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JEE • PHYSICS • ELECTRIC FIELDS AND CHARGES

Electric fields and charges practice · Medium

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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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241 Medium
241 of 385
Understanding +4 | -1
The electric field near a conducting surface having a uniform surface charge density σ\sigma is given by
View answer and explanation
Correct option: C
σε0\frac{\sigma }{{{\varepsilon _0}}} and is normal to the surface
Option C (σε0\frac{\sigma }{{\varepsilon _0}} and is normal to the surface) is correct because the electric field near a conductor is perpendicular to the surface and its magnitude is given by the formula.
242 Medium
242 of 385
Application +4 | -1
n identical mercury droplets charged to the same potential VV coalesce to form a single bigger drop. The potential of new drop will be
View answer and explanation
Correct option: D
n2/3V{n^{2/3}}V
Option D (n2/3V{n^{2/3}}V) is correct because the potential of the new drop is proportional to n2/3n^{2/3} times the potential of the small droplets.
243 Medium
243 of 385
Application +4 | -1
Two identical small conducting spheres having unequal positive charges q1q_1 and q2q_2 are separated by a distance rr. If they are now made to touch each other and then separated again to the same distance, the electrostatic force between them in this case will be
View answer and explanation
Correct option: C
more than before
Option A (1.801.80) is correct because it is calculated using the relationship between electric field, magnetic field, and the speed of light in the medium.
244 Medium
244 of 385
Analyzing +4 | -1
A soap bubble of radius RR has uniformly distributed charge QQ on it's surface. It's energy is the self-energy of charges & surface energy due to surface tension. In equilibrium, this energy is minimum. Surface tension is S\mathrm{S}. At equilibrium radius of bubble is R & pressure inside is P\mathrm{P}. Pressure outside is P0\mathrm{P}_0. Then
View answer and explanation
Correct option: A
R=(Q2128π2∈0S)1/3R=\left(\frac{Q^2}{128 \pi^2 \in_0 S}\right)^{1 / 3}
Option A (R=(Q2128π2∈0S)1/3R=\left(\frac{{Q^2}}{{128 \pi^2 \in_0 S}}\right)^{1 / 3}) is correct because it represents the equilibrium radius derived from minimizing the total energy of the bubble.
245 Medium
245 of 385
Analyzing +4 | -1
Two point charges +Q+Q and +q+q are separated by a certain distance. If +Q>+q+Q > +q then in between the charges the electric field is zero at a point:
View answer and explanation
Correct option: B
Closer to +q+q
Option B (Closer to +q+q) is correct because the electric field is zero closer to the smaller charge due to the inverse square relationship.
246 Medium
246 of 385
Understanding +4 | -1
Assertion:- Electric field inside conductor is zero in electrostatics.
Reason:- Charge appears on surface of conductor.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because while both statements are true, the charge appearing on the surface doesn't directly cause the zero electric field inside.
247 Medium
247 of 385
Understanding +4 | -1
Assertion (A): Electric lines of force cross each other.
Reason (R): The resultant electric field at a point is the superimposition of the electric fields at that point.
View answer and explanation
Correct option: D
A is false and R is true
Option D is correct because electric field lines never intersect, and the resultant electric field at a point is indeed the superposition of individual electric fields.
248 Medium
248 of 385
Analyzing +4 | -1
Assertion: A charged particle free to move in an electric field always moves along an electric field line

