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JEE • PHYSICS • ELECTRIC FIELDS AND CHARGES

Electric fields and charges practice · Mixed difficulty

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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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Showing 121-150 of 420 questions

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121 Medium
121 of 420
Application +4 | -1
Electrical force between two point charges is 200N. If we increase 10% charge on one of the charges and decrease 10% charge on the other, then electrical force between them for the same distance becomes __________ N
View answer and explanation
Correct option: A
198
Option A (198) is correct because the new force is calculated by multiplying the original force by the product of the new charges relative to the original charges.
122 Medium
122 of 420
Application +4 | -1
Two insulated charged metallic spheres PP and QQ have their centres separated by a distance of 50 cm50 \mathrm{~cm}. The radii of PP and QQ are negligible compared to the distance of separation. The mutual force of electrostatic repulsion if the charge on each is 3.2×10−7C3.2 \times 10^{-7} \mathrm{C} is:
View answer and explanation
Correct option: A
3.69×10−3 N3.69 \times 10^{-3} \mathrm{~N}
Option A (3.69×10−3 N3.69 \times 10^{-3} \mathrm{~N}) is correct because it calculates the electrostatic force using Coulomb's law with the given charges and separation distance.
123 Medium
123 of 420
Understanding +4 | -1
A comb run through one’s dry hair attracts small bits of paper. This is due to
View answer and explanation
Correct option: C
The atoms in the paper get polarised by the charged comb
Option C is correct because the charged comb polarizes the atoms in the paper, creating an attractive force.
124 Hard
124 of 420
Analyzing +4 | -1
Two equal negative charges −q-q are fixed at the points (0,a)(0, a) and (0,−a)(0,-a) on the yy-axis. A positive charge QQ is released from rest from the point (2a,0)(2 a, 0 ) on the xx-axis. The charge will
View answer and explanation
Correct option: D
Execute oscillatory motion but not S.H.M
Option D (Execute oscillatory motion but not S.H.M) is correct because the force on the charge Q is not proportional to its displacement from the origin, thus it oscillates but not in simple harmonic motion.
125 Medium
125 of 420
Application +4 | -1
FgF_g and FeF_e represents gravitational and electrostatic force respectively between electron and proton at a distance of 10 cm10 \mathrm{~cm}. The ratio of Fg/FeF_g / F_e is of the order of
View answer and explanation
Correct option: B
10−3910^{-39}
The correct answer is B (10−3910^{-39}) because the ratio of gravitational to electrostatic force is approximately 10−3910^{-39} when calculated using the respective force equations.
126 Medium
126 of 420
Analyzing +4 | -1
A positively charged thin metal ring of radius
RR is fixed in the xy-plane, with its centre at the
origin. A negatively charged particle is released
from rest at the point P(0,0,c)P (0, 0, c) where c>0.c > 0.

Consider the following statements:
(A) The electric field at PP due to the metallic ring
is always directed towards origin.

(B) The negatively charged particle is accelerated
towards origin and undergoes periodic
motion.

