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NEET • PHYSICS • ELECTRIC FIELDS AND CHARGES

Electric fields and charges practice · Medium

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Showing 31-60 of 92 questions

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31 Medium
31 of 92
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 is correct because in electrostatic equilibrium, the electric field is always perpendicular to the surface of a conductor.
32 Medium
32 of 92
Analyzing +4 | -1
Force of attraction between two point charges QQ and −Q-Q separated by d  meterd\;meter is Fe{F_e}. When these charges are placed on two identical spheres of radius R=0.3dR = 0.3d whose centres and d  meterd\;meter apart, the force of attraction between them is
View answer and explanation
Correct option: A
Greater than Fe{F_e}
Option A (Greater than Fe{F_e}) is correct because when the charges are placed on spheres, the effective distance between the charges decreases, increasing the force.
33 Medium
33 of 92
Application +4 | -1
An electron of mass mm and charge ee is accelerated from rest through a potential difference VV in vacuum. The final speed of the electron will be
View answer and explanation
Correct option: C
2eV/m\sqrt {2eV/m}
Option C (2eV/m\sqrt {2eV/m}) is correct because the kinetic energy gained by the electron is equal to the work done by the electric field, which is eVeV.
34 Medium
34 of 92
Remembering +4 | -1
Charge on α\alpha particle is
View answer and explanation
Correct option: C
3.2×10−19C3.2 \times {10^{ - 19}}C
Option C (3.2×10−19C3.2 \times {10^{ - 19}}C) is correct because an alpha particle has a charge of +2e, where e is the elementary charge.
35 Medium
35 of 92
Application +4 | -1
Two small conducting spheres of equal radius have charges +10μC+ 10\mu C and −20μC- 20\mu C respectively and placed at a distance RR from each other experience force F1{F_1}. If they are brought in contact and separated to the same distance, they experience force F2{F_2}. The ratio of F1{F_1} to F2{F_2} is
View answer and explanation
Correct option: B
−8:1- 8:1
Option B (−8:1- 8:1) is correct because the force is proportional to the product of the charges. After contact, the charges redistribute equally.
36 Medium
36 of 92
Understanding +4 | -1
Two conducting sphere of radii r1r_1 and r2r_2 are charged to the same surface charge density. The ratio of electric field near their surface is
View answer and explanation
Correct option: D
1:11\mathrm{ }:\mathrm{ }1
Option D (1:1) is correct because the electric field near the surface of a conductor depends only on the surface charge density, which is the same for both spheres.
37 Medium
37 of 92
Remembering +4 | -1
The scientist whose major contribution is
towards the measurement of electronic charge
View answer and explanation
Correct option: B
R.A. Millikan
Option B (R.A. Millikan) is correct because R.A. Millikan is famous for his oil drop experiment, which accurately measured the elementary electric charge.
38 Medium
38 of 92
Application +4 | -1
A force of 2.25 N acts on a charge of 15×10−415 \times {10^{ - 4}}. Calculate the intensity of electric field at that point
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Correct option: A
1500 NC−1\mathrm{N}{\mathrm{C}^{ - 1}}
Option A (1500 NC−1\mathrm{N}{\mathrm{C}^{ - 1}}) is correct because it is calculated by dividing the force by the charge.
39 Medium
39 of 92
Understanding +4 | -1
When a dielectric material is introduced between the plates of a charged condenser then electric field between the plates
View answer and explanation
Correct option: B
Decreases
Option B is correct because the introduction of a dielectric material reduces the electric field between the plates of a capacitor.
40 Medium
40 of 92
Analyzing +4 | -1
A body has a charge of 9.6×10−209.6 \times {10^{ - 20}} coulomb. It is
View answer and explanation
Correct option: B
Not possible
Option B (Not possible) is correct because charge is quantized, and the given charge is not an integer multiple of the elementary charge.
