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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 211-240 of 420 questions

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211 Medium
211 of 420
Application +4 | -1
An electron is liberated from the lower of the two large parallel metal plates separated by a distance of 20 mm20 \mathrm{~mm}. The upper plate has a potential of +2400 V+2400 \mathrm{~V} relative to the lower plate. How long does the electron take to reach the upper plate? Take em\frac{e}{m} of electrons 1.8×1011 C kg−11.8 \times 10^{11} \mathrm{~C} \mathrm{~kg}^{-1}.
View answer and explanation
Correct option: A
1.4×10−9 s1.4 \times 10^{-9} \mathrm{~s}
Option A (1.4×10−9 s1.4 \times 10^{-9} \mathrm{~s}) is correct because it is the time calculated using kinematics and the electric field between the plates.
212 Medium
212 of 420
Application +4 | -1
Two charges of 50 μc50\,\mu c and 100 μc100\,\mu c are separated by a distance of 0.6 m0.6\,m. The intensity of electric field of a point midway between them is ….×106vm\times {10^6}\frac{v}{m}
View answer and explanation
Correct option: B
2
Option B (2) is correct because it is the correct calculation of the electric field intensity at the midpoint between the two charges.
213 Medium
213 of 420
Application +4 | -1
The work done in placing a charge of 8×10−188 \times 10^{-18} coulomb on a condenser of capacity 100 micro-farad is
View answer and explanation
Correct option: A
32×10−3232 \times 10^{-32} Joule
Option A (32×10−3232 \times 10^{-32} Joule) is correct because the work done is equal to the energy stored in the capacitor, which can be calculated using the formula W=Q22CW = \frac{{Q^2}}{{2C}}.
214 Medium
214 of 420
Analyzing +4 | -1
A hemisphere is uniformly charged positively. The electric field at a point on a diameter away from the centre is directed
View answer and explanation
Correct option: A
perpendicular to the diameter
Option A (perpendicular to the diameter) is correct because due to symmetry, the horizontal components of the electric field cancel out, leaving only the vertical component perpendicular to the diameter.
215 Medium
215 of 420
Understanding +4 | -1
A spherical rubber balloon carries a charge, uniformly distributed over the surface. As the balloon is blown up and increases in size, the total electric flux coming out of the surface
View answer and explanation
Correct option: C
Remains unchanged
The correct answer is C (Remains unchanged) because according to Gauss's law, the total electric flux depends only on the enclosed charge, which remains constant as the balloon expands.
216 Medium
216 of 420
Analyzing +4 | -1
Two conducting spheres of radii R1{R_1} and R2{R_2} having charges Q1{Q_1} and Q2{Q_2} respectively are connected to each other. There is
View answer and explanation
Correct option: D
A decrease in the energy of the system unless  Q1R2=Q2R1{Q_1}{R_2} = {Q_2}{R_1}
Option D is correct. There is a decrease in the energy of the system unless the initial potentials of the spheres are equal, i.e., Q1R2=Q2R1{Q_1}{R_2} = {Q_2}{R_1}.
217 Medium
217 of 420
Understanding +4 | -1
The force between two charged particle placed in air at separation xx is F0F_0. Both the charged particle immerged in a medium of dielectric constant KK without changing separation between two charge, then net force on one of the particle is now
View answer and explanation
Correct option: A
F0K\frac{F_0}{K}
Option A (F0K\frac{F_0}{K}) is correct because the force between charged particles decreases by a factor of the dielectric constant when immersed in a medium.
218 Easy
218 of 420
Analyzing +4 | -1
Given below are two statements:
Statement I: The electric force changes the speed of the charged particle and hence changes its kinetic energy; whereas the magnetic force does not change the kinetic energy of the charged particle.
Statement II: The electric force accelerates the positively charged particle perpendicular to the direction of electric field. The magnetic force accelerates the moving charged particle along the direction of magnetic field.
In the light of the above statements, choose the most appropriate answer from the options given below:
View answer and explanation
Correct option: C
Statement I is correct but statement II is incorrect
The correct answer is C because statement I is correct, but statement II is incorrect. Electric force changes kinetic energy, but magnetic force doesn't. Magnetic force accelerates perpendicular, not along the field.
