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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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271 Medium
271 of 385
Analyzing +4 | -1
Assertion: At the centre of the line joining two equal and opposite charges, E=0E=0.

Reason: At the centre of the line joining two equal and similar charge, E≠0E \neq 0.
View answer and explanation
Correct option: D
If both Assertion and Reason are false.
The correct answer is D because both the Assertion and the Reason are false; the electric field is NOT zero at the center of two equal and opposite charges, and it IS zero at the center of two equal and similar charges.
272 Medium
272 of 385
Analyzing +4 | -1
Point charges +4q,−q+ 4q, - q and +4q+ 4q are kept on the xx-axis at points x=0,x=ax = 0,x = a and x=2ax = 2a respectively, then
View answer and explanation
Correct option: C
All the charges are in unstable equilibrium
Option C (All the charges are in unstable equilibrium) is correct because while the net force on each charge is zero, any displacement will cause a net force, leading to instability.
273 Medium
273 of 385
Application +4 | -1
Surface density of charge on a charged conducting sphere of radius 'RR' in terms of electric intensity 'EE' at a distance 'rr' in free space is (r>R,ε0=Permittivity of free space)\left(r>R, \varepsilon_0= \text{Permittivity of free space}\right)
View answer and explanation
Correct option: C
ε0E(rR)2\varepsilon_0 E\left(\frac{r}{R}\right)^2
Option C (ε0E(rR)2\varepsilon_0 E\left(\frac{r}{R}\right)^2) is correct because it correctly relates the surface charge density to the electric field at a distance r from the sphere.
274 Medium
274 of 385
Analyzing +4 | -1
A charge is given velocity perpendicular to uniform electric field then?

Assertion: Initial power delivered by electric field is zero.

Reason: Path of charged particle is circular.
View answer and explanation
Correct option: C
Assertion is true & Reason is false
Option C is correct. The initial power delivered is zero because the initial force and velocity are perpendicular. The path is parabolic, not circular.
275 Medium
275 of 385
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
Option B (xz plane) is correct because the electric field is in the y-direction, and equipotential surfaces are perpendicular to the electric field.
276 Medium
276 of 385
Understanding +4 | -1
Three electric dipoles of dipole moment P,2P,3P{P}, 2 {P}, 3 {P} are placed inside a hollow sphere of radius ' aa '. The total electric flux coming out of the sphere is
View answer and explanation
Correct option: B
Zero
Option B (Zero) is correct because the net charge enclosed by a hollow sphere due to dipoles is zero, resulting in zero electric flux.
277 Medium
277 of 385
Application +4 | -1
If a charge on the body is 1 nC,1 \mathrm{~nC}, then how many electrons are present on the body?
View answer and explanation
Correct option: B
6.25×1096.25 \times 10^9
Option B (6.25×1096.25 \times 10^9) is correct because it is the number of electrons calculated using the quantization of charge formula.
278 Medium
278 of 385
Application +4 | -1
Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 18×10−22 Cm−218 \times 10^{-22} \mathrm{~Cm}^{-2}. The electric field between the plates is:
View answer and explanation
Correct option: A
2.03×10−10 NC−12.03 \times 10^{-10} \mathrm{~NC}^{-1}
Option A (2.03×10−10 NC−12.03 \times 10^{-10} \mathrm{~NC}^{-1}) is correct because the electric field between two large, thin metal plates with opposite surface charge densities is given by σε0\frac{\sigma}{\varepsilon_0}.
279 Medium
279 of 385
Understanding +4 | -1
A cube of side aa is placed in a uniform electric field E=E0i^+E0j^+E0k^{E}=E_0 \hat{{i}}+E_0 \hat{{j}}+E_0 \hat{{k}}. Total electric flux passing through the cube would be
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the net electric flux through a closed surface in a uniform electric field is always zero.
280 Medium
280 of 385
Understanding +4 | -1
The electrostatic field due to a charged
conductor just outside the conductor is:
View answer and explanation
Correct option: D
Normal to the surface at every point and zero
inside the conductor
Option D (Normal to the surface at every point and zero inside the conductor) is correct because the electric field inside a conductor is zero, and the electric field just outside is normal to the surface.
281 Medium
281 of 385
Understanding +4 | -1
There are two types of electric charges positive charges and negative charges. The property which differentiates the two types of charges is:
View answer and explanation
Correct option: B
Polarity of charge
Option B is correct because the polarity of charge distinguishes positive and negative charges.
282 Medium
282 of 385
Application +4 | -1
A current in a wire is given by the equation I=2t2−3t+1I = 2{t^2} - 3t + 1 . The charge through cross section of wire in time interval t=3 s   t0t=5st = 3\,s\,\,\,{t_0}t = 5s is
View answer and explanation
Correct option: B
43.34 L
Option B (43.34 L) is correct because it is the charge calculated by integrating the current function over the given time interval.
283 Medium
283 of 385
Application +4 | -1
An infinite number of electric charges each equal to 5  nano−coulomb5\;nano - coulomb (magnitude) are placed along XX-axis at x=1  cm,    x=2  cm,    x=4  cm    x=8  cm………x = 1\;cm,\;\;x = 2\;cm,\;\;x = 4\;cm\;\;x = 8\;cm \ldots \ldots \ldots and so on. In the setup if the consecutive charges have opposite sign, then the electric field in Newton/CoulombNewton/Coulomb at X=0X = 0 is