Which of the above statements is/are correct?
View answer and explanation
Correct option: B
B only
Option B (B only) is correct. The electric field at P is along the z-axis, and the negatively charged particle accelerates towards the origin, resulting in periodic motion.
127 Medium
127 of 420
Understanding +4 | -1
Additive nature of charge means
View answer and explanation
Correct option: D
It tells about the scalar nature of charge
Option D (It tells about the scalar nature of charge) is correct because the additive nature of charge implies that charges can be added algebraically, like scalars.
128 Medium
128 of 420
Understanding +4 | -1
A capacitor of capacitance CC is charged to potential V.V. The flux of the electric field through a closed surface enclosing the capacitor is
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because the net charge enclosed by the closed surface is zero, so the net electric flux is also zero according to Gauss's law.
129 Medium
129 of 420
Application +4 | -1
A solid sphere of radius RR has a charge Q{\cal Q} distributed in its volume with a charge density ρ=kra,\mathrm{\rho } = k{r^a}, where k    and  a  k\;\;\mathrm{and\;}a\mathrm{\;} 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 a.a.
View answer and explanation
Correct option: A
2
Option A (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.
130 Medium
130 of 420
Application +4 | -1
A tin nucleus (atomic number Z = 50) has a radius of 6.6×10−15 m6.6 \times {10^{ - 15}}\,m. The potential on its surface will be (the charge on the proton =1.6×10−19 C= 1.6 \times {10^{ - 19}}\,C)
View answer and explanation
Correct option: A
1.1×107 V1.1 \times {10^7}\,\mathrm{V}
Option A (1.1×107 V1.1 \times {10^7}\,\mathrm{V}) is correct because it accurately calculates the electric potential on the surface of the tin nucleus using the formula V=14πε0⋅qrV = \frac{{1}}{{4\pi \varepsilon _0}} \cdot \frac{q}{r}.
131 Medium
131 of 420
Understanding +4 | -1
A thin metallic spherical shell contains a charge QQ on its surface. A point charge q1\mathrm{q}_1 is placed at the centre of the shell and another charge q2\mathrm{q}_2 is placed outside the shell. All the three charges are positive. Then the force on charge q1\mathrm{q}_1 is -
View answer and explanation
Correct option: C
Zero
Option C (Zero) is correct because the electric field at the center of a uniformly charged spherical shell is zero, resulting in no force on the charge q1.
132 Hard
132 of 420
Application +4 | -1
Let there be a spherically symmetric charge distribution with charge density varying as ρ(r)=ρ0[23−2rR]\rho(r)=\rho_0\left[\frac{2}{3}-\frac{2 r}{R}\right] upto r=Rr=R and ρ(r)=0\rho(r)=0 for r>Rr>R, where rr is the distance from origin. The electric field at a distance r(r<R)r(r < R) from the origin is given by:
View answer and explanation
Correct option: B
ρ0rε0[29−r2R]\frac{\rho_0 r}{\varepsilon_0}\left[\frac{2}{9}-\frac{r}{2 R}\right]
Option B is correct because the electric field is calculated using Gauss's law by integrating the charge density over a spherical volume and then applying the formula E = Kq/r2.
133 Medium
133 of 420
Understanding +4 | -1
When a body is earth connected, electrons from the earth flow into the body. This means the body is
View answer and explanation
Correct option: B
Charged positively
Option B (Charged positively) is correct because electrons flow into the body to neutralize a positive charge.
134 Medium
134 of 420
Understanding +4 | -1
Statement-1: Electric field outside the uniformly charged spherical shell (due to shell) is the same as that of a point charge at its center.
Statement-2: The centre of mass of uniform shell is at its center.
View answer and explanation
Correct option: B
Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation for Statement-1
Option B is correct because the electric field outside a uniformly charged spherical shell is the same as that of a point charge at its center, and the center of mass of a uniform shell is at its center, but the latter doesn't explain the former.
135 Medium
135 of 420
Application +4 | -1
A mass of 1 kg1 \mathrm{~kg} carrying a charge of 2 C2 \mathrm{~C} is accelerated through a potential of 1 V.1 \mathrm{~V}. The velocity acquired by it is:
View answer and explanation
Correct option: B
2 ms−12 \mathrm{~ms}^{-1}
Option B (2 ms−12 \mathrm{~ms}^{-1}) is correct because the kinetic energy gained by the charged particle is equal to the work done by the electric field, which is qV.
136 Medium
136 of 420
Analyzing +4 | -1
Five balls numbered 1 to 5 are suspended using separate threads. Pairs (1,2),(2,4)(1,2),(2,4) and (4,1)(4,1) show electrostatic attraction, while pair (2,3)(2,3) and (4,5)(4,5) show repulsion. Therefore ball 1 must be
View answer and explanation
Correct option: C
neutral
Option C is correct. Since balls 2 and 4 repel balls 3 and 5 respectively, balls 2 and 4 must have the same charge. Balls 1, 2, and 4 attract each other, which is only possible if ball 1 is neutral.
137 Medium
137 of 420
Application +4 | -1
Under the action of a given coulombic force the acceleration of an electron is 2.5×1022m/s2.2.5 \times {10^{22}}\mathrm{m}/{\mathrm{s}^2}. Then the magnitude of acceleration of a proton under the action of same force is nearly
View answer and explanation
Correct option: C
1.5×1019m/s21.5 \times {10^{19}}\mathrm{m}/{\mathrm{s}^2}
Option C (1.5×1019m/s21.5 \times {10^{19}}\mathrm{m}/{\mathrm{s}^2}) is correct because it correctly applies Newton's second law and the relationship between force, mass, and acceleration.
138 Medium
138 of 420
Understanding +4 | -1
