Electrostatic Potential and Capacitance Class 12 Physics NCERT Solutions

Step-by-step solution

1. 1Take x from the 5 × 10⁻⁸ C charge toward the –3 × 10⁻⁸ C charge; the separation is 0.16 m.
2. 2V = 0 gives k(5 × 10⁻⁸)/x + k(–3 × 10⁻⁸)/(0.16 – x) = 0 → 5/x = 3/(0.16 – x).
3. 3Between the charges: 5(0.16 – x) = 3x → 0.8 = 8x → x = 0.10 m = 10 cm from the positive charge.
4. 4Beyond the charges: 5(0.16 – x)... solving in the two outer regions gives x = 0.40 m (40 cm) on the side of the 5 × 10⁻⁸ C charge, i.e. 0.56 m from the negative charge.
5. 5Sanity: the zero-potential point must lie nearer the smaller magnitude (–3 × 10⁻⁸ C), which both points satisfy on their own side.

Step-by-step solution

1. 1For a regular hexagon the centre is 10 cm from every vertex (r = 0.10 m).
2. 2Potential adds as a scalar: V = 6 × kq/r.
3. 3V = 6 × (9 × 10⁹ × 5 × 10⁻⁶/0.10).
4. 4V = 6 × 4.5 × 10⁵ = 2.7 × 10⁶ V.

Step-by-step solution

1. 1(a) The two equal-and-opposite charges form a dipole; on the perpendicular bisector of AB every point is equidistant from both charges.
2. 2At such a point V = kq/r + k(–q)/r = 0, so the entire plane bisecting AB perpendicularly is an equipotential surface (V = 0).
3. 3(b) Field lines are always normal to equipotential surfaces.
4. 4Hence at every point of this plane the field is perpendicular to the plane, and since the field of the pair points from the +2 µC toward the –2 µC, it runs parallel to AB.

Step-by-step solution

1. 1(a) In electrostatic equilibrium the field inside a conductor is zero: E = 0.
2. 2(b) Just outside, the sphere behaves like a point charge at its centre: E = kq/R².
3. 3E = (9 × 10⁹ × 1.6 × 10⁻⁷)/(0.12)² = 1.44 × 10³/0.0144 = 1.0 × 10⁵ N/C, radially outward.
4. 4(c) At r = 0.18 m: E = kq/r² = (1.44 × 10³)/(0.18)² = 1.44 × 10³/0.0324 = 4.44 × 10⁴ N/C.
5. 5Direction is radially outward because the charge is positive.

Step-by-step solution

1. 1For a parallel-plate capacitor C = ε₀KA/d, so C ∝ K/d.
2. 2Halving the distance doubles the capacitance: C → 16 pF.
3. 3Filling with a dielectric of K = 6 multiplies it by 6: C = 16 × 6 = 96 pF.

Step-by-step solution

1. 1(a) Series: 1/C = 1/9 + 1/9 + 1/9 = 3/9 pF⁻¹ → C = 3 pF.
2. 2(b) The same charge q flows through every series capacitor: q = C_total V = (3 × 10⁻¹²)(120) = 3.6 × 10⁻¹⁰ C.
3. 3Across each capacitor: Vᵢ = q/Cᵢ = (3.6 × 10⁻¹⁰)/(9 × 10⁻¹²) = 40 V.
4. 4Check: 40 + 40 + 40 = 120 V, the supply voltage.

Step-by-step solution

1. 1(a) Parallel: C = C₁ + C₂ + C₃ = 2 + 3 + 4 = 9 pF.
2. 2(b) Every parallel capacitor sees the full 100 V: qᵢ = CᵢV.
3. 3q₁ = (2 × 10⁻¹²)(100) = 2 × 10⁻¹⁰ C; q₂ = (3 × 10⁻¹²)(100) = 3 × 10⁻¹⁰ C; q₃ = 4 × 10⁻¹⁰ C.

Step-by-step solution

1. 1C = ε₀A/d with ε₀ = 8.85 × 10⁻¹² F m⁻¹, A = 6 × 10⁻³ m², d = 3 × 10⁻³ m.
2. 2C = (8.85 × 10⁻¹² × 6 × 10⁻³)/(3 × 10⁻³) = (8.85 × 10⁻¹²)(2) = 1.77 × 10⁻¹¹ F = 17.7 pF.
3. 3Charge on each plate: q = CV = (1.77 × 10⁻¹¹)(100) = 1.77 × 10⁻⁹ C ≈ 1.8 × 10⁻⁹ C.

Step-by-step solution

1. 1With the mica, C becomes K times the air value: C = 6 × 17.7 ≈ 106 pF.
2. 2(a) With the battery still connected, V stays at 100 V.
3. 3The charge therefore rises: q = CV ≈ (1.06 × 10⁻¹⁰)(100) ≈ 1.06 × 10⁻⁸ C.
4. 4(b) After the supply is disconnected, charge is conserved at 1.77 × 10⁻⁹ C.
5. 5With C now 106 pF, V falls: V = q/C = (1.77 × 10⁻⁹)/(1.06 × 10⁻¹⁰) ≈ 16.7 V.

Step-by-step solution

1. 1Energy stored: U = ½CV².
2. 2U = ½ × (12 × 10⁻¹²) × (50)².
3. 3U = ½ × 12 × 10⁻¹² × 2500 = 1.5 × 10⁻⁸ J.

Step-by-step solution

1. 1Initial energy: Uᵢ = ½CV² = ½ × (600 × 10⁻¹²) × (200)² = 1.2 × 10⁻⁵ J.
2. 2Charge before sharing: q = CV = (600 × 10⁻¹²)(200) = 1.2 × 10⁻⁷ C.
3. 3After connection the total capacitance is 1200 pF; q is conserved.
4. 4Final energy: U_f = q²/2C_total = (1.2 × 10⁻⁷)²/(2 × 1.2 × 10⁻⁹) = 6 × 10⁻⁶ J.
5. 5Loss = Uᵢ – U_f = 1.2 × 10⁻⁵ – 6 × 10⁻⁶ = 6 × 10⁻⁶ J.