25 questions in 25 minutes on the NTA computer-based test interface, all from Electrostatic Potential and Capacitance. +4 / -1 marking, instant score, full solutions. Free, no login.
A chapter test is the fastest honest check on whether a chapter has actually landed. This one draws 25 questions from Electrostatic Potential and Capacitance, of which 22 are real NEET previous-year questions and 6 sit at the harder end of the bank. Sit it in one 25-minute block, the way you would in the hall.
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Duration: 180 minutes · 180 questions · 720 marks maximum
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How many questions are in this Electrostatic Potential and Capacitance mock test?
25 questions in 25 minutes, marked +4 for a correct answer and -1 for a wrong one, exactly like the real paper. 22 of them are NEET previous-year questions.
Is this the same interface as the real NEET CBT?
Yes. On-screen countdown, colour-coded question palette, Save & Next, Clear Response and Mark for Review, then an instant score with full solutions.
Should I take this before or after revising Electrostatic Potential and Capacitance?
Both, and that is the point. Take it cold to find out what you actually do not know, revise from the chapter notes, then retake it to confirm the gap closed.
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The complete paper in text form, with the correct option and a worked explanation for every question. Sit the timed test above first: the score is only worth having if you earn it. Then open this to revise, or to re-read a question you got wrong without replaying the whole paper.
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Physics
- Q1The electrostatic force between the metal plates of an isolated parallel plate capacitor (constant charge Q, plate area A) is:
- proportional to the square root of the distance between the plates
- inversely proportional to the distance between the plates
- linearly proportional to the distance between the plates
- independent of the distance between the plates
Answer: (D) independent of the distance between the plates
At constant charge, F=(Q²)/(2ε₀ A), which contains no d. So the force between the plates is independent of their separation.
- Q2Two capacitors C₁ = 2 µF and C₂ = 3 µF are in series across a 10 V battery. The charge on each capacitor is:
- 50 µC
- 5 µC
- 12 µC
- 30 µC
Answer: (C) 12 µC
Cs = (2×3)/(2+3) = 6/5 = 1.2 µF. Charge Q = CsV = 1.2 × 10 = 12 µC, and in series the charge is the same on each capacitor.
- Q3Two parallel metal plates having charges +Q and -Q face each other at a certain distance between them. If the plates are now dipped in a kerosene oil tank, the electric field between the plates will
- remain same
- become zero
- decrease
- increase
Answer: (C) decrease
C. Tests: what a dielectric does to the field when the charge on the plates is held fixed. Why C: the plates are isolated, so the free charge ± Q and hence the free surface charge density σ = (Q)/(A) cannot change. Inside a dielectric the field is E = (σ)/(Kε₀), because the oil polarises and its bound surface charge partly cancels the plate charge. Kerosene has K ≈ 2, which is greater than 1, so E drops by that factor: the field decreases. Why not A: the field goes to zero only if the gap is filled by a conductor, whose free electrons redistribute until the interior field is exactly cancelled; kerosene is an insulator, its bound charges shift only slightly and cancel just a fraction of σ. Why not B: an increase requires K < 1 in E = (σ)/(Kε₀), and every real dielectric has K > 1 [misconception B: ‘filling the gap with matter strengthens the field’]. Why not D: the field would stay put only if the medium did not polarise, or if a battery held V fixed instead of the charge; here the charge is fixed and kerosene does polarise, so E must change. Remember: charge fixed means E falls by K; battery connected means E stays and the charge rises by K.
