Electrostatic Potential and Capacitance 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 Electrostatic Potential and Capacitance 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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Electrostatic Potential and Capacitance previous year questions
26 questions. Tap an option to check your answer.
Q1NEET 2021
A parallel plate capacitor has a uniform electric field 'E' in the space between the plates. If the distance between the plates is 'd' and the area of each plate is 'A', the energy stored in the capacitor is : (ε₀ = permittivity of free space)
Not quite — the answer is C.
Energy density in the field is (1/2)ε₀E², and the field fills a volume Ad between the plates. So U = (1/2)ε₀E²Ad. The expression (1/2)ε₀E² alone is only energy per unit volume.
Q2NEET 2021
Two charged spherical conductors of radius R₁ and R₂ are connected by a wire. Then the ratio of surface charge densities of the spheres (σ₁/σ₂) is :
Not quite — the answer is B.
Connected conductors reach a common potential: kQ₁/R₁ = kQ₂/R₂, so Q ∝ R. Since σ = Q/(4πR²) ∝ 1/R, σ₁/σ₂ = R₂/R₁. Charge is proportional to R, but surface density is inversely proportional to R.
Q3NEET 2021
Twenty seven drops of same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.
Not quite — the answer is D.
Volume conservation: R = 27^(1/3) r = 3r; charge Q = 27q. V = kQ/R = (27/3)kq/r = 9 × 220 = 1980 V. Multiplying by 3 (660 V) uses the radius ratio instead of n^(2/3).
Q4NEET 2020 · Q152
In a certain region of space with volume 0.2 m³, the electric potential is found to be 5 V throughout. The magnitude of electric field in this region is :
Not quite — the answer is A.
E = −dV/dr, and a potential that is the same everywhere in the region has zero gradient, so the field is zero. Dividing the potential by the volume or any length has no physical meaning here.
Q5NEET 2020 · Q166
The capacitance of a parallel plate capacitor with air as medium is 6 µF. With the introduction of a dielectric medium, the capacitance becomes 30 µF. The permittivity of the medium is : (ε₀ = 8.85×10⁻¹² C² N⁻¹ m⁻²)
Not quite — the answer is C.
The dielectric constant is K = 30/6 = 5, so ε = Kε₀ = 5 × 8.85×10⁻¹² = 4.425×10⁻¹¹ ≈ 0.44×10⁻¹⁰ C² N⁻¹ m⁻². The value 5.00 is the dimensionless K, not the permittivity, and dividing ε₀ by 5 gives 1.77×10⁻¹².
Q6NEET 2020 · Q170
A short electric dipole has a dipole moment of 16×10⁻⁹ C m. The electric potential due to the dipole at a point at a distance of 0.6 m from the centre of the dipole, situated on a line making an angle of 60° with the dipole axis is : (1/4πε₀ = 9×10⁹ N m²/C²)
Not quite — the answer is B.
V = kp cos θ / r² = 9×10⁹ × 16×10⁻⁹ × cos 60° / (0.6)² = 144 × 0.5 / 0.36 = 200 V. Leaving out cos 60° gives 400 V, and zero applies only on the equatorial line (θ = 90°).
Q7NEET 2018 · Q23
The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is
Not quite — the answer is C.
Each plate sits in the uniform field σ/2ε₀ of the other, so F = Q²/(2ε₀A), which contains no separation term. For an isolated capacitor Q is fixed, so the force does not change with distance.
Q8NEET 2017 · Q152
A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system:
Not quite — the answer is B.
With the battery removed, charge Q is conserved and is shared over capacitance 2C. Energy goes from Q²/2C to Q²/(2 × 2C), i.e. it halves; the lost half is dissipated during charge flow. Energy cannot stay the same even though charge is conserved.
Q9NEET 2017 (Gujarati paper) · Q145
A parallel-plate capacitor is to be designed, using a dielectric of dielectric constant 5, so as to have a dielectric strength of 10⁹ V m⁻¹. If the voltage rating of the capacitor is 12 kV, the minimum area of each plate required to have a capacitance of 80 pF is:
Not quite — the answer is A.