219 Medium
219 of 420
Understanding +4 | -1
One metallic sphere AA is given positive charge whereas another identical sphere BB of exactly same mass as of AA is given equal amount of negative charge. Then
View answer and explanation
Correct option: D
Mass of BB increases
Option D (Mass of BB increases) is correct because the negatively charged sphere gains electrons, thereby increasing its mass.
220 Medium
220 of 420
Understanding +4 | -1
The net electric flux associated with a closed surface which encloses 1C1 \mathrm{C} positive charge is equal to (In SI unit)
View answer and explanation
Correct option: B
ε0−1\varepsilon_0^{-1}
Option B (ε0−1\varepsilon_0^{-1}) is correct because according to Gauss's law, the net electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space.
221 Medium
221 of 420
Application +4 | -1
The point charges QQ and −2Q- 2Q are placed at same distance apart. If the electric field at the location of QQ is EE, find its value at the location of −2Q- 2Q is
View answer and explanation
Correct option: C
E1=E2{E_1} = \frac{E}{2}
Option C (E1=E2{E_1} = \frac{E}{2}) is correct because the electric field due to a point charge is proportional to the charge's magnitude, and the distance is the same.
222 Medium
222 of 420
Understanding +4 | -1
For coulomb force to be operative the least side of atom will be
View answer and explanation
Correct option: B
10−15 m10^{-15} \mathrm{~m}
Option B (10−15 m10^{-15} \mathrm{~m}) is correct because Coulomb's law is effective at very short distances, such as the diameter of the nucleus, which is around 10−15 m10^{-15} \mathrm{~m}.
223 Medium
223 of 420
Understanding +4 | -1
At a point in space, the electric field points towards north. In the region surrounding this point, the rate of change of potential will be zero along
View answer and explanation
Correct option: D
east-west
Option D (east-west) is correct because the electric field points towards the north, and the electric field is the negative gradient of the potential. Therefore, the potential is constant in the direction perpendicular to the electric field, which is the east-west direction.
224 Medium
224 of 420
Application +4 | -1
A force of 16 N is acting between two charges in air. If the space between them is completely filled with glass (∈r=10),\left( {{ \in _r} = 10} \right), then new force will be
View answer and explanation
Correct option: C
1.6 N
Option C (1.6 N) is correct because the force between two charges decreases by a factor of the relative permittivity of the medium.
225 Medium
225 of 420
Understanding +4 | -1
The Electric field at a point is
View answer and explanation
Correct option: D
discontinuous if there is a charge at that point
Option D (discontinuous if there is a charge at that point) is correct because electric fields are discontinuous at the location of point charges.
226 Medium
226 of 420
Application +4 | -1
The total electric flux through a cube when a charge 8q8q is placed at one corner of the cube is
View answer and explanation
Correct option: D
qε0\frac{q}{{{\varepsilon _0}}}
Option D (qε0\frac{q}{{{\varepsilon _0}}}) is correct because it is calculated using Gauss's law, considering the charge is at the corner of a cube.
227 Medium
227 of 420
Remembering +4 | -1
In Van de Graaff generator, the process of spraying the charge is called:
View answer and explanation
Correct option: B
Corona discharge
Option B (Corona discharge) is correct because the process of spraying charge onto the belt in a Van de Graaff generator is achieved through corona discharge.
228 Hard
228 of 420
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. Statement I is correct because of Gauss's law, and Statement III is correct because a dipole has a net charge of zero.
229 Medium
229 of 420
Understanding +4 | -1
Consider the following statements and select the incorrect statement.
View answer and explanation
Correct option: D
There is no change in the energy of a charged particle moving in a magnetic field although a magnetic force is acting on it.
Option D (There is no change in the energy of a charged particle moving in a magnetic field although a magnetic force is acting on it) is the incorrect statement because while the magnetic force does work, it doesn't change the particle's speed or kinetic energy.
230 Medium
230 of 420
Understanding +4 | -1
An electric dipole is put in north-south direction in a sphere filled with water. Which statement is correct
View answer and explanation
Correct option: C
Electric flux entering into sphere and leaving the sphere are same
Option C (Electric flux entering into sphere and leaving the sphere are same) is correct because the net charge enclosed by the sphere is zero.