- Q4Three small identical bubbles of water having same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is:
- 1 : 2^(2/3)
- 1 : 3^(1/3)
- 3^(2/3) : 1
- 1 : 3^(2/3)
Answer: (D) 1 : 3^(2/3)
A. Tests: combining volume conservation on coalescence with V = (kq)/(R). Why A: the water volume is conserved, so 3×(4)/(3)π r³ = (4)/(3)π R³, giving R = 3^(1/3)r. Charge simply adds, Q = 3q. Then Vₛₘₐₗₗ = (kq)/(r) and V_(big) = (k(3q))/(3^(1/3)r) = 3^(1-1/3)(kq)/(r) = 3^(2/3)(kq)/(r), so the ratio small to big is 1 : 3^(2/3). Why not B: this is the ratio the other way round. Charge triples while the radius grows only by 3^(1/3), so the big bubble ends at the HIGHER potential [misconception B: “a bigger radius always means a lower potential”] Why not C: 1 : 3^(1/3) requires R = 3^(2/3)r, which comes from mishandling the cube root of R³ = 3r³ and squaring the exponent instead of taking it once. Why not D: 2^(2/3) is the answer for TWO bubbles coalescing; the question merges three. Remember: n identical drops merging give R = n^(1/3)r and V_(big) = n^(2/3)Vₛₘₐₗₗ.
- Q5Which one of the following is the correct dimensional formula for the capacitance in F? M, L, T and C stand for unit of mass, length, time and charge.
- [F] = [C M⁻¹ L⁻² T²]
- [F] = [C² M⁻¹ L⁻² T²]
- [F] = [C² M⁻² L² T²]
- [F] = [C M⁻² L⁻² T⁻²]
Answer: (B) [F] = [C² M⁻¹ L⁻² T²]
C. Tests: building the dimensions of capacitance from C = Q/V when charge is treated as a base quantity. Why C: potential is work per unit charge, V = W/Q, and work has dimensions [M L² T⁻²], so [V] = [M L² T⁻² C⁻¹]. Capacitance is charge per volt, so [F] = ([C])/([M L² T⁻² C⁻¹]) = [C² M⁻¹ L⁻² T²]. Why not A: it carries only one power of charge, which happens if you divide by the volt without noticing that the volt itself already hides a C⁻¹; restoring that factor turns C¹ into C² [misconception A: ‘the volt is charge free’]. Why not B: the charge is squared correctly, but M⁻² L² matches no power of [V]; substituting [V] = [M L² T⁻² C⁻¹] into Q/V can only ever give M⁻¹ L⁻², so no consistent substitution gives this. Why not D: it misses in three places at once, one power of charge instead of two, M⁻² instead of M⁻¹, and T⁻² instead of T². Multiplying by the volt instead of dividing by it gives [Q][V] = [M L² T⁻²], which is not this either, so no consistent substitution gives this. Remember: capacitance is charge per volt, and one volt already carries a C⁻¹, so charge always ends up squared.
- Q6An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity 10⁶ m/s. If the magnitude of the electric field between the plates is 9.1 V/cm, then the vertical component of velocity of electron is (mass of electron = 9.1 × 10⁻³¹ kg and charge of electron = 1.6 × 10⁻¹⁹ C)
- 0
- 1 × 10⁶ m/s
- 16 × 10⁶ m/s
- 16 × 10⁴ m/s
Answer: (C) 16 × 10⁶ m/s
B. Tests: converting a field quoted in V/cm, then chaining a = (eE)/(m) with a flight time fixed by the unchanged horizontal speed. Why B: first convert, E = 9.1 V/cm = 910 V/m. The transverse acceleration is a = (eE)/(m) = ((1.6×10⁻¹⁹)(910))/(9.1×10⁻³¹) = 1.6×10¹⁴ m/s². The horizontal speed is untouched by the vertical field, so the time inside the plates is t = (0.10)/(10⁶) = 10⁻⁷ s. Hence v_y = at = (1.6×10¹⁴)(10⁻⁷) = 16×10⁶ m/s. Why not A: 1×10⁶ m/s simply repeats the horizontal component; the vertical component is built up by the field over the flight time and has no reason to equal the entry speed. Why not C: 16×10⁴ m/s is what you get by leaving the field as 9.1 V/m instead of 910 V/m, which shrinks a to 1.6×10¹² m/s² and v_y by the same factor of 100 [misconception C: ‘V/cm and V/m are interchangeable’]. Why not D: zero would require no transverse force, but the plates are oppositely charged, so a uniform field acts on the electron for the whole 10 cm of travel and must deflect it. Remember: in a deflecting field the horizontal speed only sets the clock, t = L/vₓ, and the vertical speed is at.
