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JEE • PHYSICS • ELECTRIC FIELDS AND CHARGES

Electric fields and charges practice · Medium

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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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91 Medium
91 of 385
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.
92 Medium
92 of 385
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.
93 Medium
93 of 385
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.
94 Medium
94 of 385
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.
95 Medium
95 of 385
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.
96 Medium
96 of 385
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.
97 Medium
97 of 385
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.
98 Medium
98 of 385
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.
99 Medium
99 of 385
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.
100 Medium
100 of 385
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.
101 Medium
101 of 385
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.
102 Medium
102 of 385
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.
103 Medium
103 of 385
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.
104 Medium
104 of 385
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.
105 Medium
105 of 385
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.
106 Medium
106 of 385
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.
107 Medium
107 of 385
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.
108 Medium
108 of 385
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.
109 Medium
109 of 385
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.
110 Medium
110 of 385
Application +4 | -1
Electrical force between two point charges is 200N. If we increase 10% charge on one of the charges and decrease 10% charge on the other, then electrical force between them for the same distance becomes __________ N
View answer and explanation
Correct option: A
198
Option A (198) is correct because the new force is calculated by multiplying the original force by the product of the new charges relative to the original charges.
111 Medium
111 of 385
Application +4 | -1
Two insulated charged metallic spheres PP and QQ have their centres separated by a distance of 50 cm50 \mathrm{~cm}. The radii of PP and QQ are negligible compared to the distance of separation. The mutual force of electrostatic repulsion if the charge on each is 3.2×10−7C3.2 \times 10^{-7} \mathrm{C} is:
View answer and explanation
Correct option: A
3.69×10−3 N3.69 \times 10^{-3} \mathrm{~N}
Option A (3.69×10−3 N3.69 \times 10^{-3} \mathrm{~N}) is correct because it calculates the electrostatic force using Coulomb's law with the given charges and separation distance.
112 Medium
112 of 385
Understanding +4 | -1
A comb run through one’s dry hair attracts small bits of paper. This is due to
View answer and explanation
Correct option: C
The atoms in the paper get polarised by the charged comb
Option C is correct because the charged comb polarizes the atoms in the paper, creating an attractive force.
113 Medium
113 of 385
Application +4 | -1
FgF_g and FeF_e represents gravitational and electrostatic force respectively between electron and proton at a distance of 10 cm10 \mathrm{~cm}. The ratio of Fg/FeF_g / F_e is of the order of
View answer and explanation
Correct option: B
10−3910^{-39}
The correct answer is B (10−3910^{-39}) because the ratio of gravitational to electrostatic force is approximately 10−3910^{-39} when calculated using the respective force equations.
114 Medium
114 of 385
Analyzing +4 | -1
A positively charged thin metal ring of radius
RR is fixed in the xy-plane, with its centre at the
origin. A negatively charged particle is released
from rest at the point P(0,0,c)P (0, 0, c) where c>0.c > 0.

Consider the following statements:
(A) The electric field at PP due to the metallic ring
is always directed towards origin.

(B) The negatively charged particle is accelerated
towards origin and undergoes periodic
motion.

Which of the above statements is/are correct?
View answer and explanation
Correct option: B
B only
Option B (B only) is correct. The electric field at P is along the z-axis, and the negatively charged particle accelerates towards the origin, resulting in periodic motion.
115 Medium
115 of 385
Understanding +4 | -1
Additive nature of charge means
View answer and explanation
Correct option: D
It tells about the scalar nature of charge
Option D (It tells about the scalar nature of charge) is correct because the additive nature of charge implies that charges can be added algebraically, like scalars.
116 Medium
116 of 385
Understanding +4 | -1
A capacitor of capacitance CC is charged to potential V.V. The flux of the electric field through a closed surface enclosing the capacitor is
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because the net charge enclosed by the closed surface is zero, so the net electric flux is also zero according to Gauss's law.
117 Medium
117 of 385
Application +4 | -1
A solid sphere of radius RR has a charge Q{\cal Q} distributed in its volume with a charge density ρ=kra,\mathrm{\rho } = k{r^a}, where k    and  a  k\;\;\mathrm{and\;}a\mathrm{\;} are constants and rr is the distance from its centre. If the electric field at r=R2r = \frac{R}{2} is 18\frac{1}{8} times that at r=R,r = R, find the value of a.a.
View answer and explanation
Correct option: A
2
Option A (2) is correct because by applying Gauss's law and comparing the electric fields at r = R/2 and r = R, we can solve for the value of a.
118 Medium
118 of 385
Application +4 | -1
A tin nucleus (atomic number Z = 50) has a radius of 6.6×10−15 m6.6 \times {10^{ - 15}}\,m. The potential on its surface will be (the charge on the proton =1.6×10−19 C= 1.6 \times {10^{ - 19}}\,C)
View answer and explanation
Correct option: A
1.1×107 V1.1 \times {10^7}\,\mathrm{V}
Option A (1.1×107 V1.1 \times {10^7}\,\mathrm{V}) is correct because it accurately calculates the electric potential on the surface of the tin nucleus using the formula V=14πε0⋅qrV = \frac{{1}}{{4\pi \varepsilon _0}} \cdot \frac{q}{r}.
119 Medium
119 of 385
Understanding +4 | -1
A thin metallic spherical shell contains a charge QQ on its surface. A point charge q1\mathrm{q}_1 is placed at the centre of the shell and another charge q2\mathrm{q}_2 is placed outside the shell. All the three charges are positive. Then the force on charge q1\mathrm{q}_1 is -
View answer and explanation
Correct option: C
Zero
Option C (Zero) is correct because the electric field at the center of a uniformly charged spherical shell is zero, resulting in no force on the charge q1.
120 Medium
120 of 385
Understanding +4 | -1
When a body is earth connected, electrons from the earth flow into the body. This means the body is
View answer and explanation
Correct option: B
Charged positively
Option B (Charged positively) is correct because electrons flow into the body to neutralize a positive charge.
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