[14πε0=9×109N−m2/c2]\left[ {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}N - {m^2}/{c^2}} \right]
View answer and explanation
Correct option: C
36×10436 \times {10^4}
Option C (36×10436 \times {10^4}) is correct because the electric field is calculated by summing the contributions of all charges, considering their alternating signs and positions.
284 Medium
284 of 385
Application +4 | -1
Two similarly equally charged identical metal spheres AA and BB repel each other with a force of 2.0×10−5 N2.0 \times 10^{-5} \mathrm{~N}. AA third identical uncharged sphere CC is touched to AA, then placed at the midpoint between AA and BB. Calculate the net electrostatic force on CC.
View answer and explanation
Correct option: B
2.0×10−5 N2.0 \times 10^{-5} \mathrm{~N}
Option B (2.0×10−5 N2.0 \times 10^{-5} \mathrm{~N}) is correct because it is the net electrostatic force on sphere C, calculated by considering the forces due to spheres A and B.
285 Medium
285 of 385
Application +4 | -1
Two particles, each having a mass of 5 g5 \mathrm{~g} and charge 1.0×10−7C1.0 \times 10^{-7} \mathrm{C}, stay in limiting equilibrium on a horizontal table with a separation of 10 cm10 \mathrm{~cm} between them. The coefficient of friction between each particle and the table is the same. Find the value of this coefficient
View answer and explanation
Correct option: B
0.18
Option B (0.18) is correct because the coefficient of friction is found by equating the electric force and the frictional force in limiting equilibrium.
286 Medium
286 of 385
Application +4 | -1
Calculate the net charge on a substance consisting of a combination of 7×10137 \times 10^{13} protons and 4×10134 \times 10^{13} electrons.
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Correct option: A
+4.8 μC+4.8\,\mu \mathrm{C}
Option A (+4.8 μC) is correct because it is calculated by finding the difference between the total positive charge of protons and the total negative charge of electrons.
287 Medium
287 of 385
Understanding +4 | -1
Assertion: Force between two point charges at rest is not changed by the presence of third point charge between them.