- Q7Two circular plates each of radius ‘r‘ are kept parallel to each other distance ‘d‘ apart. The capacitance of the capacitor formed is ‘C₁‘. If the radius of each of the plates is increased to √(3) times the earlier radius and their distance of separation decreased to half the initial value, the capacitance now becomes ‘C₂‘. The ratio C₁ : C₂ is
- 1 : 6
- 1 : 2
- 1 : 4
- 6 : 1
Answer: (A) 1 : 6
C. Tests: scaling C = (ε₀ A)/(d) when both radius and separation change. Why C: for circular plates C = (ε₀ π r²)/(d), so C is proportional to (r²)/(d). Multiplying r by √(3) multiplies the area by 3, and halving d multiplies the capacitance by another 2. Hence C₂ = 6C₁ and C₁ : C₂ = 1 : 6. Why not A: 1 : 2 uses only the halved separation and ignores the area growth altogether, since √(3) on the radius becomes a factor 3 on the area [misconception: “radius times √(3) means area times √(3)“]. Why not B: no consistent substitution gives 1 : 4; the area factor is exactly 3 and the separation factor exactly 2, and no product, quotient or sum of those two lands on 4 [misconception: “round √(3) to 2 and multiply”]. Why not D: 6 : 1 has the correct factor but the wrong order; the second arrangement has the larger plates and the smaller gap, so C₂ is the bigger capacitance [misconception: “the order of a ratio does not matter”]. Remember: C is proportional to (r²)/(d), so scaling r by √(3) and d by (1)/(2) gives 3 times 2 = 6.
- Q8To produce an instantaneous displacement current of 2 mA in the space between the parallel plates of a capacitor of capacitance 4 ~μF, the rate of change of applied variable potential difference ((dV)/(dt)) must be :-
- 400 V/s
- 800 V/s
- 200 V/s
- 500 V/s
Answer: (D) 500 V/s
B. Tests: the displacement current relation i = C(dV)/(dt) between capacitor plates. Why B: the displacement current in the gap equals the conduction current in the leads, i = (dq)/(dt) = C(dV)/(dt) since C is fixed. So (dV)/(dt) = (i)/(C) = (2 × 10⁻³)/(4 × 10⁻⁶) = 500 V/s. Why not A: 800 V/s would require C = (i)/(dV/dt) = (2 × 10⁻³)/(800) = 2.5 μF, not the 4 μF given [misconception: “the capacitance value can be adjusted to suit the arithmetic”]. Why not C: 200 V/s implies C = (2 × 10⁻³)/(200) = 10 μF, again not the stated 4 μF [misconception: “any convenient round division of 2 mA works”]. Why not D: 400 V/s implies C = (2 × 10⁻³)/(400) = 5 μF, the result of dividing 2 by 5 instead of 2 by 4 [misconception: “a slipped digit in the capacitance”]. Remember: displacement current obeys the very same i = C(dV)/(dt) as the current in the wires, which is why the current appears continuous.
- Q9The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is
- Inversely proportional to the distance between the plates
- Independent of the distance between the plates
- Linearly proportional to the distance between the plates
- Proportional to the square root of the distance between the plates
Answer: (B) Independent of the distance between the plates
A. Tests: the force between capacitor plates when the charge, not the voltage, is held fixed. Why A: isolated means Q cannot change, so σ = (Q)/(A) is fixed. One plate sits in the field made by the other, (σ)/(2ε₀), a value independent of separation because the field of a large charged sheet does not fall off with distance. Hence F = Q×(σ)/(2ε₀) = (Q²)/(2ε₀A), an expression containing no d at all. Why not B: a force growing with d would need the sheet field to grow with distance; (σ)/(2ε₀) is uniform, so nothing in the derivation can produce a factor of d. Why not C: a square-root dependence has no source here, since F = (Q²)/(2ε₀A) involves only Q and A and no consistent substitution introduces √(d). Why not D: an inverse dependence on d comes from treating the plates as two point charges obeying an inverse-power law, but a plate is an extended sheet whose field near the surface does not depend on distance [misconception D: “plates attract each other like point charges”] Remember: isolated capacitor means Q is fixed, and F = (Q²)/(2ε₀A) never mentions the gap.
