Electric Charges and Fields NEET PYQReal NEET & AIPMT 2006–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Electric Charges and Fields question from the NEET & AIPMT papers on this site (2006–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Electric Charges and Fields previous year questions
16 questions. Tap an option to check your answer.
Q1NEET 2021
Polar molecules are the molecules :
Not quite — the answer is D.
In polar molecules such as H₂O and HCl the centres of positive and negative charge do not coincide, so they have a permanent dipole moment even without an external field. Acquiring a dipole only in an applied field describes non-polar molecules (induced dipoles).
Q2NEET 2020 · Q151
A spherical conductor of radius 10 cm has a charge of 3.2×10⁻⁷ C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere ? (1/4πε₀ = 9×10⁹ N m²/C²)
Not quite — the answer is B.
Outside the sphere it acts like a point charge: E = kq/r² = 9×10⁹ × 3.2×10⁻⁷ / (0.15)² = 2880/0.0225 = 1.28×10⁵ N/C. Using 10 cm (the radius) instead of 15 cm, or forgetting to convert cm to m, gives the other powers of ten.
Q3NEET 2019 · Q25
A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre:
Not quite — the answer is B.
By Gauss's law the field inside a charged conducting shell is zero, and outside it falls as kQ/r². A field increasing outside the sphere would contradict the inverse-square dependence.
Q4NEET 2019 · Q36
Two parallel infinite line charges with linear charge densities +λ C/m and −λ C/m are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges?
Not quite — the answer is C.
Each line gives E = λ/(2πε₀R) at distance R. At the midpoint the field of +λ points away from it and that of −λ points toward it, i.e. in the same direction, so they add: λ/(πε₀R). Zero would hold only for like charges.
Q5NEET 2019 · Q37
Two point charges A and B, having charges +Q and −Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes:
Not quite — the answer is B.
Transferring Q/4 from A to B leaves A with +3Q/4 and B with −Q + Q/4 = −3Q/4. Force ∝ product of charges, so F′ = (9/16)F. 16F/9 inverts the ratio.
Q6NEET 2018 · Q22
An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of 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
Not quite — the answer is C.
In both cases the electric force is downward with magnitude eE, so each accelerates at a = eE/m (plus g). Since t = √(2h/a) and the electron's mass is far smaller, its acceleration is much larger and its time of fall is smaller. The times are not equal because the masses differ.
Q7NEET 2016 Phase 2 · Q23
An electric dipole is placed at an angle of 30° with an electric field intensity 2 × 10⁵ N/C. It experiences a torque equal to 4 N m. The charge on the dipole, if the dipole length is 2 cm, is
Not quite — the answer is B.
τ = pE sinθ = q(2a)E sinθ, so 4 = q × 0.02 × 2 × 10⁵ × 0.5 = 2000q, giving q = 2 × 10⁻³ C = 2 mC. Using sin30° as 1 by mistake gives 1 mC, and forgetting to convert cm to m gives a far smaller charge.
Q8NEET 2016 Phase 1 · Q9
Two identical charged spheres suspended from a common point by two massless strings of lengths l, are initially at a distance d (d << l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then v varies as a function of the distance x between the spheres, as:
Not quite — the answer is C.
For small angles, equilibrium gives kq²/x² = mg·(x/2l), so q² ∝ x³ and q ∝ x^(3/2). Differentiating, dq/dt ∝ x^(1/2)·dx/dt; since dq/dt is constant, v = dx/dt ∝ x^(−1/2). The speed therefore increases as the spheres come closer, ruling out v ∝ x^(1/2) and v ∝ x.
Q9AIPMT 2015
The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sphere of radius 'a' centred at the origin of the field, will be given by:
Not quite — the answer is D.
At r = a the field is Aa, radial, so the flux through the sphere is Aa × 4πa² = 4πAa³. By Gauss's law q = ε₀ × flux = 4πε₀Aa³. Forgetting the 4π gives ε₀Aa³.
Q10NEET 2013
Two pith balls carrying equal charges are suspended from a common point by strings of equal length; the equilibrium separation between them is r. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now becomes
Not quite — the answer is B.
For a ball at equilibrium, tan θ = F/mg; with small angles tan θ ≈ (r/2)/L, so (r/2)/L = kq²/(r²mg), giving r³ ∝ L. Clamping at half the height halves the effective length, so r′³ = r³/2 and r′ = r/∛2.
Q11AIPMT 2012 Prelims
What is the flux through a cube of side a if a point charge of q is at one of its corners?
Not quite — the answer is C.
A charge at a corner can be enclosed symmetrically by 8 identical cubes, so each cube receives 1/8 of the total flux q/ε₀, i.e. q/(8ε₀). q/ε₀ would apply only if the charge were inside the cube.
Q12AIPMT 2011 Prelims
A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will
Not quite — the answer is D.
By Gauss's law, the total flux through any closed surface is q(enclosed)/ε₀. The enclosed charge is still Q when the radius doubles, so the flux remains the same; the field falls as 1/r² but the area grows as r².
Q13AIPMT 2010 Mains · Q64
The electric field at a distance 3R/2 from the centre of a charged conducting spherical shell of radius R is E. The electric field at a distance R/2 from the centre of the sphere is
Not quite — the answer is A.
All the charge of a conducting shell sits on its surface, and by Gauss's law the enclosed charge for any point inside (r = R/2) is zero, so the field there is zero. The value E at 3R/2 only describes points outside the shell and does not scale down inside.
Q14AIPMT 2010 Prelims · Q25
Two positive ions, each carrying a charge q, are separated by a distance d. If F is the force of repulsion between the ions, the number of electrons missing from each ion will be (e being the charge on an electron)
Not quite — the answer is D.
Each ion has charge q = ne. F = (ne)²/(4πε₀d²), so n² = 4πε₀Fd²/e² and n = √(4πε₀Fd²/e²). The same expression without the square root gives n², not n.
Q15AIPMT 2010 Prelims · Q26
A square surface of side L meter in the plane of the paper is placed in a uniform electric field E (volt/m) acting along the same plane at an angle θ with the horizontal side of the square. The electric flux linked to the surface, in units of volt-m, is
Not quite — the answer is A.
Flux = E·A cos(angle between E and the area vector). The area vector is normal to the paper, while E lies in the plane of the paper, so the angle is 90° and the flux is zero for any θ. EL² cos θ wrongly treats θ as the angle with the normal.
Q16AIPMT 2006 Prelims · Q43
An electric dipole of moment p⃗ is lying along a uniform electric field E⃗. The work done in rotating the dipole by 90° is :
Not quite — the answer is D.
Work done in rotating a dipole in a uniform field is W = pE(cos θ₁ − cos θ₂). From θ = 0° to 90°, W = pE(1 − 0) = pE. The value 2pE applies to rotating it through 180°.
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