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CBSE Class 12 Physics · 11 questions · 26 marks
Fields tell you the force on a charge; potential tells you the energy it carries, and that scalar view makes hard problems easy. Here you will learn to add potentials without worrying about direction, to see why every point of a conductor sits at one voltage, and to compute how much charge a capacitor can store. Almost every circuit you meet later is built on the ideas in this chapter.
A charge of 5 μC is moved from one point to another along an equipotential surface of a 200 V conductor. The work done is:
Answer
Zero — on an equipotential surface every point has the same potential, so the potential difference between start and end is zero. Since W = q(V_A - V_B) = q × 0, no work is done regardless of the path length or the size of the charge.
A parallel plate capacitor is charged and then disconnected from the battery. A dielectric slab of constant K is now slipped in to completely fill the gap. Which quantity remains unchanged?
Answer
The charge on the plates — once disconnected, the charge has nowhere to flow, so Q is fixed. The capacitance rises to KC, so V = Q/C falls to V/K, the field falls to E/K, and the stored energy Q²/2C falls to U/K. Only Q stays the same.
Three capacitors of 6 μF each are joined in series and the combination is connected across a 60 V supply. The charge on each capacitor is:
Answer
120 μC — in series 1/C = 1/6 + 1/6 + 1/6 = 3/6, so C = 2 μF. The charge is the same on every capacitor in series: Q = CV = 2 μF × 60 V = 120 μC.
Two point charges of +4 μC and -4 μC are held 20 cm apart. The electrostatic potential at the midpoint of the line joining them is:
Answer
Zero — potential is a scalar, so the two contributions simply add: V = kq/r + k(-q)/r with equal r = 0.10 m. The positive and negative terms are equal in magnitude and cancel exactly, giving V = 0, even though the electric field there is large and non-zero.
Assertion (A): The electrostatic potential at a point can be zero while the electric field at that same point is not zero. Reason (R): Potential is a scalar that adds algebraically, so opposite contributions can cancel, whereas field contributions add as vectors and may reinforce.
Answer
Both A and R are true and R is the correct explanation of A — at the midpoint of a dipole the potentials of +q and -q cancel to zero, but their fields both point from +q towards -q and add up to a maximum. The scalar-versus-vector distinction is precisely why this happens.
Define an equipotential surface. Sketch in words the shape of the equipotential surfaces around an isolated point charge and explain why they cannot intersect.
Answer
An equipotential surface is a surface on which the electrostatic potential has the same value at every point, so no work is needed to move a charge anywhere on it. Around an isolated point charge the equipotential surfaces are a family of concentric spheres centred on the charge, since V = kq/r depends only on r; they get further apart as r increases because V falls more slowly there. Two equipotential surfaces can never intersect, because at a point of intersection the potential would have to take two different values simultaneously, which is impossible.
A 12 μF capacitor is charged to 200 V and then disconnected. It is connected in parallel to an identical uncharged 12 μF capacitor. Find the common potential and the total energy stored after connection, and comment on the change in energy.
Answer
Initial charge Q = CV = 12×10⁻⁶ × 200 = 2.4×10⁻³ C. After connection, total capacitance = 12 + 12 = 24 μF, and total charge is conserved at 2.4×10⁻³ C. Common potential V' = Q/C(total) = 2.4×10⁻³/24×10⁻⁶ = 100 V. Final energy U' = ½C(total)V'² = ½ × 24×10⁻⁶ × (100)² = 0.12 J. Initial energy U = ½ × 12×10⁻⁶ × (200)² = 0.24 J. So half the energy, 0.12 J, is lost. It is dissipated as heat in the connecting wires (and a little as electromagnetic radiation) during the transient flow of charge.
Show that the electric potential energy of a dipole placed at angle θ in a uniform field E is U = -pE cosθ, and state the orientations of stable and unstable equilibrium.
Answer
The torque needed to rotate the dipole against the field is τ = pE sinθ. The work done in turning it from the reference position θ = 90° (where U is taken as zero) to angle θ is U = ∫pE sinθ dθ from 90° to θ = pE[-cosθ] from 90° to θ = -pE cosθ. Stable equilibrium: θ = 0°, where U = -pE is minimum and the dipole is aligned with the field. Unstable equilibrium: θ = 180°, where U = +pE is maximum and the dipole is anti-parallel to the field.
Derive an expression for the capacitance of a parallel plate capacitor with a dielectric slab of thickness t (t < d) and dielectric constant K inserted between plates separated by d. Hence calculate the capacitance when A = 250 cm², d = 5.0 mm, t = 3.0 mm and K = 5.