- Q10A condenser of capacity C₁ is charged to potential V₁ and then disconnected. An uncharged capacitor of capacity C₂ is connected in parallel with C₁. The resultant potential V₂ is
- (C₁ V₁)/(C₂)
- (C₁ V₁)/(C₁ + C₂)
- (C₂ V₁)/(C₁ + C₂)
- (C₂ V₁)/(C₁)
Answer: (B) (C₁ V₁)/(C₁ + C₂)
B. Tests: charge conservation when a charged capacitor is paralleled with an uncharged one. Why B: the disconnected capacitor carries Q = C₁ V₁, and since the pair is isolated that total charge cannot change. In parallel the two settle at one common potential V₂, and parallel capacitances add to C₁ + C₂. So V₂ = (Qₜₒₜₐₗ)/(Cₜₒₜₐₗ) = (C₁ V₁)/(C₁ + C₂). Why not A: (C₁ V₁)/(C₂) divides the conserved charge by C₂ alone, as though C₁ dropped out of the combination; it also diverges as C₂ arrow 0, the limit in which nothing was connected and the answer must return to V₁. Why not C: (C₂ V₁)/(C₁) puts the uncharged capacitor on top, so it predicts V₂ = 0 when C₂ = 0, i.e. the charged capacitor loses all its potential by having nothing attached to it. Why not D: the denominator is right but the numerator is not; the stored charge is C₁ V₁ and C₂ arrived empty, so C₂ contributes capacitance but no charge [misconception D: ‘the added capacitor supplies the charge’]. Remember: paralleling capacitors conserves charge, so the common potential is total charge over total capacitance.
- Q11A parallel plate capacitor with air between the plates has a capacitance of 15 pF. The separation between the plates becomes twice and the space between them is filled with a medium of dielectric constant 3.5. Then the capacitance becomes x/4 pF. The value of x is
- 111
- 109
- 115
- 105
Answer: (D) 105
A. Tests: combining a change of separation with a dielectric fill in C = (Kε₀A)/(d). Why A: doubling d halves the capacitance and the dielectric multiplies it by K = 3.5, so the net factor is (K)/(2) = 1.75 and C’ = 15×1.75 = 26.25 pF. Setting (x)/(4) = 26.25 gives x = 105. Why not B: (109)/(4) = 27.25 pF, and no consistent substitution of K = 3.5 with d → 2d into 15×(K)/(2) produces that value. Why not C: (111)/(4) = 27.75 pF, again unreachable from 15×1.75 by any consistent route; it is a near miss placed to catch a slip in the multiplication. Why not D: (115)/(4) = 28.75 pF, which would need a net factor of about 1.92 rather than the (3.5)/(2) = 1.75 that the data give. Remember: a dielectric multiplies by K and doubling the gap divides by 2, so the net factor is simply (K)/(2).
- Q12Electric field in a region is given by E = Axi + Byj, where A = 10 V/m² and B = 5 V/m². If the electric potential at a point (10,20) is 500 V, then the electric potential at origin is ____ V.
