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NEET • PHYSICS • ELECTRIC FIELDS AND CHARGES

Electric fields and charges practice · Mixed difficulty

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61 Medium
61 of 156
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.
62 Hard
62 of 156
Application +4 | -1
If volume change density inside a solid sphere of radius R{R} varies with distance rr from centre of sphere as ρ=ρ0rR;ρ0=\rho=\rho_0 \frac{r}{R} ; \rho_0= constant then find electric field intensity at distance r(<R){r}(<{R}) from centre is
View answer and explanation
Correct option: A
ρ0r24∈0R\frac{\rho_0 r^2}{4 \in_0 R}
Option A (ρ0r24∈0R\frac{\rho_0 r^2}{4 \in_0 R}) is correct because it is derived from Gauss's law, considering the varying charge density within the sphere.
63 Medium
63 of 156
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.
64 Medium
64 of 156
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.
65 Hard
65 of 156
Application +4 | -1
If the electric flux entering and leaving an enclosed surface respectively is ϕ1{\phi _1} and ϕ2{\phi _2} the electric charge inside the surface will be
View answer and explanation
Correct option: B
(ϕ2−ϕ1)ε0\left( {{\phi _2} - {\phi _1}} \right){\varepsilon _0}
Option B ((ϕ2−ϕ1{\phi _2} - {\phi _1}){ε _0}) is correct because Gauss's law relates the net electric flux through a closed surface to the enclosed charge.
66 Easy
66 of 156
Application +4 | -1
Two charges qq and 3q3 q are separated by a distance ' rr ' in air. At a distance xx from charge qq, the resultant electric field is zero. The value of xx is :
View answer and explanation
Correct option: C
r(1+3)\frac{r}{(1+\sqrt{3})}
Option C is correct because the distance x is calculated by equating the magnitudes of the electric fields due to both charges at that point.
67 Medium
67 of 156
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.
68 Easy
68 of 156
Understanding +4 | -1
A sphere of radius 1 m encloses a charge of 5μC.5\mu C. Another charge of −5μC-5 \mu C is placed inside the sphere.
The net electric flux would be
View answer and explanation
Correct option: C
Zero
Option C (Zero) is correct because the net charge enclosed by the sphere is zero, resulting in zero net electric flux.
69 Hard
69 of 156
Analyzing +4 | -1
Given below are two statements: one is labelled
as Assertion (A) and the other is labelled as
Reason (R).

Assertion (A): Net dipole moment of a polar
linear isotropic dielectric substance is not zero
even in the absence of an external electric field.

Reason (R): In absence of an external electric
field, the different permanent dipoles of a polar
dielectric substance are oriented in random
directions.

In the light of the above statements, choose the
most appropriate answer from the options given
below:
View answer and explanation
Correct option: D
(A) is not correct but (R) is correct
The correct answer is D because the assertion is false; a polar dielectric has zero net dipole moment in the absence of an external field due to random orientation of dipoles. The reason is correct.
70 Easy
70 of 156
Remembering +4 | -1
Electric lines of forces are
View answer and explanation
Correct option: A
Open curves
Option A is correct because electric field lines originate from positive charges and terminate at negative charges, thus forming open curves.
71 Medium
71 of 156
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
View answer and explanation
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.
72 Hard
72 of 156
Application +4 | -1
A uniformly charged disc of radius RR having surface charge density σ\sigma is placed in the xy- plane with its centre at the origin. Find the electric field intensity along the zz –axis at a distance ZZ from origin
View answer and explanation
Correct option: A
E=σ2ε0(1−Z(Z2+R2))E = \frac{\sigma }{{2{\varepsilon _0}}}\left( {1 - \frac{\mathrm{Z}}{{\sqrt {({Z^2} + {R^2})} }}} \right)
Option A (E=σ2ε0(1−Z(Z2+R2))E = \frac{\sigma }{{2{\varepsilon _0}}}\left( {1 - \frac{\mathrm{Z}}{{\sqrt {({Z^2} + {R^2})} }}} \right)) is correct because it represents the electric field due to a charged disc along its axis, derived from integrating the contributions of infinitesimal rings.
73 Medium
73 of 156
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.
74 Medium
74 of 156
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.
75 Medium
75 of 156
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.
76 Medium
76 of 156
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.
77 Easy
77 of 156
Application +4 | -1
Two uniformly charged spherical conductors AA and BB of radii 5 mm5 \mathrm{~mm} and 10 mm10 \mathrm{~mm} are separated by a distance of 2 cm2 \mathrm{~cm}. If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitude of the electric fields at the surface of the sphere AA and BB will be
View answer and explanation
Correct option: B
2:12: 1
Option B is correct because when connected by a wire, the spheres have the same potential, and the electric field ratio is inversely proportional to the radii ratio.
78 Medium
78 of 156
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.
79 Medium
79 of 156
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.
80 Medium
80 of 156
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.
81 Hard
81 of 156
Understanding +4 | -1
Select the correct statements from the following:

I. The electric field due to a charge outside the Gaussian surface contributes zero net flux through the surface.

II. Total flux linked with a closed body, enclosing any charges will be zero.

III. Total electric flux, if a dipole is enclosed by a surface is zero.
View answer and explanation
Correct option: C
I and III
Option C (I and III) is correct because a charge outside a Gaussian surface contributes zero net flux, and the net flux through a surface enclosing a dipole is zero.
82 Easy
82 of 156
Application +4 | -1
If two point charges 5 C and 10 C are separated by a distance of 10 cm in air. What is electrostatic force between them is
View answer and explanation
Correct option: A
45×1012N45 \times {10^{12}}N
Option A (45×1012N45 \times {10^{12}}N) is correct because it is the electrostatic force calculated using Coulomb's law with the given charges and distance.
83 Easy
83 of 156
Analyzing +4 | -1
Assertion:- For a charged shell, potential inside shell depends only on the field inside shell.
Reason:- Potential of shell may be determined only by the field outside.
View answer and explanation
Correct option: D
If both Assertion & Reason are False
Option D (If both Assertion & Reason are False) is correct because the potential inside a charged shell is constant and depends on the charge and radius of the shell, not just the field inside, and it can be determined by the field outside.
84 Medium
84 of 156
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.
85 Medium
85 of 156
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).
86 Medium
86 of 156
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.
87 Medium
87 of 156
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.
88 Medium
88 of 156
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.
89 Medium
89 of 156
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.
90 Medium
90 of 156
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.
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