The minimum plate separation is d = V/E_max = 12 × 10³ / 10⁹ = 1.2 × 10⁻⁵ m. Then A = Cd/(Kε₀) = (80 × 10⁻¹² × 1.2 × 10⁻⁵)/(5 × 8.85 × 10⁻¹²) ≈ 2.17 × 10⁻⁵ m² = 21.7 × 10⁻⁶ m². Leaving out the dielectric constant would make the area five times larger.
Q10AIPMT 2015
A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?
Not quite — the answer is A.
After disconnection the plates are isolated, so the charge stays constant; the statement that charge is not conserved is the incorrect one. With Q fixed, V = Q/KC falls K times and U = Q²/2KC falls K times, so the change is ½CV²(1/K − 1).
Q11AIPMT 2014
A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the centre of the sphere respectively are:
Not quite — the answer is A.
The charge sits on the surface and the field inside a conductor is zero, so the field at the centre is zero. Since E = 0 inside, the potential is constant throughout and equals the surface value Q/(4πε₀R); a zero potential at the centre is the common trap.
Q12AIPMT 2014
In a region, the potential is represented by V(x, y, z) = 6x − 8xy − 8y + 6yz, where V is in volts and x, y, z are in meters. The electric force experienced by a charge of 2 coulomb situated at point (1, 1, 1) is:
Not quite — the answer is C.
E = −∇V: Ex = −(6 − 8y) = 2, Ey = −(−8x − 8 + 6z) = 10, Ez = −6y = −6 at (1, 1, 1). |E| = √(4 + 100 + 36) = √140 = 2√35 V/m, so F = qE = 4√35 N.
Q13AIPMT 2012 Mains
A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and area of each plate is A, the energy stored in the capacitor is
Not quite — the answer is C.
Energy density between the plates is ½ε₀E² and the field occupies volume Ad, so U = ½ε₀E²Ad. The expression ½ε₀E² alone is energy per unit volume, not total energy.
Q14AIPMT 2012 Mains
Two metallic spheres of radii 1 cm and 3 cm are given charges of −1 × 10⁻² C and 5 × 10⁻² C, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is
Not quite — the answer is B.
Total charge = −1 × 10⁻² + 5 × 10⁻² = 4 × 10⁻² C. When connected the spheres reach the same potential, so charge divides in the ratio of radii 1 : 3, giving the bigger sphere ¾ × 4 × 10⁻² = 3 × 10⁻² C.
Q15AIPMT 2012 Prelims
An electric dipole of moment p is placed in an electric field of intensity E. The dipole acquires a position such that the axis of the dipole makes an angle θ with the direction of the field. Assuming that the potential energy of the dipole to be zero when θ = 90°, the torque and the potential energy of the dipole will respectively be
Not quite — the answer is B.
Torque on a dipole is τ = pE sin θ, and with U taken as zero at θ = 90°, the potential energy is U = −pE cos θ. The forms with 2pE cos θ wrongly double the energy, and pE cos θ is not the torque.
Q16AIPMT 2012 Prelims
Four point charges −Q, −q, 2q and 2Q are placed, one at each corner of a square. The relation between Q and q for which the potential at the centre of the square is zero is
Not quite — the answer is B.
All four charges are at the same distance from the centre, so the potential is proportional to the sum of charges: −Q − q + 2q + 2Q = Q + q. For zero potential, Q = −q. Q = q would make the sum 2q, not zero.
Q17AIPMT 2011 Mains
The electric potential V at any point (x, y, z), in meters in space is given by V = 4x² volt. The electric field at the point (1, 0, 2) in volt/meter, is
Not quite — the answer is B.
E = −dV/dx = −8x, which at x = 1 m is −8 V/m, i.e. 8 V/m along the negative x-axis. The value 16 V/m comes from wrong differentiation, and the positive direction ignores the minus sign in E = −∇V.
