Skip to main content
Back to ELECTRIC FIELDS AND CHARGES

Your Progress

0 / 385 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • PHYSICS • ELECTRIC FIELDS AND CHARGES

Electric fields and charges practice · Medium

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

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.

Take a chapter quiz →
Showing 181-210 of 385 questions

Preparing practice and saved progress…

181 Medium
181 of 385
Application +4 | -1
Two point charges +3μC+ 3\mu C and +8μC+ 8\mu C repel each other with a force of 40NN. If a charge of −5μC- 5\mu C is added to each of them, then the force between them will become
View answer and explanation
Correct option: B
+10N+ 10N
Option B (+10N+ 10N) is correct because the force is proportional to the product of the charges, and adding the same charge to both changes the magnitude and sign of the force.
182 Medium
182 of 385
Application +4 | -1
An electron moves a distance of 6 cm6 \mathrm{~cm} when accelerated from rest by an electric field of strength 2×104 NC−12 \times 10^4 \mathrm{~NC}^{-1}. Calculate the time of travel. The mass and charge of electron are 9×10−31 kg9 \times 10^{-31} \mathrm{~kg} and 1.6×10−19C1.6 \times 10^{-19} \mathrm{C} respectively.
View answer and explanation
Correct option: C
0.585×10−8 s0.585 \times 10^{-8} \mathrm{~s}
Option C (0.585×10−8 s0.585 \times 10^{-8} \mathrm{~s}) is correct because it is calculated using kinematics equations with the acceleration derived from the electric field force.
183 Medium
183 of 385
Application +4 | -1
Electric charge is uniformly distributed along a long straight wire of radius 1mm. The charge per cm length of the wire is Q{\cal Q} coulomb. Another cylindrical surface of radius 50 cm and length 1m symmetrically encloses the wire. The total electric flux passing through the cylindrical surface is
View answer and explanation
Correct option: B
100Qε0\frac{{100{\cal Q}}}{{{\varepsilon _0}}}
Option B (100Qε0\frac{{100{\cal Q}}}{{{\varepsilon _0}}}) is correct because it applies Gauss's law to find the electric flux through the cylindrical surface, considering the charge enclosed within the 1-meter length.
184 Medium
184 of 385
Application +4 | -1
Calculate the ratio of electric to gravitational force between two electrons.
View answer and explanation
Correct option: A
4.17×10424.17 \times 10^{42}
Option A (4.17×10424.17 \times 10^{42}) is correct because it is calculated by dividing the electric force by the gravitational force between two electrons, using the respective formulas.
185 Medium
185 of 385
Application +4 | -1
A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field 100V  m−1.100V\;{m^{ - 1}}. If the mass of the drop is 1.6×10−3g,1.6 \times {10^{ - 3}}g, the number of electrons carried by the drop is (g=10ms−2)\left( {g = 10m{s^{ - 2}}} \right)
View answer and explanation
Correct option: D
1012{10^{12}}
Option D (1012{10^{12}}) is correct because the number of electrons is determined by equating the electric force to the gravitational force and then dividing the charge by the elementary charge.
186 Medium
186 of 385
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.
187 Medium
187 of 385
Analyzing +4 | -1
Statement-1: Total work done by non-uniform electric field on a charged particle starting from rest till any time is non-negative. (assume no other forces act on the charged particle).
Statement-2: The angle between electrostatic force an velocity of the charged particle released from rest in non-uniform electric field is always acute. (assume no other forces act on the charged particle).
View answer and explanation
Correct option: C
Statement-1 is true, Statement-2 is false
Option C is correct because Statement-1 is true (work done is non-negative), but Statement-2 is false (the angle can be obtuse).
188 Medium
188 of 385
Application +4 | -1
If the electrostatic force on a small sphere of charge 0.4μC0.4 \mu \mathrm{C} due to another small sphere of charge −0.8μC-0.8 \mu \mathrm{C} in air is 0.2 N.0.2 \mathrm{~N}. The separation of the spheres is (k=9×109 N m2/C2)\left(\mathrm{k}=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}^2\right)
View answer and explanation
Correct option: D
12 cm12 \mathrm{~cm}
Option D (12 cm12 \mathrm{~cm}) is correct because it is the calculated separation using Coulomb's law with the given charges and force.
189 Medium
189 of 385
Analyzing +4 | -1
Assertion: Two charges q1{q}_1 and q2{q}_2 are placed at separation rr. Then magnitude of force on each charge is FF.

Reason: Now a third charge q3{q}_3 is placed near q1{q}_1 and q2{q}_2 Then force on q1{q}_1 due to q2{q}_2 remains F{F}.
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 the force between two charges is independent of the presence of other charges, although the net force on a charge will be affected.
190 Medium
190 of 385
Understanding +4 | -1
The electric field at a distance 3R2\frac{{3R}}{2} from the centre of a charged conducting spherical shell of radius R  is  ER\;\mathrm{is}\;E. The electric field at a distance R2\frac{R}{2} from the centre of the sphere is
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because the electric field inside a charged conducting spherical shell is always zero.
191 Medium
191 of 385
Understanding +4 | -1
Select the correct statements from the following
(i) inside a charged or neutral conductor, electrostatic field is zero
(ii) the electrostatic field at the surface of the charged conductor must be tangential to the surface at any point
(iii) there is no net charge at any point inside the conductor
View answer and explanation
Correct option: B
(i) and (iii)
Option B ((i) and (iii)) is correct because inside a charged or neutral conductor, the electrostatic field is zero, and there is no net charge at any point inside the conductor.
192 Medium
192 of 385
Application +4 | -1
Two identical conducting balls AA and BB have positive charges q1{q_1} and q2{q_2} respectively but q1≠q2.{q_1} \ne {q_2}. The balls are brought together so that they touch each other and then kept in their original positions. The force between them is
View answer and explanation
Correct option: B
Greater than that before the balls touched
Option B (Greater than that before the balls touched) is correct because after touching, the charges redistribute, resulting in a larger product of charges and thus a greater force.
193 Medium
193 of 385
Application +4 | -1
A 10 μC10\,\mu C charge is divided into two equal parts and kept at 1 cm distance. Find repulsion between charges.
View answer and explanation
Correct option: C
2250 N
Option C (2250 N) is correct because it is calculated using Coulomb's law with the given charge and distance.
194 Medium
194 of 385
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.
195 Medium
195 of 385
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.
196 Medium
196 of 385
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}}.
197 Medium
197 of 385
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.
198 Medium
198 of 385
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.
199 Medium
199 of 385
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}.
200 Medium
200 of 385
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.
201 Medium
201 of 385
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.
202 Medium
202 of 385
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.
203 Medium
203 of 385
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.
204 Medium
204 of 385
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}.
205 Medium
205 of 385
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.
206 Medium
206 of 385
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.
207 Medium
207 of 385
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.
208 Medium
208 of 385
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.
209 Medium
209 of 385
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.
210 Medium
210 of 385
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.
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 181-210 of 385 questions

Your practice

0 / 385 answered · 385 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.