Skip to main content
Back to ELECTRIC FIELDS AND CHARGES

Your Progress

0 / 156 Answered
0
Correct
0
Incorrect
0%
Accuracy
NEET • PHYSICS • ELECTRIC FIELDS AND CHARGES

Electric fields and charges practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

The practice player uses single-correct-answer MCQs. Chapter quizzes help you review a topic; they are not full-length NEET mock papers.

Take a chapter quiz →
Showing 91-120 of 156 questions

Preparing practice and saved progress…

91 Easy
91 of 156
Application +4 | -1
There is an electric field E{E} in xx-direction. If the work done on moving a charge of 0.2C0.2 \mathrm{C} through a distance of 2 m2 \mathrm{~m} along a line making an angle 60∘60^{\circ} with xx-axis is 1 J1 \mathrm{~J}, then what is the value of EE?
View answer and explanation
Correct option: A
5
Option A is correct because the electric field is calculated using the work done formula, considering the angle between the force and displacement.
92 Easy
92 of 156
Application +4 | -1
If an electron is accelerated by 8.8×1014m/s28.8 \times 10^{14} m / s ^2, then electric field required for acceleration is (given specific charge of the electron =1.76×1011Ckg−1=1.76 \times 10^{11} Ckg ^{-1} )
View answer and explanation
Correct option: B
50 V cm−150~ V~ cm ^{-1}
Option B (50 V cm−150~ V~ cm ^{-1}) is correct because the electric field is calculated using E=a/qE = a/q, where aa is the acceleration and qq is the specific charge.
93 Medium
93 of 156
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.
94 Easy
94 of 156
Application +4 | -1
The electric field intensity which would be just sufficient to balance the weight of a charge 100 μC100\, \mu \mathrm{C} and mass 10 g10 \mathrm{~g} is _____ N/C. (Take g=10 m s−2g=10 \mathrm{~m} \mathrm{~s}^{-2} )
View answer and explanation
Correct option: A
1000
Option A is correct because the electric field required to balance the weight is calculated by equating the electric force and gravitational force.
95 Easy
95 of 156
Remembering +4 | -1
The magnitude of electric force, F is
View answer and explanation
Correct option: A
directly proportional to the multiplication of both charges.
Option A is correct because Coulomb's Law states that the electric force is directly proportional to the product of the magnitudes of the charges.
96 Easy
96 of 156
Analyzing +4 | -1
A negatively charged object X is repelled by another charged object Y. However an object Z is attracted to object

Y. Which of the following is the most possibility for the charge of object Z?
View answer and explanation
Correct option: B
neutral or positively charged
Option B is correct because a neutral object can be attracted to a charged object due to polarization, and a positively charged object will be attracted to a negatively charged object.
97 Easy
97 of 156
Analyzing +4 | -1
Statement-1: An electric dipole placed in the electric field of a point charge can never experience zero resultant force.
Statement-2: Electric field of a point charge is non uniform.
View answer and explanation
Correct option: A
Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1
Option A (Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1) is correct because the non-uniform field causes unequal forces on the dipole's charges, resulting in a net force.
98 Easy
98 of 156
Understanding +4 | -1
Assertion: On disturbing an electric dipole in stable equilibrium in an electric field, it returns back to its stable equilibrium orientation.
Reason: A restoring torque acts on the dipole on being disturbed from its stable equilibrium.
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 the assertion and reason are both true, and the reason correctly explains why the dipole returns to its stable equilibrium.
99 Medium
99 of 156
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}}.
100 Easy
100 of 156
Application +4 | -1
The electric field required to keep a water drop of mass m just to remain suspended, when charged with one
electron, is
View answer and explanation
Correct option: B
mge\frac{{mg}}{e}
Option B is correct because the electric field is calculated by equating the electric force on the electron to the gravitational force on the water drop.
101 Medium
101 of 156
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.
102 Medium
102 of 156
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.
103 Medium
103 of 156
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.
104 Medium
104 of 156
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.
105 Medium
105 of 156
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.
106 Medium
106 of 156
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.
107 Medium
107 of 156
Analyzing +4 | -1
Two point charges QQ each are placed at a distance dd apart. A third point charge qq is placed at a distance xx from mid-point on the perpendicular bisector. The value of xx at which charge qq will experience the maximum Coulombs force is :
View answer and explanation
Correct option: D
x=d22x=\frac{d}{2 \sqrt{2}}
The maximum Coulomb force occurs at x=d22x=\frac{d}{2 \sqrt{2}}. This is found by differentiating the force expression with respect to x and setting it to zero to find the maximum.
108 Medium
108 of 156
Understanding +4 | -1
A particle of ′m′'m' and charge ′q′'q' is accelerated through a potential difference of V  volt,V\;volt, its energy will be
View answer and explanation
Correct option: A
qVqV
The correct answer is qV because the energy gained by a charge accelerated through a potential difference is the product of the charge and the potential difference.
109 Medium
109 of 156
Analyzing +4 | -1
For a uniformly charged ring of radius R,R, the electric field on its axis has the largest magnitude at a distance hh from its centre. Then, value of hh is
View answer and explanation
Correct option: A
R2\frac{R}{{\sqrt 2 }}
Option A (R2\frac{R}{{\sqrt 2 }}) is correct because the electric field on the axis of a ring is maximum at a distance of R/sqrt(2) from the center.
110 Medium
110 of 156
Understanding +4 | -1
If a unit positive charge is taken from one point to another over an equipotential surface, then
View answer and explanation
Correct option: D
No work is done
Option D is correct because the potential difference between any two points on an equipotential surface is zero, hence no work is done.
111 Medium
111 of 156
Remembering +4 | -1
The maximum electric field that a dielectric medium of a capacitor can withstand without break
down (of its insulating property) is called its :
View answer and explanation
Correct option: C
dielectric strength
Option C (dielectric strength) is correct because it defines the maximum electric field a dielectric can withstand before breakdown.
112 Easy
112 of 156
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 (If both Assertion and Reason are correct and Reason is the correct explanation of Assertion) is correct because in a uniform electric field, the net flux through any closed surface is zero, implying no net charge within the volume.
113 Medium
113 of 156
Application +4 | -1
Force acting upon a charged particle kept between the plates of a charged condenser is F.F. If one plate of the condenser is removed, then the force acting on the same particle will become
View answer and explanation
Correct option: B
F/2F/2
Option B (F/2F/2) is correct because removing one plate halves the electric field strength, and thus halves the force on the charged particle.
114 Medium
114 of 156
Application +4 | -1
A sphere of radius R has a uniform distribution of electric charge (q) in its volume. At a distance x from its centre, the electric field (at x=R2x = \frac{R}{2}) will be
View answer and explanation
Correct option: B
q8πε0R2\frac{q}{{8\pi {\varepsilon _0}{R^2}}}
Option B (q8πε0R2\frac{q}{{8\pi {\varepsilon _0}{R^2}}}) is correct because it is derived from Gauss's law considering the charge enclosed within the Gaussian surface at x = R/2.
115 Medium
115 of 156
Analyzing +4 | -1
Statement-I: A charge qq is lying at the centre of the line joining two similar charges QQ each. The system will be in equilibrium if q=Q/4q=Q / 4

