Electric Charges and Fields NEET PYQs (2005–2026) with Solutions

10 solved Electric Charges and Fields PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2025)

   NEET 2025MediumElectric Charge & Coulomb's Law

   Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is F. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres A and B (Radius of A and B are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as :

   1. AOption A: 3F/5
   2. BOption B: 2F/3
   3. COption C: F/2
   4. DOption D: 3F/8

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   Correct answer: (D) 3F/8

   Explanation

   Initially: $$F = \frac{kq^2}{r^2}$$ When the uncharged sphere touches A, charge distributes equally: A = q/2, third sphere = q/2 Then third sphere touches B (which has q): Total charge = q + q/2 = 3q/2 After equal distribution: B = 3q/4 Thus final charges are: $$q_A = q/2, \quad q_B = 3q/4$$ New force: $$F' = \frac{k}{r^2} \times \frac{q}{2} \times \frac{3q}{4} = \frac{3}{8}F$$

2. Question 2 (NEET 2025)

   NEET 2025MediumElectric Dipole

   An electric dipole with dipole moment 5 × 10⁻⁶ Cm is aligned with the direction of a uniform electric field of magnitude 4 × 10⁵ N/C. The dipole is then rotated through an angle of 60° with respect to the electric field. The change in the potential energy of the dipole is :

   1. AOption A: 0.8 J
   2. BOption B: 1.0 J
   3. COption C: 1.2 J
   4. DOption D: 1.5 J

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   Correct answer: (B) 1.0 J

   Explanation

   Potential energy of a dipole in an electric field is: U = -pEcosθ Initially, θ₁ = 0° U₁ = -pE Finally, θ₂ = 60° U₂ = -pEcos60° = -pE/2 Change in potential energy: ΔU = U₂ - U₁ = (-pE/2) - (-pE) = pE/2 Substituting values: ΔU = (5 × 10⁻⁶ × 4 × 10⁵)/2 = 1 J Therefore, option (B) is correct.

3. Question 3 (NEET 2024)

   NEET 2024EasyElectric Dipole

   Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The potential $V$ at any axial point, at $2\,m$ distance $(r)$ from the centre of the dipole of dipole moment vector $\vec{P}$ of magnitude $4\times10^{-6}\,C\,m$, is $\pm 9\times10^{3}\,V$. (Take: $\frac{1}{4\pi\varepsilon_0}=9\times10^9\;\text{SI units)}$ Reason R: $V = \pm \frac{2P}{4\pi\varepsilon_0 r^2}$ where r is the distance of any axial point situated at 2 m from the centre of the dipole. In the light of the above statements, choose the correct answer from the options given below:

   1. AOption A: A is true but R is false.
   2. BOption B: A is false but R is true.
   3. COption C: Both A and R are true and R is the correct explanation of A.
   4. DOption D: Both A and R are true and R is NOT the correct explanation of A.

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   Correct answer: (A) A is true but R is false.

   Explanation

   For an electric dipole, potential at an axial point is: $$V = \pm \frac{1}{4\pi\varepsilon_0}\frac{P}{r^2}$$ The given reason incorrectly uses: $$V = \pm \frac{2P}{4\pi\varepsilon_0 r^2}$$ Hence Reason R is false. Calculating Assertion A: $$V = \frac{9\times10^9\times4\times10^{-6}}{(2)^2}$$ $$V = 9\times10^3\;V$$ Thus Assertion A is true. Therefore, the correct option is: **(1) A is true but R is false.**

4. Question 4 (NEET 2023)

   NEET 2023MediumGauss's Law

   If $\oint_S \vec{E} \cdot d\vec{S} = 0$ over a surface, then:

   1. AOption A: the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
   2. BOption B: the magnitude of electric field on the surface is constant.
   3. COption C: all the charges must necessarily be inside the surface.
   4. DOption D: the electric field inside the surface is necessarily uniform.

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   Correct answer: (A) the number of flux lines entering the surface must be equal to the number of flux lines leaving it.

   Explanation

   Zero net electric flux means the number of electric field lines entering the surface equals the number leaving the surface.

5. Question 5 (NEET 2023)

   NEET 2023EasyGauss's Law

   The temperature of a gas is -50° C. To what temperature the gas should be heated so that the rms speed is increased by 3 times?

   1. AOption A: 669°C
   2. BOption B: 3295°C
   3. COption C: 3097 K
   4. DOption D: 223K

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   Correct answer: (B) 3295°C

   Explanation

   According to Gauss's law: $$\oint_S \vec{E} \cdot d\vec{S} = \frac{q_{\text{enc}}}{\varepsilon_0}$$ Given that net electric flux through the closed surface is zero: $$\oint_S \vec{E} \cdot d\vec{S} = 0 \Rightarrow q_{\text{enc}} = 0$$ This means the net number of electric field lines entering the surface equals the number leaving it. Hence option (A) is correct. From the E1 answer key, question 23 corresponds to answer code **2**, which maps to option **B** in the provided JSON format. Chapter and topic were mapped from Class 12 Physics syllabus under **Electrostatics → Gauss theorem**.