The magnitude of electric field intensity at a point outside uniformly charged thin infinite plane sheet is independent of
View answer and explanation
Correct option: D
Distance of the point from plane sheet
Option D (Distance of the point from plane sheet) is correct because the electric field due to an infinite charged sheet is uniform and independent of the distance from the sheet.
139 Medium
139 of 420
Application +4 | -1
A point charge qq produces an electric field of magnitude 22 NC−1\mathrm{N}{\mathrm{C}^{ - 1}} at a point distance 0.25  m0.25\;\mathrm{ }m from it. What is the value of charge?
View answer and explanation
Correct option: A
1.39×10−111.39 \times {10^{ - 11}} C
Option A (1.39×10−111.39 \times {10^{ - 11}} C) is correct because it is calculated directly from the electric field formula using the given magnitude and distance.
140 Medium
140 of 420
Analyzing +4 | -1
Statement-1: When a charge is placed at rest in an electric field its path will always be along electric line of force.
Statement-2: The force on the charge is along the tangent drawn on electric field line.
View answer and explanation
Correct option: D
Statement-1 is false, Statement-2 is true
Option D (Statement-1 is false, Statement-2 is true) is correct because while the force on a charge is along the tangent to the electric field line, the path of a charge placed at rest isn't always along the field line if the field line is curved.
141 Medium
141 of 420
Analyzing +4 | -1
A certain charge QQ is divided into two parts let’s say q1q_1 and q2q_2 then these two point charges are placed at a certain distance apart. The ratio q1q2,\frac{{{q_1}}}{{{q_2}}}, for which the electrostatic force between them will be maximum
View answer and explanation
Correct option: C
1
Option C is correct because the electrostatic force is maximized when the charges are divided equally, i.e., q1 = q2.
142 Medium
142 of 420
Understanding +4 | -1
Four metal conductors having different shapes 1. A sphere 2. Cylindrical 3. Pear 4. Lightning conductor Are mounted on insulating stands and charged. The one which is best suited to retain the charges for a longer time is
View answer and explanation
Correct option: A
1
Option A (1) is correct because a sphere distributes charge uniformly, minimizing charge leakage compared to other shapes.
143 Medium
143 of 420
Remembering +4 | -1
The electrostatic force between two point charges is directly proportional to the
View answer and explanation
Correct option: D
Product of the charges
Option D (Product of the charges) is correct because Coulomb's law states that the electrostatic force is directly proportional to the product of the magnitudes of the charges.
144 Easy
144 of 420
Understanding +4 | -1
A sphere of radius 1 m encloses a charge of 5μC.5\mu C. Another charge of −5μC-5 \mu C is placed inside the sphere.
The net electric flux would be
View answer and explanation
Correct option: C
Zero
Option C is correct because according to Gauss's law, the net electric flux through a closed surface depends only on the net charge enclosed. Since the total charge inside the sphere is zero (+5μC - 5μC = 0), the net electric flux is zero.
145 Medium
145 of 420
Understanding +4 | -1
The number of electric lines that emerge from a finite charge +q+q is
View answer and explanation
Correct option: C
Proportional to the charge
Option C (Proportional to the charge) is correct because Gauss's law states that the electric flux is proportional to the enclosed charge.
146 Medium
146 of 420
Understanding +4 | -1
Assertion:- Electric field inside a metal surface is zero in electrostatic condition.
Reason:-Electric charges exist only on the 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 is correct because both the assertion and the reason are true, but the reason doesn't correctly explain the assertion. The electric field inside a metal is zero due to charge redistribution, not just the location of charges.
147 Medium
147 of 420
Analyzing +4 | -1
Which is not the unit of electric field :
View answer and explanation
Correct option: C
Joule / Coloumb
Option C (Joule / Coloumb) is correct because electric field is force per unit charge (N/C) or potential gradient (V/m), and J/C is Volt, not electric field.
148 Medium
148 of 420
Application +4 | -1
The force of interaction between two charges q1=6μ  {q_1} = 6\mu \; C and q2=2μ{q_2} = 2\mu C is 12 N. If charge q=−2μq = - 2\mu C is added to each of the charges, then the new force of interaction is
View answer and explanation
Correct option: B
Zero
Option B (Zero) is correct because adding the same charge to both original charges results in the product of the new charges being zero, thus the force is zero.
149 Medium
149 of 420
Understanding +4 | -1
One moving electron when comes closer to
other stationary electron, then its kinetic
energy and potential energy respectively _____
and _____
View answer and explanation
Correct option: C
Decreases, increases
The correct answer is C because as the moving electron approaches the stationary electron, its kinetic energy decreases due to repulsion, and its potential energy increases.
150 Medium
150 of 420
Understanding +4 | -1
A point charge +q+ q is placed at the midpoint of a cube of side aa. The electric flux emerging from the cube is
View answer and explanation
Correct option: C
qε0\frac{q}{{{\varepsilon _0}}}
The correct answer is C because according to Gauss's law, the electric flux emerging from a closed surface enclosing a charge q is q/ε₀, regardless of the shape or size of the surface.
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