- 0
- 2000
- 500
- 1000
Answer: (B) 2000
C. Tests: recovering potential from a position-dependent field by integrating dV = -E· dr. Why C: V(x,y) = V(0,0) – ∫₀ˣ Ax dx – ∫₀ʸ By dy = V(0,0) – (Ax²)/(2) – (By²)/(2). Substituting the given point, 500 = V(0,0) – (10×(10)²)/(2) – (5×(20)²)/(2) = V(0,0) – 500 – 1000, so V(0,0) = 500 + 1500 = 2000 V. Why not A: 1000 V is 500 + 500, which keeps only the x contribution and drops the y contribution of (5×400)/(2) = 1000 V [misconception A: “only the field component along one axis matters”] Why not B: 0 V assumes the origin is automatically the zero of potential. The zero here is already fixed by the stated 500 V at (10,20), so the origin value follows from the integral, not from convention. Why not D: 500 V says nothing changed between the two points, which would need E = 0 along the whole path; the given field is non-zero everywhere except at the origin itself. Remember: when E ∝ x, the potential drop goes as (x²)/(2), so integrate, never multiply E by the displacement.
- Q13A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the centre of the sphere are respectively:
- zero and (Q)/(4πε₀ R²)
- (Q)/(4πε₀ R) and zero
- both are zero
- (Q)/(4πε₀ R) and (Q)/(4πε₀ R²)
Answer: (B) (Q)/(4πε₀ R) and zero
Inside a conductor the field is zero, so E=0 at the centre. The potential is constant throughout the conductor and equals its surface value (Q)/(4πε₀ R).
- Q14A parallel plate capacitor has capacitance C₀ with air between the plates. A dielectric slab of constant K=3 is inserted so that it fills exactly half of the plate area (covering the full gap d over that half). The new capacitance is:
- 3C₀
- (4)/(3)C₀
- (3)/(2)C₀
- 2C₀
Answer: (D) 2C₀
Half-filled by area acts as two capacitors in parallel, each of area A/2: C=(ε₀(A/2))/(d)+(Kε₀(A/2))/(d)=((1+K))/(2)·(ε₀ A)/(d)=(1+3)/(2)C₀=2C₀.
- Q15A 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
- decreases by a factor of 2
- remains the same
- increases by a factor of 2
- increases by a factor of 4
Answer: (A) decreases by a factor of 2
A. Tests: what happens to stored energy when charge is shared at constant total charge. Why A: initially Uᵢ = (Q²)/(2C). With the battery removed the charge Q is fixed, and adding an identical capacitor in parallel makes the capacitance 2C, so U_f = (Q²)/(2(2C)) = (Q²)/(4C) = (Uᵢ)/(2). The energy halves, and the missing half is dissipated as heat and radiation in the connecting wires while the charge redistributes. Why not B: charge is conserved in the sharing, energy is not; any redistribution of charge through a real connecting path dissipates energy [misconception B: “conserved charge implies conserved energy”] Why not C: doubling would need the voltage to stay at its original value, which happens only if the battery remains connected to supply extra charge. Here it has been removed and the common voltage falls to (V)/(2). Why not D: with Q fixed, U = (Q²)/(2C) can only FALL as C rises, so no increase, let alone a fourfold one, is possible. Remember: disconnect the battery and share charge with an identical capacitor, and the stored energy always drops to exactly half.
- Q16If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then A. the charge stored in it, increases. B. the energy stored in it, decreases. C. its capacitance increases. D. the ratio of charge to its potential remains the same. E. the product of charge and voltage increases. Choose the most appropriate answer from the options given below:
- B, D and E only
- A, B and C only
- A, C and E only
- A, B and E only
Answer: (C) A, C and E only
B. Tests: tracking every capacitor quantity when d shrinks with the battery still connected. Why B: the battery is connected, so V is held fixed. Then C = (ε₀ A)/(d) rises as d falls, making C true. Charge Q = CV rises with C, making A true. Energy U = (1)/(2)CV² rises as well, so B is false. The ratio of charge to potential is (Q)/(V) = C, which changes, so D is false. The product QV rises because Q rises at fixed V, making E true. The true set is A, C and E. Why not A: it keeps statement B, but bringing the plates closer at fixed V raises the stored energy rather than lowering it [misconception: “closing the gap releases stored energy”]. Why not C: it keeps both false statements B and D and drops the true A and C; the ratio (Q)/(V) is the capacitance itself, and that is precisely what changed [misconception: “with a battery attached the ratio of charge to potential is locked”]. Why not D: it keeps the false B and drops the true E, though QV must rise once Q rises at constant V [misconception: “the product QV can hold steady while Q grows”]. Remember: battery attached means V is the constant, and C, Q, QV and U all rise together as d shrinks.
