Q2.1 Two charges 5 × 10⁻⁸ C and −3 × 10⁻⁸ C are placed 16 cm apart. At what points is the electric potential zero?
Given:
q₁ = 5 × 10⁻⁸ C
q₂ = −3 × 10⁻⁸ C
Distance = 16 cm = 0.16 m
Let the zero potential point be at distance r from positive charge.
Using superposition principle:
V = k (q₁/r + q₂/(0.16 − r))
For V = 0,
5/r = 3/(0.16 − r)
Solving:
5(0.16 − r) = 3r
0.8 − 5r = 3r
0.8 = 8r
r = 0.1 m = 10 cm
For outer region:
5/s = 3/(s − 0.16)
Solving gives:
s = 0.4 m = 40 cm
Electric potential is zero at 10 cm between the charges and 40 cm outside the system.
Q2.2 A regular hexagon of side 10 cm has 5 µC charge at each vertex. Find the electric potential at its centre.
Given:
Side = 10 cm = 0.1 m
Charge at each vertex q = 5 µC = 5 × 10⁻⁶ C
Number of charges = 6
Distance from centre to each vertex = 0.1 m
Electric potential due to one charge:
V₁ = kq/r
Total potential:
V = 6 × (9 × 10⁹ × 5 × 10⁻⁶) / 0.1
V = 2.7 × 10⁶ V
Electric potential at centre = 2.7 × 10⁶ V
Q2.3 Two charges 2 µC and −2 µC are placed 6 cm apart. (a) Identify equipotential surface. (b) What is direction of electric field?
(a) Since charges are equal and opposite, the perpendicular bisector of the line joining charges forms an equipotential surface.
At every point on this plane:
V = 0
(b) Electric field direction is from positive to negative charge along dipole axis.
Equipotential surface: Perpendicular bisector plane.
Electric field direction: From + to − along axis.
Q2.4 A spherical conductor of radius 12 cm carries charge 1.6 × 10⁻⁷ C. Find electric field (i) inside (ii) just outside (iii) at 18 cm.
Given:
R = 12 cm = 0.12 m
Q = 1.6 × 10⁻⁷ C
k = 9 × 10⁹ Nm²/C²
(i) Inside conductor:
E = 0
(ii) Just outside:
E = kQ/R²
E = (9 × 10⁹ × 1.6 × 10⁻⁷) / (0.12)²
E = 10⁵ N/C
(iii) At 18 cm (0.18 m):
E = kQ/r²
E = (9 × 10⁹ × 1.6 × 10⁻⁷) / (0.18)²
E = 4.4 × 10⁴ N/C
Inside = 0
Just outside = 10⁵ N/C
At 18 cm = 4.4 × 10⁴ N/C
Q2.5 A parallel plate capacitor with air has capacitance 8 pF. If distance is halved and dielectric constant 6 inserted, find new capacitance.
Original:
C = ε₀A/d
New:
C₁ = 6ε₀A/(d/2)
C₁ = 12C
C₁ = 96 pF
New capacitance = 96 pF
Q2.6 Three capacitors of 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the equivalent capacitance? (b) If the combination is connected to a 100 V supply, find the charge on each capacitor.
Given:
C₁ = 2 pF
C₂ = 3 pF
C₃ = 4 pF
V = 100 V
(a) For parallel combination:
Ceq = C₁ + C₂ + C₃
Ceq = 2 + 3 + 4 = 9 pF
(b) Charge on each capacitor:
Q = CV
For 2 pF:
Q₁ = 2 × 100 = 200 pC = 2 × 10⁻¹⁰ C
For 3 pF:
Q₂ = 3 × 100 = 300 pC = 3 × 10⁻¹⁰ C
For 4 pF:
Q₃ = 4 × 100 = 400 pC = 4 × 10⁻¹⁰ C
Equivalent capacitance = 9 pF
Charges = 2×10⁻¹⁰ C, 3×10⁻¹⁰ C, 4×10⁻¹⁰ C
Q2.7 Three capacitors of 2 pF, 3 pF and 4 pF are connected in series. Find the equivalent capacitance.
For series combination:
1/Ceq = 1/2 + 1/3 + 1/4
LCM = 12
1/Ceq = (6 + 4 + 3)/12
1/Ceq = 13/12
Ceq = 12/13 pF
Equivalent capacitance = 12/13 pF ≈ 0.92 pF
Q2.8 A parallel plate capacitor has area 6 × 10⁻³ m² and plate separation 3 mm. Find its capacitance and the charge stored when connected to 100 V supply.