Reason: The electric field lines diverge from a negative charge and converge at a positive charge.
View answer and explanation
Correct option: D
Assertion is false and Reason is false
Option D (Assertion is false and Reason is false) is correct because a charged particle's motion in an electric field depends on its initial conditions, and electric field lines diverge from positive charges and converge at negative charges.
249 Medium
249 of 385
Application +4 | -1
Two positive point charges are 3 m3 \mathrm{~m} apart and their combined charge is 20μC20 \mu \mathrm{C}. If the force between them is 0.075 N0.075 \mathrm{~N}, then the charges are:
View answer and explanation
Correct option: B
15μC,5μC15 \mu \mathrm{C}, 5 \mu \mathrm{C}
Option B (15μC,5μC15 \mu \mathrm{C}, 5 \mu \mathrm{C}) is correct because these charges satisfy both the combined charge and the force between them as given in the problem.
250 Medium
250 of 385
Understanding +4 | -1
Assertion (A): When bodies are charged through friction, there is transfer of charge from one body to another. No charge is created or destroyed.
Reason (R): This is according to the law of conservation of electric charge.
View answer and explanation
Correct option: A
Both A and R are true and R is the correct explanation of A
Option A (Both A and R are true and R is the correct explanation of A) is correct because the transfer of charge during friction is governed by the law of conservation of electric charge.
251 Medium
251 of 385
Application +4 | -1
A particle of mass m\text{m} and charge q\text{q} is thrown perpendicular to an electric field of intensity E\text{E} with an initial velocity v.\text{v}. The particle moves a distance xx perpendicular to the field and a distance y\text{y} along the direction of the field. If y=αx2\text{y}=\alpha \text{x}^2 then the α\alpha is given by:
View answer and explanation
Correct option: D
qE2mv2\frac{\mathrm{qE}}{2 \mathrm{mv}^2}
Option D (qE2mv2\frac{\mathrm{qE}}{2 \mathrm{mv}^2}) is correct because it is derived from the equations of motion under constant acceleration due to the electric field.
252 Medium
252 of 385
Application +4 | -1
Two concentric conducting thin shells of radius rr and 2r2 r carry charges +2q+2 q and +6q+6 q, respectively. The magnitude of electric field at a distance xx outside and inside from the surface of outer sphere is same, then the volume of xx is
View answer and explanation
Correct option: B
2r3\frac{2 r}{3}
Option B (2r3\frac{2 r}{3}) is correct because it represents the distance x where the electric fields due to the charged shells are equal outside and inside the outer sphere.
253 Medium
253 of 385
Analyzing +4 | -1
Assertion: If there exists coulomb attraction between two bodies, both of them must be charged.
Reason: In coulomb attraction two bodies are positively charged.
View answer and explanation
Correct option: D
If both assertion and reason are false
Option D (If both assertion and reason are false) is correct because Coulomb attraction can exist between a charged and an uncharged body due to induction, and the reason incorrectly states that both bodies must be positively charged.
254 Medium
254 of 385
Application +4 | -1
A conductor has been given a charge −3×10−7C- 3 \times {10^{ - 7}}C by transferring electron. Mass increase (in kgkg) of the conductor and the number of electron added to the conductor are respectively
View answer and explanation
Correct option: D
2×10−182 \times {10^{ - 18}} and 2×10122 \times {10^{12}}
Option D (2×10−182 \times {10^{ - 18}} and 2×10122 \times {10^{12}}) is correct because it accurately calculates the mass increase and the number of electrons transferred.
255 Medium
255 of 385
Understanding +4 | -1
Which of the following is correct regarding electric charge?
(i) If a body is having positive charge, then there is shortage of electrons.
(ii) If a body is having negative charge, then there is excess of electrons.
(iii) Minimum possible charge =±1.6×10−19C= \pm 1.6 \times 10^{-19} \mathrm{C}.
(iv) Charge is quantised, i.e. Q=±neQ= \pm n e, where n=1,2,3,4,…n=1,2,3,4, \ldots
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because all the statements regarding electric charge are accurate and fundamental principles.
256 Medium
256 of 385
Understanding +4 | -1
Assertion : Total flux through a closed surface is zero if no charge is enclosed by the surface.

Reason : Gauss law is true for any closed surface, no matter what its shape or size is.
View answer and explanation
Correct option: B
Both assertion and reason are true, but reason is not the correct explanation of assertion
Option B is correct because both the assertion and reason are true, but the reason doesn't directly explain why the flux is zero when no charge is enclosed.
257 Medium
257 of 385
Understanding +4 | -1
Assertion:- Two charged particles at rest experiences only electrical force.
Reason:- Charges at rest can only produce electric field.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because stationary charges only produce electric fields, and thus experience only electrical forces.
258 Medium
258 of 385
Analyzing +4 | -1
Assertion: A charged conductor may have charged particle inside it.