- Q17A parallel plate capacitor of capacitance 1 μF is charged to a potential difference of 20 V. The distance between plates is 1 μm. The energy density between plates of capacitor is:
- 2×10² J/m³
- 2×10⁻⁴ J/m³
- 1.8×10⁵ J/m³
- 1.8×10³ J/m³
Answer: (D) 1.8×10³ J/m³
A. Tests: energy density u = (1)/(2)ε₀E², with E obtained from the plate voltage and separation. Why A: E = (V)/(d) = (20)/(10⁻⁶) = 2×10⁷ V/m. Then u = (1)/(2)ε₀E² = (1)/(2)×8.85×10⁻¹²×(2×10⁷)² = (1)/(2)×8.85×10⁻¹²×4×10¹⁴ = 1.77×10³ ≈ 1.8×10³ J/m³. Why not B: 2×10² comes from taking the total stored energy U = (1)/(2)CV² = (1)/(2)×10⁻⁶×400 = 2×10⁻⁴ J and dividing it by the separation 10⁻⁶ m alone. The volume is Ad, not d [misconception B: “divide the energy by d to get a density”] Why not C: 2×10⁻⁴ is the TOTAL stored energy (1)/(2)CV² in joules, relabelled as a density without ever being divided by a volume. Why not D: 1.8×10⁵ follows from d = 0.1 μm, which gives E = 2×10⁸ V/m and u = 1.77×10⁵ J/m³. The stated separation is 1 μm, so the field is ten times smaller and u a hundred times smaller. Remember: energy density is (1)/(2)ε₀E² with E = (V)/(d); the total energy is that density times the plate volume Ad.
- Q18If charge +q is taken from one point to another over an equipotential surface, then
- no work is done.
- work is done by the charge.
- work done on the charge continuously increases.
- work done on the charge is constant.
Answer: (A) no work is done.
D. Tests: the defining property of an equipotential surface, Δ V = 0, and its consequence for work. Why D: work done in moving a charge between two points is W = q(Vᵢ – V_f) = qΔ V. On an equipotential surface every point has the same potential, so Δ V = 0 and W = 0 for any path and any charge. Equivalently, E is everywhere perpendicular to an equipotential, so F· ds = 0 at every step of the journey. Why not A: work cannot pile up along the way, because each infinitesimal step contributes qE· ds = 0; the running total stays pinned at zero rather than increasing. Why not B: work done by the charge would require the field to have a component along the motion, but E is normal to an equipotential surface, so there is no such component to push along. Why not C: it stops one step short. Zero is technically a constant, but the physics fixes the exact value, and only D states it; saying merely constant leaves room for a nonzero amount, which Δ V = 0 forbids [misconception C: ‘same potential means the same work, without pinning it to zero’]. Remember: field lines cut equipotentials at right angles, so travelling along one costs exactly nothing.