Given:
A = 6 × 10⁻³ m²
d = 3 mm = 3 × 10⁻³ m
V = 100 V
ε₀ = 8.85 × 10⁻¹² F/m
Capacitance:
C = ε₀A/d
C = (8.85×10⁻¹² × 6×10⁻³) / (3×10⁻³)
C = 17.7 × 10⁻¹² F = 17.7 pF
Charge:
Q = CV
Q = 17.7×10⁻¹² × 100
Q = 1.77 × 10⁻⁹ C
Capacitance = 17.7 pF
Charge = 1.77 × 10⁻⁹ C
Q2.9 In Question 2.8, a dielectric slab of thickness 3 mm and dielectric constant 6 is inserted between the plates. Find the new capacitance and the charge stored (battery connected).
Dielectric constant, K = 6
New capacitance:
C₁ = K C
C₁ = 6 × 17.7 pF
C₁ = 106.2 pF
Charge (battery connected → V constant):
Q₁ = C₁V
Q₁ = 106.2×10⁻¹² × 100
Q₁ = 1.062 × 10⁻⁸ C
New capacitance = 106.2 pF
New charge = 1.062 × 10⁻⁸ C
Q2.10 A 12 pF capacitor is connected to a 50 V battery. Calculate the electrostatic energy stored.
Given:
C = 12 × 10⁻¹² F
V = 50 V
Energy stored:
U = ½ CV²
U = ½ × 12×10⁻¹² × (50)²
U = 1.5 × 10⁻⁸ J
Energy stored = 1.5 × 10⁻⁸ J
Q2.11 A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
Given:
C₁ = 600 pF = 600 × 10⁻¹² F
V = 200 V
Initial energy:
U₁ = ½ C₁ V²
U₁ = ½ × 600×10⁻¹² × (200)²
U₁ = 1.2 × 10⁻⁵ J
After connection with identical capacitor:
Total capacitance:
C_total = 600 + 600 = 1200 pF
Final voltage (charge shared equally):
V' = V/2 = 100 V
Final energy:
U₂ = ½ × 1200×10⁻¹² × (100)²
U₂ = 6 × 10⁻⁶ J
Energy lost:
ΔU = U₁ − U₂
ΔU = 6 × 10⁻⁶ J
Electrostatic energy lost = 6 × 10⁻⁶ J
Q2.12 A charge of 8 mC is located at the origin. Calculate the work done in moving a charge of −2 × 10⁻⁹ C from point P(0,0,3 cm) to Q(0,4 cm,0).
Given:
Q = 8 × 10⁻³ C
q = −2 × 10⁻⁹ C
OP = 0.03 m
OQ = 0.04 m
Potential at P:
V₁ = kQ / 0.03
Potential at Q:
V₂ = kQ / 0.04
Work done:
W = q (V₂ − V₁)
After substitution:
W = 1.2 J
Work done = 1.2 J
Q2.13 A cube of side b has a charge q at each of its vertices. Determine the electric potential and electric field at the centre of the cube.
Distance from centre to each vertex:
r = (√3 / 2) b
Potential due to one charge:
V₁ = kq/r
Total potential (8 charges):
V = 8kq / r
V = 8kq / ( (√3/2) b )
V = (16kq)/(√3 b)
Due to symmetry, electric field vectors cancel.
E = 0
Potential = (16kq)/(√3 b)
Electric field = 0
Q2.14 Two charges 1.5 µC and 2.5 µC are located 30 cm apart. Find the electric potential and electric field at the midpoint.
Distance from midpoint:
r = 0.15 m
Potential:
V = k(q₁ + q₂)/r
V = 2.4 × 10⁵ V
Electric field:
E = k(q₂ − q₁)/r²
E = 4 × 10⁵ N/C
Potential = 2.4 × 10⁵ V
Electric field = 4 × 10⁵ N/C
Q2.15 A spherical conducting shell of inner radius r₁ and outer radius r₂ has a net charge Q. If a point charge q is placed at its centre, find the surface charge density on inner and outer surfaces.
Induced charge on inner surface:
−q
Surface charge density (inner):
σ₁ = −q / 4πr₁²
Charge on outer surface:
Q + q
Surface charge density (outer):
σ₂ = (Q + q) / 4πr₂²
σ₁ = −q / 4πr₁²
σ₂ = (Q + q) / 4πr₂²
Q2.16 Show that the normal component of the electric field has a discontinuity of magnitude σ/ε₀ across a charged surface.
Using Gauss’s law, consider a small Gaussian pillbox across the charged surface.
Total flux:
Φ = (E₂ − E₁) A
Charge enclosed:
q = σA
From Gauss law:
(E₂ − E₁) A = σA / ε₀
E₂ − E₁ = σ / ε₀
Normal component of electric field changes by σ/ε₀ across surface.
Q2.17 A long charged cylinder of linear charge density λ is surrounded by a hollow coaxial conducting cylinder. Find the electric field in the region between the cylinders.
Using Gauss law:
Consider cylindrical Gaussian surface of radius r and length L.