Q18AIPMT 2011 Prelims
A parallel plate condenser has a uniform electric field E (V/m) in the space between the plates. If the distance between the plates is d (m) and area of each plate is A (m²) the energy (joules) stored in the condenser is
Not quite — the answer is A.
The energy density in the field is (1/2)ε₀E² and the field fills a volume Ad, so U = (1/2)ε₀E²Ad. The expression (1/2)ε₀E² alone is just the energy per unit volume, not the total energy.
Q19AIPMT 2010 Mains · Q63
Two parallel metal plates having charges +Q and –Q face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will
Not quite — the answer is C.
With fixed charges ±Q the field between the plates is σ/(Kε₀) in a dielectric instead of σ/ε₀ in air. Kerosene has K > 1, so the field decreases. The charge on isolated plates does not change, so the field cannot remain the same.
Q20AIPMT 2010 Prelims · Q27
A series combination of n₁ capacitors, each of value C₁, is charged by a source of potential difference 4V. When another parallel combination of n₂ capacitors, each of value C₂, is charged by a source of potential difference V, it has the same (total) energy stored in it, as the first combination has. The value of C₂, in terms of C₁, is then
Not quite — the answer is A.
Series: C_eq = C₁/n₁, energy = ½(C₁/n₁)(4V)² = 8C₁V²/n₁. Parallel: C_eq = n₂C₂, energy = ½n₂C₂V². Equating gives C₂ = 16C₁/(n₁n₂).
Q21AIPMT 2009 Prelims · Q26
Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities σ, −σ and σ respectively. If VA, VB and VC denote the potentials of the three shells, then for c = a + b, we have:
Not quite — the answer is D.
VA = (σ/ε₀)(a − b + c) = 2aσ/ε₀ for c = a + b. VC = (σ/ε₀)(a² − b² + c²)/c = (σ/ε₀)(2a² + 2ab)/(a + b) = 2aσ/ε₀. VB = (σ/ε₀)(a²/b − b + c) = (σ/ε₀)(a²/b + a), which differs from 2aσ/ε₀ unless a = b, so VC = VA ≠ VB.
Q22AIPMT 2009 Prelims · Q38
Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will be:
Not quite — the answer is B.
Capacitors in series: 1/C_eq = 3/C, so C_eq = C/3. The same charge sits on each, so the applied voltage divides equally and each can take V before breakdown: total 3V.
Q23AIPMT 2009 Prelims · Q45
The electric potential at a point (x, y, z) is given by V = −x²y − xz³ + 4. The electric field E⃗ at that point is:
Not quite — the answer is D.
E⃗ = −∇V. −∂V/∂x = 2xy + z³, −∂V/∂y = x², −∂V/∂z = 3xz². So E⃗ = î(2xy + z³) + ĵx² + k̂3xz². A sign slip on the z³ term gives the (2xy − z³) form.
Q24AIPMT 2008 Prelims · Q25
The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field between the plates is E, is
Not quite — the answer is A.
Energy = (1/2)CV² with C = ε₀A/d and V = Ed, giving (1/2)(ε₀A/d)(Ed)² = (1/2)ε₀E²Ad, i.e. energy density (1/2)ε₀E² times the volume Ad. ε₀E²Ad misses the factor 1/2, and the forms divided by Ad have the wrong dimensions.
Q25AIPMT 2008 Prelims · Q27
The electric potential at a point in free space due to a charge Q coulomb is Q × 10¹¹ volts. The electric field at that point is
Not quite — the answer is B.
V = Q/(4πε₀r) = Q × 10¹¹ gives r = 1/(4πε₀ × 10¹¹). Then E = V/r = Q × 10¹¹ × 4πε₀ × 10¹¹ = 4πε₀Q × 10²² V/m. The 10²⁰ forms come from mishandling the power of ten.
Q26AIPMT 2006 Prelims · Q44
A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates :
Not quite — the answer is D.
With the battery disconnected, the charge Q stays constant. Increasing the plate separation lowers C = ε₀A/d, so V = Q/C increases. The p.d. would stay constant only if the battery remained connected.
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