- Q19Two parallel plate air capacitors are connected in parallel. Each capacitor has plate area A/2 and separation between the plates is d and 2d respectively. The equivalent capacity of the combination is (ε₀ = absolute permittivity of free space)
- (Aε₀)/(4d)
- (3Aε₀)/(4d)
- (Aε₀)/(d)
- (2Aε₀)/(3d)
Answer: (B) (3Aε₀)/(4d)
B. Tests: parallel capacitors add, with each one computed from its OWN area and separation. Why B: C₁ = (ε₀(A/2))/(d) = (ε₀A)/(2d) and C₂ = (ε₀(A/2))/(2d) = (ε₀A)/(4d). Being in parallel they simply add: C = (ε₀A)/(2d) + (ε₀A)/(4d) = (2ε₀A + ε₀A)/(4d) = (3Aε₀)/(4d). Why not A: (Aε₀)/(d) equals 2×(ε₀A)/(2d), that is using separation d for BOTH capacitors and ignoring that the second one has a 2d gap. Why not C: (2Aε₀)/(3d) is what you get by merging the two into ONE capacitor of full area A with an averaged gap 1.5d, since (ε₀A)/(1.5d) = (2Aε₀)/(3d). Gaps cannot be averaged; each capacitor is a separate element of area (A)/(2) evaluated at its own d [misconception C: “unequal plate gaps can be replaced by their average”] Why not D: (Aε₀)/(4d) is the second capacitor alone, with the first simply left out of the sum [misconception D: “the smaller capacitor limits the combination, as it does in series”] Remember: parallel means add, so evaluate (ε₀Aᵢ)/(dᵢ) separately before summing.
- Q20A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates. A sheet having thickness ((1)/(3))ʳᵈ of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is :
- (3 C K²)/((2 K+1)²)
- (3 K C)/(2 K+1)
- (4 K C)/(3 K-1)
- (C K)/(2+K)
Answer: (B) (3 K C)/(2 K+1)
D. Tests: the partially filled capacitor formula and the K = 1 sanity check. Why D: with t = (d)/(3), the effective gap is d – t + (t)/(K) = (2d)/(3) + (d)/(3K) = (d(2K+1))/(3K). Therefore C’ = (ε₀ A)/(d(2K+1)/(3K)) = (ε₀ A)/(d) · (3K)/(2K+1) = (3KC)/(2K+1). It passes both checks: at K = 1 it returns C, and as K grows it tends to (3C)/(2), the value for a gap shrunk to (2d)/(3). Why not B: at K = 1 this gives (4C)/(2) = 2C, but with no dielectric the capacitance must return to C, so the formula is already wrong before any algebra; the minus sign also makes it blow up at K = (1)/(3) [misconception: “the empty part of the gap subtracts from the filled part”]. Why not A: at K = 1 this gives (3C)/(9) = (C)/(3), again failing the no dielectric check; squaring the denominator has no basis, since the layers add in series only once [misconception: “series layers square the denominator”]. Why not C: at K = 1 this gives (C)/(3) as well, and for large K it tends to C rather than to (3C)/(2), so it fails at both ends [misconception: “any expression with K on top and a sum below will do”]. Remember: test every dielectric formula at K = 1; it must collapse back to the empty capacitor value.
- Q21The electric potential due to a point charge varies with distance r as:
- 1/r³
- 1/r²
- r
- 1/r
Answer: (D) 1/r
V = (1/4πε₀)(q/r) ∝ 1/r. The field varies as 1/r², but potential varies as 1/r.
- Q22Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. V. If one of the capacitors is inserted in a liquid of dielectric constant K, then the potential difference of the other capacitor will become
- (KV)/(K+1)
- (K-1)/(KV)
- (KV)/(K-1)
- (K+1)/(KV)
Answer: (A) (KV)/(K+1)
A. Tests: series capacitors carry a common charge, so the applied voltage divides in inverse proportion to capacitance. Why A: let each capacitor be C. After the liquid fills one of them it becomes KC, and the pair KC and C sits in series across V. The same charge Q is on both, so the other capacitor holds V₂ = (Q)/(C) and the immersed one holds (Q)/(KC). Their sum is the battery e.m.f.: (Q)/(C)(1 + (1)/(K)) = V, giving (Q)/(C) = (KV)/(K+1). That is exactly V₂ = (KV)/(K+1). Why not B: (KV)/(K-1) comes from subtracting the reciprocal capacitances instead of adding them; it diverges at K = 1 and in fact stays above the battery e.m.f. for every K > 1, since (K)/(K-1) > 1 always, which no series element can do [misconception B: ‘series combines with a minus sign’]. Why not C: (K+1)/(KV) is the reciprocal of a voltage, so it carries units of 1/volt and cannot be a potential difference at all; it is the correct expression flipped upside down. Why not D: (K-1)/(KV) has the same unit defect and additionally vanishes at K = 1, whereas two identical capacitors in series must each hold (V)/(2). Remember: series capacitors share charge, so the smaller capacitance takes the larger share of the voltage.