6. Question 6 (NEET 2023)

   NEET 2023MediumElectric Dipole

   An electric dipole is placed at an angle of $30^\circ$ with an electric field of intensity $2 \times 10^5\, \text{N C}^{-1}$. It experiences a torque equal to $4\, \text{N m}$. Calculate the magnitude of charge on the dipole, if the dipole length is $2\, \text{cm}$.

   1. AOption A: 8 mC
   2. BOption B: 6 mC
   3. COption C: 4 mC
   4. DOption D: 2 mC

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   Correct answer: (D) 2 mC

   Explanation

   Torque on an electric dipole is given by: $$\tau = pE\sin\theta$$ where: $$p = q \times (2l)$$ Given: $$\tau = 4\, \text{N m}$$ $$E = 2 \times 10^5\, \text{N C}^{-1}$$ $$\theta = 30^\circ$$ $$2l = 2\, \text{cm} = 0.02\, \text{m}$$ Substituting: $$4 = q(0.02)(2 \times 10^5)\sin 30^\circ$$ $$4 = q(0.02)(2 \times 10^5)\left(\frac{1}{2}\right)$$ $$4 = q \times 2000$$ $$q = \frac{4}{2000} = 2 \times 10^{-3}\, \text{C}$$ $$q = 2\, \text{mC}$$ Hence, option (D) is correct.

7. Question 7 (NEET 2023)

   NEET 2023MediumElectric Dipole

   An electric dipole is placed as shown in the figure. The electric potential (in 10² V) at point P due to the dipole is (ε₀ = permittivity of free space and 1/4π ε₀ = K) :

   1. AOption A: (3/8) qK
   2. BOption B: (5/8) qK
   3. COption C: (8/5) qK
   4. DOption D: (8/3) qK

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   Correct answer: (A) (3/8) qK

   Explanation

   From the figure: Distance of point P from centre O is 5 cm. Dipole charges are separated symmetrically with: $$O(-q)=3\text{ cm}$$ Hence, $$(-q)P=5+3=8\text{ cm}$$ and $$(+q)P=5-3=2\text{ cm}$$ Electric potential at P: $$V=K\left(\frac{q}{2}-\frac{q}{8}\right)$$ $$V=Kq\left(\frac{4-1}{8}\right)$$ $$V=\frac{3}{8}qK$$ Hence, the correct option is (1).

8. Question 8 (NEET 2022)

   NEET 2022EasyElectric Field & Field Lines

   The angle between the electric lines of force and the equipotential surface is:

   1. AOption A: 180°
   2. BOption B: 0°
   3. COption C: 45°
   4. DOption D: 90°

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   Correct answer: (D) 90°

   Explanation

   Electric field lines are always perpendicular to equipotential surfaces because no work is done in moving along an equipotential surface. Hence, the angle between them is 90°. Therefore option (D) is correct.

9. Question 9 (NEET 2020)

   NEET 2020MediumApplications of Gauss's Law

   A spherical conductor of radius 10 cm has a charge of $3.2 \times 10^{-7}$ C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere? Take: $$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\,\text{Nm}^2\text{/C}^2$$

   1. AOption A: $1.28 \times 10^4\,\text{N/C}$
   2. BOption B: $1.28 \times 10^5\,\text{N/C}$
   3. COption C: $1.28 \times 10^6\,\text{N/C}$
   4. DOption D: $1.28 \times 10^7\,\text{N/C}$

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   Correct answer: (B) $1.28 \times 10^5\,\text{N/C}$

   Explanation

   For a point outside a charged spherical conductor: $$E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$$ Substituting: $$E = \frac{9 \times 10^9 \times 3.2 \times 10^{-7}}{(0.15)^2}$$ $$E = 1.28 \times 10^5\,\text{N/C}$$

10. Question 10 (NEET 2018)

    NEET 2018EasyMotion of Charged Particles in Electric Field

    An electron falls from rest through a vertical distance $h$ in a uniform and vertically upward directed electric field $E$. The direction of the electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance $h$. The time of fall of the electron, in comparison to the time of fall of the proton is:

    1. AOption A: Smaller
    2. BOption B: 5 times greater
    3. COption C: Equal
    4. DOption D: 10 times greater

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    Correct answer: (A) Smaller

    Explanation

    For motion under constant acceleration: $$h = \frac{1}{2}at^2$$ In an electric field: $$a = \frac{eE}{m}$$ Therefore, $$t = \sqrt{\frac{2hm}{eE}}$$ Thus, $$t \propto \sqrt{m}$$ Since the mass of the electron is much smaller than the mass of the proton, the electron takes less time to fall. Hence, the correct answer is: $$\boxed{\text{Smaller}}$$