Reason: There can't exist electric field lines inside the conductor.
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. A charged conductor can have charged particles inside, and there are no electric field lines inside a conductor, but the latter doesn't explain the former.
259 Medium
259 of 385
Application +4 | -1
How much positive and negative charge is there in a cup of water ?
View answer and explanation
Correct option: C
1.33×107C1.33 \times 10^7 \mathrm{C}
Option C (1.33×107C1.33 \times 10^7 \mathrm{C}) is correct because it is the total positive or negative charge calculated based on the number of water molecules in a cup and the charge of each proton and electron.
260 Medium
260 of 385
Understanding +4 | -1
In finding the electric field using Gauss law the formula ∣E∣=qencε0∣A∣\left| E \right| = \frac{{{q_{\mathrm{enc}}}}}{{{\varepsilon _0}\left| A \right|}} is applicable. In the formula, ε0{\varepsilon _0} is permittivity of free space, AA is the area of Gaussian surface and qenc{q_{\mathrm{enc}}} is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation?
View answer and explanation
Correct option: A
Only when the Gaussian surface is an equipotential surface and ∣E∣|E| is constant on the surface.
Option A (Only when the Gaussian surface is an equipotential surface and ∣E∣|E| is constant on the surface) is correct because Gauss's law is easily applied when the electric field is constant and perpendicular to the Gaussian surface.
261 Medium
261 of 385
Application +4 | -1
A free pith-ball AA of 8 g8 \mathrm{~g} carries a positive charge of 5×10−8C5 \times 10^{-8} \mathrm{C}. What must be the nature and magnitude of charge that should be given to a second pith-ball BB fixed 5 cm5 \mathrm{~cm} below the former ball so that the upper ball is stationary?
View answer and explanation
Correct option: A
4.36×10−7C4.36 \times 10^{-7} \mathrm{C}
Option A (4.36×10−7C4.36 \times 10^{-7} \mathrm{C}) is correct because the electrostatic force of repulsion must balance the weight of the upper pith-ball.
262 Medium
262 of 385
Application +4 | -1
Radius of a hollow sphere is RR and a charge qq is placed at the centre of hollow sphere. If the radius of sphere becomes half and charge also becomes half, then the value of emergent total flux from the surface of sphere is
View answer and explanation
Correct option: C
q/2ε0q / 2 \varepsilon_0
Option C (q/2ε0q / 2 \varepsilon_0) is correct because the flux depends only on the enclosed charge, which is halved.
263 Medium
263 of 385
Understanding +4 | -1
A electric dipole is placed at an angle of 45∘45^\circ to a non – uniform electric field. The dipole will experience
View answer and explanation
Correct option: D
a torque as well as a translational force
Option D (a torque as well as a translational force) is correct because in a non-uniform electric field, a dipole experiences both a torque and a net force.
264 Medium
264 of 385
Application +4 | -1
A solid ball of radius RR has a charge density ρ\rho given by ρ=ρ0(1−rR)\rho=\rho_0\left(1-\frac{r}{R}\right) for 0≤r≤R0 \leq r \leq R. The electric field cutside the ball is
View answer and explanation
Correct option: A
ρ0R312ε0r2\frac{\rho_0 R^3}{12 \varepsilon_0 r^2}
The correct answer is A. The electric field outside the ball is calculated using Gauss's law, integrating the charge density over the volume of the sphere.
265 Medium
265 of 385
Application +4 | -1
A solid sphere of radius RR has a charge QQ distributed in its volume with a charge densityρ=k  ra\rho = k\;{r^a}, where kk and aa are constants and rr is the distance from its centre. If the electric field at r=R2r = \frac{R}{2} is 18\frac{1}{8} times that at r=R,r = R, find the value of aa
View answer and explanation
Correct option: C
2
Option C (2) is correct because by applying Gauss's law and comparing the electric fields at r = R/2 and r = R, we can solve for the value of a.
266 Medium
266 of 385
Analyzing +4 | -1
Statement-1: There is an isolated system of two charged conducting spheres AA and BB. The resultant electric field at point PP is the sum of electric field at PP due to charged sphere AA only (that is, assuming sphere BB and all its effects to be absent) and the electric field at PP only due to sphere BB (that is, assuming sphere AA and all its effects to be absent).
Statement-2: Superposition theorem for electric field due to point charges states that resultant electric field at a point due to point charges is the sum of electric field at that point due to individual charges.
View answer and explanation
Correct option: D
Statement-1 is false, Statement-2 is true
Option D is correct. Superposition requires the original charge distribution to remain unchanged, which isn't the case when conductors are present.
267 Medium
267 of 385
Application +4 | -1
Two points are at distances a\mathrm{a} and b(a<b)\mathrm{b}(\mathrm{a}<\mathrm{b}) from a long string of charge per unit length λ\lambda. The potential difference between the points is proportional to -
View answer and explanation
Correct option: D
log⁡(b/a)\log (b / a)
Option A (332.5) is correct because it is the calculated speed of sound in air at standard temperature and pressure using the corrected Newton's formula, considering the mass of one mole of air and the adiabatic index.
268 Medium
268 of 385
Remembering +4 | -1
Dielectric constant for metal is
View answer and explanation
Correct option: B
Infinite
Option B (Infinite) is correct because metals are conductors and have an infinite ability to store electrical energy, resulting in an infinite dielectric constant.
269 Medium
269 of 385
Application +4 | -1
The charge present in a doubly ionized helium atom is
View answer and explanation
Correct option: D
3.2×10−19C3.2 \times 10^{-19} \mathrm{C}
Option D (3.2×10−19C3.2 \times 10^{-19} \mathrm{C}) is correct because a doubly ionized helium atom has a charge of +2e.
270 Medium
270 of 385
Application +4 | -1
Three equal charges are placed on the three corners of a square. If the force between q1{q_1} and q2{q_2} is F12{F_{12}} and that between&nbsp; q1{q_1} and q3{q_3} is F13{F_{13}}, the ratio of magnitudes F12F13\frac{{{F_{12}}}}{{{F_{13}}}} is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the ratio of forces is inversely proportional to the square of the distance between the charges.
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