B
6×10−3 N6 \times 10^{-3} \mathrm{~N}
Option B (6×10−3 N6 \times 10^{-3} \text{ N}) is correct because it is the electrostatic force calculated using Coulomb's law.
106 Medium
106 of 420
Application +4 | -1
A positive charge of 17.7μC17.7 \mu \mathrm{C} is placed at the centre of a hollow sphere of radius 0.5 m0.5 \mathrm{~m}. Calculate the flux density through the surface of the sphere.
View answer and explanation
Correct option: A
6.4×105 NC−16.4 \times 10^5 \mathrm{~NC}^{-1}
Option A (6.4 x 10⁵ NC⁻¹) is correct because it is calculated using Gauss's law, which relates the electric flux to the enclosed charge.
107 Medium
107 of 420
Understanding +4 | -1
The electric field inside a spherical shell of uniform surface charge density is
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because, according to Gauss's law, the electric field inside a uniformly charged spherical shell is zero as there is no enclosed charge.
108 Medium
108 of 420
Understanding +4 | -1
Two spheres of copper of the same radii, one hollow and other solid, are charged to the same potential. The hollow sphere has
View answer and explanation
Correct option: D
Same charge as the solid sphere.
Option D is correct because the charge on a sphere depends on its capacitance and potential, and since both spheres have the same radius and potential, they have the same charge.
109 Medium
109 of 420
Understanding +4 | -1
The force between two point charges placed in a material medium of dielectric constant εr\varepsilon_\text{r} is F.\text{F}. If the material is removed, then the force between them becomes
View answer and explanation
Correct option: A
εrF\varepsilon_{\mathrm{r}} \mathrm{F}
Option A (εrF\varepsilon_{\mathrm{r}} \mathrm{F}) is correct because removing the dielectric medium increases the force by a factor of the dielectric constant.
110 Medium
110 of 420
Analyzing +4 | -1
A conducting sphere of radius RR, carrying charge QQ, lies inside an uncharged conducting shell of radius 2R2R. If they are joined by a metal wire,

(A) Q/3Q/3 amount of charge will flow from the sphere to the shell

(B) 2Q/32Q/3 amount of charge will flow from the sphere to the shell

(C) QQ amount of charge will flow from the sphere to the shell

(D) amount of heat will be produced KQ2/4RK Q^2 / 4 R
View answer and explanation
Correct option: C
CC and DD are correct
Option C (C and D are correct) is correct because when the spheres are connected, they reach the same potential, causing all the charge to flow to the outer sphere, and heat is generated due to the change in electrostatic energy.
111 Medium
111 of 420
Remembering +4 | -1
Top of the stratosphere has an electric field EE (in units of V/mV/m) nearly equal to
View answer and explanation
Correct option: C
100
Option C (100) is the correct order of magnitude for the electric field at the top of the stratosphere.
112 Medium
112 of 420
Understanding +4 | -1
The intensity of electric field at the surface of conducting hollow sphere is 10 NC−110\, \mathrm{NC}^{-1} and its radius is 10 cm10 \mathrm{~cm}. The value of electric field at the centre of sphere is
View answer and explanation
Correct option: A
zero
Option A (zero) is correct because the electric field inside a conducting hollow sphere is always zero, regardless of the charge or radius.
113 Medium
113 of 420
Analyzing +4 | -1
Two identical conducting spheres carrying different charges attract each other with a force FF when placed in air medium at a distance dd apart. The spheres are brought into contact and then taken to their original positions. Now the two spheres repel each other with a force whose magnitude is equal to that of the initial attractive force. The ratio between initial charges on the spheres is
View answer and explanation
Correct option: C
−3(3+√8)or(−3+√8)- 3\left( {3 + \surd 8} \right)\mathrm{or}\left( { - 3 + \surd 8} \right)
Option A (−(3+√8)- \left( {3 + \surd 8} \right) only) is incorrect. Option B (−3+√8- 3 + \surd 8 only) is incorrect. Option D (+3+ \sqrt 3) is incorrect. Option C (−3(3+√8)or(−3+√8)- 3\left( {3 + \surd 8} \right)\mathrm{or}\left( { - 3 + \surd 8} \right)) is correct because it is derived from equating the magnitudes of the attractive and repulsive forces after the spheres are brought into contact.
114 Medium
114 of 420
Remembering +4 | -1
Electric charge between two bodies can be produced by
View answer and explanation
Correct option: B
rubbing
Option B (rubbing) is correct because rubbing two dissimilar materials together causes charge transfer due to the triboelectric effect.
115 Medium
115 of 420
Analyzing +4 | -1
Consider E1→=xi^+j^\overrightarrow {{E_1}} = x\hat i + \hat j and E2→=xy2i^+x2y  j^;\overrightarrow {{E_2}} = x{y^2}\hat i + {x^2}y\;\hat j; then
View answer and explanation
Correct option: C
Both are electrostatic
Option C (Both are electrostatic) is correct because for an electrostatic field, ∇×E⃗=0\nabla \times \vec{E} = 0, and both given fields satisfy this condition.
116 Medium
116 of 420
Understanding +4 | -1
Select the correct statements. Coulomb's law correctly describes the electric force that:
(i) binds the electrons of an atom to its nucleus.
(ii) binds the protons and neutrons in the nucleus of an atom.
(iii) binds atoms together to form molecules.
View answer and explanation
Correct option: B
(i) and (iii)
Option B ((i) and (iii)) is correct because Coulomb's law describes the electrostatic forces between charged particles, such as electrons and the nucleus, and between atoms forming molecules.
117 Medium
117 of 420
Understanding +4 | -1
A Gaussian sphere encloses an electric dipole within it. The total flux across the sphere is
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because the net charge enclosed by the Gaussian surface is zero for an electric dipole.
118 Medium
118 of 420
Application +4 | -1
A simple pendulum of period TT has a metal bob which is negatively charged. If it is allowed to oscillate above a positively charged metal plate, its period will
View answer and explanation
Correct option: B
Less than TT
Option B (Less than TT) is correct because the effective acceleration due to gravity increases, causing the period to decrease.
119 Medium
119 of 420
Analyzing +4 | -1
A proton and an electron are released from
an infinite distance apart and they get
attracted towards each other. Which of the
following statement about their kinetic
energy is true?
View answer and explanation
Correct option: A
Kinetic energy of electron is more than
that proton
Option A (Kinetic energy of electron is more than that proton) is correct because, due to conservation of momentum, the lighter electron will have a higher kinetic energy than the heavier proton.
120 Easy
120 of 420
Analyzing +4 | -1
Assertion (A): Using Gauss law, it is possible to find the electric field at any point.
Reason (R): Gauss law is applicable for any type of charge distribution.
View answer and explanation
Correct option: C
A is true but R is false
Option C (A is true but R is false) is correct because Gauss's law can be used to find the electric field at any point, but it's only practical for charge distributions with sufficient symmetry.
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