Reason: Force depends on the magnitude of the first two charges and seperation between them.
View answer and explanation
Correct option: A
Both Assertion & Reason are true. Reason is the correct explanation for assertion
Option A (Both Assertion & Reason are true. Reason is the correct explanation for assertion) is correct because the force between two point charges depends only on their magnitudes and the separation between them, as described by Coulomb's law.
288 Medium
288 of 385
Application +4 | -1
Two charges of equal magnitudes and at a distance rr exert a force FF on each other. If the charges are halved and distance between them is doubled, then the new force acting on each charge is
View answer and explanation
Correct option: D
F/16F/16
Option D (F/16F/16) is correct because the force is proportional to the square of the charge and inversely proportional to the square of the distance.
289 Medium
289 of 385
Understanding +4 | -1
A point charge + q+\,q, is placed at a distance d from an isolated conducting plane. The field at a point PP on the other side of the plane is
View answer and explanation
Correct option: A
directed perpendicular to the plane and away from the plane.
Option A (directed perpendicular to the plane and away from the plane) is correct because the electric field lines from the charge induce an opposite charge on the conducting plane, resulting in a field perpendicular and away from the plane on the other side.
290 Medium
290 of 385
Understanding +4 | -1
Two bodies are charged by rubbing one against the other. During the process, one becomes positively charged while the other becomes negatively charged. Then
View answer and explanation
Correct option: D
Mass of each body changes slightly but the total mass remains the same.
Option D (Mass of each body changes slightly but the total mass remains the same) is correct because electrons are transferred, changing individual masses, but the total mass is conserved.
291 Medium
291 of 385
Application +4 | -1
A point charge placed at any point on the axis of an electric dipole at some large distance experiences a force FF. Find the force acting on the point charge when it's distance from the dipole is doubled.
View answer and explanation
Correct option: C
F′=F8F^{\prime}=\frac{F}{8}
Option C (F′=F8F^{\prime}=\frac{F}{8}) is correct because the force on a point charge due to an electric dipole varies inversely with the cube of the distance.
292 Medium
292 of 385
Application +4 | -1
An electron of mass Me,  {M_e},\; initially at rest, moves through a certain distance in a uniform electric field in time t1.  {t_1}.\; A proton of mass Mp{M_p} also initially at rest, takes time t2{t_2} to move through an equal distance in this uniform electric field, neglecting the effect of gravity, the ratio t1/t2  {t_1}/{t_{2\;}} is nearly equal to
View answer and explanation
Correct option: B
Mp/Me\sqrt {{M_p}/{M_e}}
Option B (Mp/Me\sqrt {{M_p}/{M_e}}) is correct because it is derived from the relationship between time, distance, mass, and acceleration in a uniform electric field.
293 Medium
293 of 385
Understanding +4 | -1
An electric charge does not have which of the
following properties?
View answer and explanation
Correct option: D
Circular line of force
Option D (Circular line of force) is correct because electric field lines originate from positive charges and terminate at negative charges, and they are not circular.
294 Medium
294 of 385
Application +4 | -1
A charged particle of mass 0.003 g0.003 \mathrm{~g} is held stationary is space by placing it in a downward direction of electric field of 6×104 N/C6 \times 10^4 \mathrm{~N} / \mathrm{C}. Then the magnitude of the charge is
View answer and explanation
Correct option: B
5×10−10C5 \times 10^{-10} \mathrm{C}
Option B (5×10−10C5 \times 10^{-10} \mathrm{C}) is correct because the charge is calculated by equating the electric force to the gravitational force.
295 Medium
295 of 385
Understanding +4 | -1
When we touch a pith ball with an electrified plastic rod, some of the negative charges on the rod are transferred to the pith ball and it also gets charged. Thus, pith ball is charged by
View answer and explanation
Correct option: B
Contact
Option B (Contact) is correct because the pith ball acquires charge through direct physical contact with the charged rod.
296 Medium
296 of 385
Understanding +4 | -1
In nature, the electric charge of any system is always equal to
View answer and explanation
Correct option: D
Integral multiple of the least amount of charge
Option D is correct because electric charge is quantized, meaning it exists only in integer multiples of the elementary charge.
297 Medium
297 of 385
Application +4 | -1
A charge of 1 μ\mathrm{\mu }C is divided into two parts such that their charges are in the ratio of 2:3. These two charges are kept at a distance 1 m apart in vacuum. Then, the electric force between them (in N) is
View answer and explanation
Correct option: B
0.00216
Option B (0.00216) is correct because the electric force is calculated using Coulomb's law with the given charges and distance.
298 Medium
298 of 385
Application +4 | -1
Two conducting spheres have radii R1=0.5 mR_1=0.5 \mathrm{~m} and R2=1 mR_2=1 \mathrm{~m}. Each of them has 300 μC300 ~\mu \mathrm{C} charge on its surface. If the spheres are joined by a conducting wire then the amount of charge transferred from one body to the other is
View answer and explanation
Correct option: A
100 μC100 ~\mu \mathrm{C}
Option A (100 μC100 ~\mu \mathrm{C}) is correct because it is the amount of charge transferred to equalize the potential after connecting the spheres.
299 Medium
299 of 385
Understanding +4 | -1
Assertion: Two particles of same charge projected with different velocity normal to electric field experience same force.

Reason: A charged particle experiences force, independent of velocity in electric field.
View answer and explanation
Correct option: A
Both Assertion & Reason are true. Reason is the correct explanation for assertion
Option A (Both Assertion & Reason are true. Reason is the correct explanation for assertion) is correct because the force on a charged particle in an electric field is independent of its velocity, and this explains why particles with the same charge experience the same force.
300 Medium
300 of 385
Analyzing +4 | -1
Assertion : When we rub a glass rod with silk, the rod gets negatively charged and the silk gets positively charged.

Reason : On rubbing, electrons from silk cloth move to the glass rod.
View answer and explanation
Correct option: D
Both assertion and reason are false
Option D (Both assertion and reason are false) is correct because when a glass rod is rubbed with silk, the rod becomes positively charged and the silk becomes negatively charged due to electron transfer from the glass to the silk.
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