41 Medium
41 of 92
Application +4 | -1
Two point charges repel each other with a force of 100 N. One of the charges is increased by 10% and other is reduced by 10%. The new force of repulsion at the same distance would be
View answer and explanation
Correct option: C
99 N
Option C (99 N) is correct. The new force is calculated by considering the percentage increase and decrease in the charges using Coulomb's Law.
42 Medium
42 of 92
Analyzing +4 | -1
Assertion: Induced charge does not contribute to electric field or potential at a given point.

Reason: A point chagre q0q_0 is kept outside a solid metallic sphere, the elcetric field inside the sphere is Zero.
View answer and explanation
Correct option: D
Assertion is false and Reason is false
Assertion is false and Reason is true. Induced charge does contribute to the electric field and potential. The electric field inside a solid metallic sphere is zero.
43 Medium
43 of 92
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 Earth has a net negative surface charge density.
44 Medium
44 of 92
Application +4 | -1
Infinite charges of magnitude q each are lying at x=1,2,4,8..x = 1,2,4,8.. metre on X-axis. The value of intensity of electric field at point x=0x = 0 due to these charges will be
View answer and explanation
Correct option: A
12×109  q  NC−112 \times {10^9}\mathrm{\;}q\mathrm{\;N}{\mathrm{C}^{ - 1}}
Option A (12×109  q  NC−112 \times {10^9}\mathrm{\;}q\mathrm{\;N}{\mathrm{C}^{ - 1}}) is correct because it is the sum of an infinite geometric series representing the electric field due to each charge.
45 Medium
45 of 92
Understanding +4 | -1
If the electric flux entering and leaving an enclosed surface respectively is φ1{\varphi _1} and φ2{\varphi _2} the electric charge inside the surface will be
View answer and explanation
Correct option: B
(φ2−φ1)ε0\left( {{\varphi _2} - {\varphi _1}} \right){\varepsilon _0}
Option B (φ2−φ1)ε0\left( {{\varphi _2} - {\varphi _1}} \right){\varepsilon _0} is correct because Gauss's law relates the net electric flux through a closed surface to the enclosed charge.
46 Medium
46 of 92
Understanding +4 | -1
The ratio of the forces between two small spheres with constant charge (a)\left( a \right) in air (b)\left( b \right) in a medium of dielectric constant KK is
View answer and explanation
Correct option: B
K  :1K\;:1
Option B (K:1K:1) is correct because the force between two charges decreases by a factor of K when placed in a medium with dielectric constant K.
47 Medium
47 of 92
Remembering +4 | -1
The dimensional formula of electric charge is:
View answer and explanation
Correct option: A
[M0L0T1A1]\left[\mathrm{M}^0 \mathrm{L}^0 \mathrm{T}^1 \mathrm{A}^1\right]
Option A ([M0L0T1A1]\left[\mathrm{M}^0 \mathrm{L}^0 \mathrm{T}^1 \mathrm{A}^1\right]) is correct because electric charge is defined as current multiplied by time, and current is a fundamental dimension (Ampere).
48 Medium
48 of 92
Understanding +4 | -1
Identify the wrong statement in the following. Coulomb’s law correctly described 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 correct answer because Coulomb's law does not describe the nuclear force.
49 Medium
49 of 92
Application +4 | -1
A charged particle of mass m and charge q is released from rest in a uniform electric field E. The KE of particle after time t is:
View answer and explanation
Correct option: B
q2E2t22m\frac{{{q^2}{E^2}{t^2}}}{{2m}}
Option B (q2E2t22m\frac{{{q^2}{E^2}{t^2}}}{{2m}}) is correct because it is derived from calculating the work done by the electric field on the charged particle, which equals the kinetic energy gained.
50 Medium
50 of 92
Remembering +4 | -1
Electric charge between two bodies can be produced by
View answer and explanation
Correct option: B
rubbing
Option B is correct because rubbing two materials together can transfer electrons from one material to the other, creating a charge imbalance.