E(2πrL) = λL / ε₀
Cancel L:
E = λ / (2πε₀ r)
Electric field between cylinders: E = λ / (2πε₀ r)
Q2.18 In a hydrogen atom, electron and proton are separated by 0.53 Å. Find the electrostatic potential energy of the system and the minimum work required to separate them.
Distance:
r = 0.53 Å = 0.53 × 10⁻¹⁰ m
Potential energy:
U = − k e² / r
Substituting values:
U = − 27.2 eV
Minimum work required:
W = 13.6 eV
Potential energy = −27.2 eV
Minimum work required = 13.6 eV
Q2.19 Calculate the electrostatic potential energy of the hydrogen molecular ion H₂⁺.
Total potential energy is sum of interactions:
U = k [ (q₁q₂/r₁₂) + (q₁q₃/r₁₃) + (q₂q₃/r₂₃) ]
After substituting numerical values from given distances:
U = − 19.2 eV
Electrostatic potential energy = −19.2 eV
Q2.20 Two conducting spheres of radii a and b are connected by a wire. Find the ratio of electric fields at their surfaces.
Since connected by wire, potentials are equal:
kQ₁/a = kQ₂/b
Q₁/Q₂ = a/b
Electric field at surface:
E = kQ/R²
Therefore:
E₁/E₂ = b/a
Ratio of electric fields = b/a
Q2.21 Two charges –q and +q are located at (0,0,–a) and (0,0,a). Find the electric potential at (i) axial point (0,0,z) and (ii) equatorial plane point (x,y,0).
This system forms an electric dipole.
(i) At axial point (0,0,z):
V = (1/4πε₀) [ q/(z−a) − q/(z+a) ]
For large distance (z >> a):
V ≈ (1/4πε₀) × (2qa / z²)
(ii) At equatorial plane (x,y,0):
Distances from both charges are equal.
V = 0
Axial potential: Dipole expression
Equatorial potential: 0
Q2.22 For an electric quadrupole, show that electric potential varies as 1/r³ for r >> a.
Using binomial expansion for r >> a:
Potential due to quadrupole:
V ∝ 1/r³
Comparison:
Monopole → 1/r
Dipole → 1/r²
Quadrupole → 1/r³
Quadrupole potential decreases as 1/r³.
Q2.23 You need a 2 µF capacitor rated 1 kV. Available capacitors are 1 µF rated 400 V. How many are required?
To withstand 1000 V:
1000 / 400 ≈ 2.5
Therefore 3 capacitors in series are required.
Series capacitance:
Cseries = 1/3 µF
To get 2 µF:
n/3 = 2
n = 6 branches
Total capacitors:
6 × 3 = 18
Minimum number of capacitors required = 18
Q2.24 Find the area of plates of a 2 F capacitor if separation is 0.5 cm.
Given:
C = 2 F
d = 0.5 cm = 5 × 10⁻³ m
ε₀ = 8.85 × 10⁻¹² F/m
Using:
C = ε₀A/d
A = Cd/ε₀
A = (2 × 5×10⁻³) / (8.85×10⁻¹²)
A = 1.13 × 10⁹ m²
Area required = 1.13 × 10⁹ m²
Q2.25 Find equivalent capacitance of the given capacitor network and charge on each capacitor when connected to 300 V supply.
From circuit reduction:
Two capacitors in series:
C' = 100 pF
Parallel with third:
C'' = 200 pF
Series with fourth:
Ceq = 200/3 pF
Total charge:
Q = CV
Q = 2 × 10⁻⁸ C
Equivalent capacitance = 200/3 pF
Total charge = 2 × 10⁻⁸ C
Q2.26 A parallel plate capacitor has area 90 cm² and separation 2.5 mm. It is connected to 400 V supply. Find energy stored and energy density.
Convert:
A = 90 × 10⁻⁴ m²
d = 2.5 × 10⁻³ m
Capacitance:
C = ε₀A/d
Energy:
U = ½ CV²
U = 2.55 × 10⁻⁶ J
Energy density:
u = ½ ε₀E²
Energy stored = 2.55 × 10⁻⁶ J
Energy density = ½ ε₀E²
Q2.27 A 4 µF capacitor charged to 200 V is connected to a 2 µF uncharged capacitor. Find energy lost.
Initial energy:
U₁ = ½ × 4×10⁻⁶ × (200)²
U₁ = 0.08 J
Final voltage:
V = 400/3 V
Final energy:
U₂ = 5.33 × 10⁻² J
Energy lost:
ΔU = 2.67 × 10⁻² J
Energy lost = 2.67 × 10⁻² J
Q2.28 Show that the force on each plate of a parallel plate capacitor is (½)QE.
Energy density:
u = ½ ε₀E²
Force = Energy density × Area
Using Q = ε₀EA:
F = ½ QE
The factor ½ appears because plate does not exert force on itself.
Force on each plate = ½ QE