Statement-II: For QQ to be in equilibrium, sum of the forces on QQ due to rest of the two charges must zero.
View answer and explanation
Correct option: A
Statement-I is True, statement-II is True
Statement-I is false, Statement-II is true. The system will be in equilibrium if q = -Q/4. Statement II is true because the sum of forces on Q due to other charges must be zero for equilibrium.
116 Medium
116 of 156
Application +4 | -1
Charge q2{q_2} of mass mm revolves around a stationary charge q1{q_1} in a circular orbit of radius rr The orbital periodic time of q2{q_2} would be
View answer and explanation
Correct option: A
[4π3mr2kq1q2]1/2{\left[ {\frac{{4{\pi ^3}m{r^2}}}{{k{q_1}{q_2}}}} \right]^{1/2}}
Option A ([4π3mr2kq1q2]1/2{\left[ {\frac{{4{\pi ^3}m{r^2}}}{{k{q_1}{q_2}}}} \right]^{1/2}}) is correct because the centripetal force is provided by the electrostatic force, and equating them allows solving for the period.
117 Easy
117 of 156
Application +4 | -1
Two charges placed in air repel each other by a force of 10−4 N10^{-4} {~N}. When oil is introduced between the charges, the force becomes 5×10−5 N5 \times 10^{-5} {~N}. The dielectric constant of oil is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the dielectric constant is the ratio of the force in a vacuum (or air) to the force in the medium.
118 Medium
118 of 156
Analyzing +4 | -1
Assertion (A): The basic difference between magnetic lines of force and electric lines of force is electric lines of force are discontinuous and magnetic lines of force are continuous.
Reason (R): Magnetic lines of force exist in a magnet but no electric lines of force exists in a charged body.
View answer and explanation
Correct option: A
Both A and R are true and R is the correct explanation of A
Option A is correct because electric lines of force are discontinuous due to the existence of isolated charges, while magnetic lines of force are continuous as magnetic monopoles do not exist. The reason correctly explains this difference.
119 Medium
119 of 156
Analyzing +4 | -1
The bob of simple pendulum is hanging vertically down from a fixed identical bob by means of string of length ll. If both bobs are charged with a charge with a charge qq each, time period of the pendulum is (ignore the radii of the bobs)
View answer and explanation
Correct option: C
2πlg2\pi \sqrt {\frac{l}{\mathrm{g}}}
The time period remains 2πlg2\pi \sqrt{\frac{l}{g}} because the electrostatic force is horizontal and does not affect the vertical component of gravity, which determines the pendulum's period.
120 Medium
120 of 156
Understanding +4 | -1
Which of the following is deflected by electric field
View answer and explanation
Correct option: D
α\alpha-particles
The correct answer is D because alpha particles are charged particles and experience a force in an electric field, causing them to be deflected.
Revision notes and key takeaways

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

Showing 91-120 of 156 questions

Your practice

0 / 156 answered · 156 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.