51 Medium
51 of 92
Analyzing +4 | -1
A glass rod rubbed with silk is used to charge a gold leaf electroscope and the leaves are observed to diverge. The electroscope thus charged is exposed to XX-rays for a short period. Then
View answer and explanation
Correct option: B
The leaves will diverge further
Option B is correct because X-rays ionize the air around the electroscope, leading to neutralization of the charges on the leaves and thus further divergence.
52 Medium
52 of 92
Application +4 | -1
A conducting sphere of radius 10cm10cm is charged 10μ  C.10\mu \;C. Another uncharged sphere of radius 20  cm20\;cm is allowed to touch it for some time. After that if the sphere are separated, then surface density of charges, on the spheres will be in the ratio of
View answer and explanation
Correct option: C
2:12\mathrm{ }:\mathrm{ }1
Option C (2:12\mathrm{ }:\mathrm{ }1) is correct. The ratio of surface charge densities is inversely proportional to the ratio of the radii of the spheres after they are separated.
53 Medium
53 of 92
Application +4 | -1
A polythene piece, rubbed with wool, is found to have negative charge of 4×10−74 \times {10^{ - 7}}C. the number of electrons transferred from wool to polythene is
View answer and explanation
Correct option: B
2.5×10122.5 \times {10^{12}}
Option B (2.5×10122.5 \times {10^{12}}) is correct because the number of electrons transferred is calculated by dividing the total charge by the charge of a single electron.
54 Medium
54 of 92
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}}
According to Gauss's law, the flux coming out from a unit positive charge enclosed in air is the charge divided by the permittivity of free space, which is 1ε0=(ε0)−1\frac{1}{\varepsilon_0} = {\left( {{\varepsilon _0}} \right)^{ - 1}}.
55 Medium
55 of 92
Application +4 | -1
A charged particle of mass 5×10−5kg5 \times {10^{ - 5}}kg is held stationary in space by placing it in an electric field of strength 107NC−1{10^7}N{C^{ - 1}} directed vertically downwards. The charge on the particle is
View answer and explanation
Correct option: B
−5×10−5μC- 5 \times {10^{ - 5}}\mu C
Option B (−5×10−5μC- 5 \times {10^{ - 5}}\mu C) is correct because the charge is calculated by equating the electric force to the gravitational force and considering the direction of the electric field.
56 Medium
56 of 92
Understanding +4 | -1
According to Gauss’ Theorem, electric field of an infinitely long straight wire is proportional to
View answer and explanation
Correct option: D
1/r1/r
Option D (1/r1/r) is correct because the electric field of an infinitely long straight wire decreases inversely with distance.
57 Medium
57 of 92
Understanding +4 | -1
Electric field in a region is E⃗=3j^ N/C\vec E = 3\hat j\,N/C. Equipotential surfaces are parallel to
View answer and explanation
Correct option: B
xz plane
The correct answer is xz plane because the electric field is in the y-direction, and equipotential surfaces are always perpendicular to the electric field.
58 Medium
58 of 92
Remembering +4 | -1
A current carrying wire in the neighborhood produces
View answer and explanation
Correct option: B
Magnetic field only
Option B is correct because a current-carrying wire generates a magnetic field around it, as described by Ampere's Law.
59 Medium
59 of 92
Application +4 | -1
A force FF acts between sodium and chlorine ions of salt (sodium chloride) when put 1cm apart in air. The permittivity of air and dielectric constant of water are ε0{\varepsilon _0} and KK respectively. When a piece of salt is put in water electrical force acting between sodium and chlorine ions 1cm1cm apart is
View answer and explanation
Correct option: A
FK\frac{F}{K}
Option A (FK\frac{F}{K}) is correct because the force between charges in a medium is reduced by the dielectric constant.
60 Medium
60 of 92
Remembering +4 | -1
The law, governing the force between electric charge is known as
View answer and explanation
Correct option: D
Coulomb’s law
The correct answer is Coulomb's law because it governs the force between electric charges, stating that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
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