Answer
Derivation: Let the plates carry surface charge density σ = Q/A. Between the plates, in the free-space gap of total thickness (d - t), the field is E₀ = σ/ε₀. Inside the dielectric slab the field is reduced by the factor K, so E = E₀/K = σ/Kε₀. The potential difference across the plates is the sum of the drops across the two regions: V = E₀(d - t) + E(t) V = (σ/ε₀)(d - t) + (σ/Kε₀)t V = (σ/ε₀)[(d - t) + t/K] Substituting σ = Q/A: V = (Q/ε₀A)[(d - t) + t/K] Therefore C = Q/V = ε₀A/[(d - t) + t/K]. Note the checks: with t = 0 this gives C = ε₀A/d, and with t = d it gives C = Kε₀A/d, both as expected. Numerical: A = 250 cm² = 250×10⁻⁴ = 2.5×10⁻² m², d = 5.0×10⁻³ m, t = 3.0×10⁻³ m, K = 5. Denominator = (5.0×10⁻³ - 3.0×10⁻³) + 3.0×10⁻³/5 = 2.0×10⁻³ + 0.6×10⁻³ = 2.6×10⁻³ m. C = 8.85×10⁻¹² × 2.5×10⁻²/2.6×10⁻³ C = 2.2125×10⁻¹³/2.6×10⁻³ = 8.51×10⁻¹¹ F ≈ 85 pF.
(a) Derive an expression for the energy stored in a charged capacitor and hence obtain the energy density of the electric field between the plates. (b) A 4.0 μF capacitor and a 6.0 μF capacitor are joined in parallel and the combination is charged by a 50 V battery. Find the total energy stored and the charge on each capacitor.
Answer
(a) Suppose at some instant the capacitor holds charge q, so its potential difference is V' = q/C. To move a further small charge dq from the negative plate to the positive plate, the work required is dW = V' dq = (q/C) dq. Total work in charging from 0 to Q: W = ∫(q/C)dq from 0 to Q = Q²/2C. This work is stored as electrostatic potential energy: U = Q²/2C = ½CV² = ½QV (using Q = CV). Energy density: for a parallel plate capacitor C = ε₀A/d and V = Ed, so U = ½ × (ε₀A/d) × (Ed)² = ½ε₀E² × (Ad). Since Ad is the volume between the plates, the energy stored per unit volume is u = U/(Ad) = ½ε₀E². This result is general: wherever there is an electric field E, energy of density ½ε₀E² is stored in that region of space. (b) In parallel, C(total) = 4.0 + 6.0 = 10.0 μF, and each capacitor has the full 50 V across it. Total energy U = ½CV² = ½ × 10.0×10⁻⁶ × (50)² = ½ × 10.0×10⁻⁶ × 2500 = 1.25×10⁻² J = 12.5 mJ. Charge on the 4.0 μF capacitor: Q₁ = C₁V = 4.0×10⁻⁶ × 50 = 2.0×10⁻⁴ C = 200 μC. Charge on the 6.0 μF capacitor: Q₂ = C₂V = 6.0×10⁻⁶ × 50 = 3.0×10⁻⁴ C = 300 μC. Total charge drawn = 500 μC, consistent with Q = C(total)V = 10 μF × 50 V.
Read the passage and answer the questions that follow: A capacitive touch screen works by storing a tiny amount of charge between a transparent conducting layer and a grid of electrodes, separated by a thin insulating film. When a fingertip approaches, it acts like a nearby conductor and changes the local capacitance by a few picofarads. The controller detects this change and reports a touch. In one test rig an engineer models a single touch cell as a parallel plate capacitor with plate area 0.50 mm² and separation 0.10 mm, filled with an insulating film of dielectric constant 3.0. (a) Calculate the capacitance of this touch cell. (b) The cell is held at 3.0 V. Find the charge stored on it. (c) Find the energy stored in the cell at 3.0 V. (d) Explain what happens to the capacitance and the stored charge if the insulating film is replaced by one of dielectric constant 6.0, the voltage being held constant.
Answer
(a) A = 0.50 mm² = 0.50×10⁻⁶ m², d = 0.10 mm = 1.0×10⁻⁴ m, K = 3.0. C = Kε₀A/d = 3.0 × 8.85×10⁻¹² × 0.50×10⁻⁶/1.0×10⁻⁴ C = 1.3275×10⁻¹⁷/1.0×10⁻⁴ = 1.33×10⁻¹³ F ≈ 0.13 pF. (b) Q = CV = 1.33×10⁻¹³ × 3.0 = 4.0×10⁻¹³ C, that is about 0.40 pC. (c) U = ½CV² = ½ × 1.33×10⁻¹³ × (3.0)² = ½ × 1.33×10⁻¹³ × 9.0 = 6.0×10⁻¹³ J. (d) Capacitance is directly proportional to K, so doubling K from 3.0 to 6.0 doubles the capacitance to about 0.27 pF. Because the cell is held at a constant 3.0 V by the controller, the charge Q = CV also doubles, to about 0.80 pC, with the extra charge supplied by the source. The stored energy ½CV² likewise doubles to about 1.2×10⁻¹² J.
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