- Q23Three capacitors each of capacity 4 μ F are to be connected in such a way that the effective capacitance is 6 μ F. This can be done by
- connecting two in series and one in parallel
- connecting all of them in series
- connecting two in parallel and one in series
- connecting them in parallel
Answer: (A) connecting two in series and one in parallel
C. Tests: mixing series and parallel groupings to reach a target equivalent capacitance. Why C: take two of the capacitors in series, (1)/(Cₛ) = (1)/(4) + (1)/(4) = (1)/(2), so Cₛ = 2 μ F. Now connect the third capacitor, 4 μ F, in parallel across that pair: C = 2 + 4 = 6 μ F, exactly the required value. Why not A: all three in series give (1)/(C) = (3)/(4), so C = (4)/(3) ≈ 1.33 μ F, far below 6 μ F; series can never exceed the smallest capacitor in the chain. Why not B: all three in parallel give 4+4+4 = 12 μ F, twice the target; parallel simply adds, so it overshoots here. Why not D: two in parallel give 8 μ F, and that in series with the remaining 4 μ F gives (8×4)/(8+4) = (8)/(3) ≈ 2.67 μ F, not 6 μ F [misconception D: ‘any mixed arrangement can be tuned to whatever value you need’]. Remember: series always lands below the smallest capacitor, parallel always above the largest, so a target in between needs a mix.
- Q24The electric potential as a function of x, y is given by V = 5(x² – y²) V. The electric field at a point (2,3) m is ____ V/m.
- (20 i + 45 j)
- (20 i – 30 j)
- (-4 i + 6 j)
- (-20 i + 30 j)
Answer: (D) (-20 i + 30 j)
A. Tests: recovering the field from a potential function through E = -(dV)/(dr) component by component. Why A: Eₓ = -(dV)/(dx) = -(d)/(dx)[5(x² – y²)] = -10x, and E_y = -(dV)/(dy) = -(-10y) = +10y. At (2,3) this gives Eₓ = -10(2) = -20 and E_y = +10(3) = 30, so the field is (-20i + 30j) V/m. Why not B: this is the exact negative of the answer, obtained by writing E = +(dV)/(dr) and dropping the minus sign that defines the field as the downhill direction of potential [misconception: “the field points from low to high potential”]. Why not C: (20i + 45j) is V’s own terms evaluated at the point instead of its derivatives: 5x² = 5(4) = 20 and 5y² = 5(9) = 45. Reading numbers straight out of V skips both the differentiation and the minus sign [misconception: “the value of the potential at a point is the field at that point”]. Why not D: (-4i + 6j) is the answer divided by 5, the gradient of (x² – y²) with the multiplier 5 in V left out [misconception: “a constant multiplier can be ignored when differentiating”]. Remember: take the derivative of V with respect to each coordinate, then flip every sign.
- Q25Charges +q and -q are placed at points A and B which are a distance 2L apart; C is the midpoint of AB. A charge +Q is taken from C to a point D that lies on the line AB extended beyond B, with BD=L (so AD=3L, BD=L). The work done is:
- (qQ)/(2πε₀ L)
- (qQ)/(4πε₀ L)
- (qQ)/(6πε₀ L)
- -(qQ)/(6πε₀ L)
Answer: (D) -(qQ)/(6πε₀ L)
At C (distance L from each charge): V_C=(kq)/(L)-(kq)/(L)=0. At D (AD=3L, BD=L): V_D=(kq)/(3L)-(kq)/(L)=-(2kq)/(3L). Work =Q(V_D-V_C)=-(2kqQ)/(3L)=-(qQ)/(6πε₀ L) (using k=(1)/(4πε₀)).