18 NEET previous-year questions on Electric Charges and Fields, each with the correct answer and a step-by-step solution. Sourced directly from official NEET papers across every booklet code.
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Zero and 2 0 R 4 Q ∈ π
R 4 Q 0 ∈ π and Zero
R 4 Q 0 ∈ π and 2 0 R 4 Q ∈ π
Both are zero
Solution
For a conducting sphere, the electric potential at the center is , and the electric field inside the conductor is zero. Therefore, option (b) is correct.
N 5 6
30N
24 N
N 35 4
Solution
The electric field is given by . For , . At , . The force . The magnitude of is . The closest option is (d) .
−(3î+5ĵ+3k̂)
−(6î+5ĵ+2k̂)
−(2î+3ĵ+k̂)
−(6î+9ĵ+k̂)
Solution
V=6xy−y+ 24z E⃗⃗ =(∂V ∂xî+∂V ∂yĵ∂V ∂zk̂) E⃗⃗ =[(6y)î+(6x−1+ 2z)ĵ+(2y)k̂] E⃗⃗ =−(6î+5ĵ+2k̂) (1,1,0) 104. A remote-sensing satellite of earth revolves in a circular orbit at a height of 0.25×106 m above the surface of earth. If earth’s radius is 6.38×106 m and g=9.8 ms−2, then the orbital speed of the satellite is: (1) 7.76 km s−1 (2) 8.56 km s−1 (3) 9.13 km s−1 (4) 6.67 km s−1 Sol ution: (1) V0=√GM r =√GM R2.R2 r =√9.8×6.38×6.38 6.63×106 =√60×106 m/sec =7.76 km/sec
10 –20 C
10 –23 C
10 –37 C
10 –47 C
Solution
Fe = Fg 22 22 0 1 4 eG m dd Δ =πε 9 × 109 (Δe2) = 6.67 × 10 –11 × 1.67 × 10–27 × 1.67 × 10 –27 27 46.67 1.67 1.67 109e -××Δ= × 3710e -Δ≈
0.1°
0.266°
0.15°
0.05°
Solution
θ = λ 𝑑 θ ′= λ η×1 𝑑 θ θ ′= η 1, θ ′ =0.2 η = θ ′ θ ′ =0.15°
decreases as r increases for 𝑟<𝑅 and for 𝑟> 𝑅
increases as r increases for 𝑟<𝑅 and for 𝑟>𝑅
zero as r increases for 𝑟<𝑅, decreases as r increase for 𝑟>𝑅
zero as r increases for 𝑟<𝑅, increases as r increases for 𝑟>𝑅
Solution
𝐸 for a metal thin spherical shell 𝐸𝑖𝑛=0 𝑟<𝑅 𝐸0= 1 4πε0 .𝑄 𝑟 𝑟 >𝑅
C / NR02πε λ
Zero
C / NR0 2 πε λ
C / NR0πε λ
Solution
EP = λ 2πε0𝑅+ λ 2πε0𝑅= λ πε0𝑅
1.28 × 10 4 N/C
1.28 × 10 5 N/C
1.28 × 10 6 N/C
1.28 × 10 7 N/C
Solution
The electric field outside a charged spherical conductor is given by . Substituting the values, , so option (b) is correct.
towards the left as its potential energy will increase.
towards the right as its potential energy will decrease.
towards the left as its potential energy will decrease.
towards the right as its potential energy will increase.
Solution
A dipole in an electric field aligns such that it moves towards a region of lower potential energy. Since the potential energy decreases when the dipole aligns with the field, it will move towards the right, so option (b) is correct.
1 2 R R
2 1 R R
1 2 R R
2 2 1 2 R R
Solution
When two charged spherical conductors are connected by a wire, the potential on both spheres becomes equal. The surface charge density is related to the radius by . Therefore, , so option (b) is correct.
2 1 R
3 1 R
4 1 R
6 1 R
Solution
For R >> L, arrangement is an electric dipole 3 0 2 ; 4 pE R = πε where p = qL 3 1E R ∝ - 18 - NEET (UG)-2022 (Code-Q1)
the electric field inside the surface is necessarily uniform.
the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
the magnitude of electric field on the surface is constant.
all the charges must necessarily be inside the surface.
Solution
If over a surface, it implies that the net electric flux through the surface is zero. This means the number of flux lines entering the surface is equal to the number of flux lines leaving it, as per Gauss's Law. Option (b) is correct.
2 mC
8 mC
6 mC
4 mC
Solution
The torque on an electric dipole is given by , where . Substituting the values, . Thus, option (a) is correct.
(8/3)qK
(3/8)qK
(5/8)qK
(8/5)qK
Solution
The electric potential at point P due to a dipole is given by , where and . Substituting, . However, the correct formula for the perpendicular bisector is . Thus, the correct answer is , so option (b) is correct.
Both A and R are true and R is the correct explanation of A.
Both A and R are true and R is NOT the correct explanation of A.
A is true but R is false.
A is false but R is true.
Solution
The potential V at any point, at distance r from centre of dipole 2 cosθ= KP r At axial point where θ = 0°, 96 3 22 9 10 4 10 9 10 V 2 −× × ×= = = ×KPV r At axial point where θ = 180°, 3 2 9 10 V−= = − ×KPV r
3 × 105
1 × 105
0.5 × 105
Zero
Solution
For uniformly charged spherical shell, = kqV R (For r ≤ R) ∴ VC = VP VC – VP = Zero
Solution
Sol.
After contact:
Charge on A becomes , charge on B becomes
0.8 J
1.0 J
1.2 J
1.5 J
Solution
Sol. Given
and
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