231 Medium
231 of 420
Understanding +4 | -1
Electric field intensity at a point outside uniformly charged thin infinite plane sheet is 'E1\text{E}_1'. The electric field intensity at a point near and outside the surface of a positively charged conductor of any shape is 'E2\mathrm{E}_2'. The relation between magnitude of E1\text{E}_1 and E2\text{E}_2 is (assumed air as the medium)
View answer and explanation
Correct option: A
2E1=E22 \mathrm{E}_1=\mathrm{E}_2
Option A is correct because the electric field intensity near a charged conductor is twice the electric field intensity of an infinite plane sheet with the same charge density.
232 Medium
232 of 420
Understanding +4 | -1
The electric flux through a closed surface area SS enclosing charge Q{\cal Q} is ϕ\phi. If the surface area is doubled, then the flux is
View answer and explanation
Correct option: A
2ϕ2\phi
Option D (ϕ\phi) is incorrect. The electric flux through a closed surface depends only on the enclosed charge, not the surface area. Since the enclosed charge remains the same, the flux remains the same.
233 Medium
233 of 420
Understanding +4 | -1
When a glass rod is rubbed with a silk cloth, charge
appears on both. This observation is consistent with
law of conservation of charge as
View answer and explanation
Correct option: C
Charges appearing on both bodies are equal and opposite
Option C is correct because the charges appearing on both bodies are equal in magnitude but opposite in sign, conserving the total charge.
234 Medium
234 of 420
Understanding +4 | -1
In general, metallic ropes are suspended from the
carriers to the ground which take inflammable material.
The reason is
View answer and explanation
Correct option: C
To keep the body of the carrier in contact with the earth
Option C (To keep the body of the carrier in contact with the earth) is correct because the ropes provide a path for static electricity to discharge, preventing sparks.
235 Medium
235 of 420
Application +4 | -1
The number of electric lines of force emerging from 1C1C of positive charge in vaccum is ______ × 1011\times \,{10^{11}}
View answer and explanation
Correct option: A
1.13
Option A (1.13) is correct because the number of electric lines of force is calculated using the formula φ = q/ε_0.
236 Medium
236 of 420
Application +4 | -1
A cubical Gaussian surface encloses 24 C of charge. The electric flux through each surface of the cube is
View answer and explanation
Correct option: B
4 WB
Option B (4 WB) is correct because the electric flux through each surface of the cube is the total flux divided by the number of surfaces (6).
237 Medium
237 of 420
Understanding +4 | -1
Consider a conductor with a spherical cavity in it. A point charge q0q_0 is placed at the centre of cavity and a point charge QQ is placed outside conductor.

Statement-1: Total charge induced on cavity wall is equal and opposite to the charge inside.

Statement-2: If cavity is surrounded by a Gaussian surface, where all parts of Gaussian surface are located inside the conductor, ∫E.dA=0;\int E.d A=0; hence qinduced=−q0q_{\text {induced}}=-q_0
View answer and explanation
Correct option: A
Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1
Option A (Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1) is correct because the electric field inside a conductor is zero, leading to induced charge on the cavity wall.
238 Medium
238 of 420
Understanding +4 | -1
Assertion: An electrostatic field line never form closed loop

Reason: Electrostatic field is a conservative field
View answer and explanation
Correct option: B
Both Assertion & Reason are true. Reason is not the correct explanation for assertion
Option B (Both Assertion & Reason are true. Reason is not the correct explanation for assertion) is correct because both the assertion and the reason are true, but the reason does not explain why electrostatic field lines never form closed loops.
239 Medium
239 of 420
Application +4 | -1
64 drops each having the capacity CC and potential VV are combined to form a big drop. If the charge on the small drop is q,q, then the charge on the big drop will be
View answer and explanation
Correct option: D
64q64q
Option D (64q) is correct because the total charge on the big drop is the sum of the charges on the 64 smaller drops.
240 Medium
240 of 420
Understanding +4 | -1
When a positively charged body is earthed electrons from the earth flow into the body. This means the body is
View answer and explanation
Correct option: A
Uncharged
Option A (Uncharged) is correct because when a positively charged body is earthed, electrons flow from the earth to neutralize the positive charge, making the body uncharged.
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