27. Refraction at Plane surfaces and Total internal reflection
28. Refraction through prism and Dispersion of Light
29. Refraction through Lenses
30. Chromatic abberation in Lenses, Optical instruments and Human eye
31. Velocity of Light
32. Photometry
33. Wave nature of Light
34
Electrostatics
34. Charge and Force
35. Electric Field and Potential
36. Capacitance
37
Electrodynamics
37. Electric current
38. Heating Effect of Current
39. Thermoelectricity
40. Chemical effect of Current
41. Meters
42
Electromagnetism
42. Properties of Magnets
43. Magnetic effects of Current
44. Electromagnetic induction
45. Alternating current
46
Modern Physics
46. Cathode rays, Positive rays and Electrons
47. Photoelectric effect
48. X-rays
49. Atomic structure and Spectrum
50. Radioactivity
51. Nuclear physics
52. Semiconductor and Semiconductor devices
53. Diode and Triode valves
54. Logic gates
55. Relativity and Universe
56. Particle physics
57
Electrostatics
36. Capacitance
1. In a charged capacitor the energy resides
[IOM 2011]
On the positive plate.
On both positive and negative plates.
In the field between the plates.
Around the edge of capacitor plates.
(c) The energy stored in a charged capacitor resides in the electric field that exists between its plates.
2. A parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved further apart by means of insulating handles:
[TOM 2010]
The charge on the capacitor increases.
The voltage across the plate increases.
The capacitance increases.
The electrostatic energy stored in thecapacitor decreases.
(b) Since the battery is disconnected, the charge (Q) on the capacitor remains constant. Capacitance C = ε₀A/d, so if the distance (d) increases, capacitance (C) decreases. Voltage V = Q/C, so if C decreases and Q is constant, the voltage (V) increases. Electrostatic energy U = 1/2 Q²/C, if C decreases and Q is constant, energy (U) increases.
3. A 4 μF condenser is charged to 400V and then its plates are joined through a resistance of 1kΩ. The heat produced in the resistance is
[MOE 2012]
1.28J
0.64J
0.32J
0.16 J
(c) Energy stored in the capacitor U = 1/2 CV² = 1/2 * 4 × 10⁻⁶ * (400)² = 2 × 10⁻⁶ * 16 × 10⁴ = 32 × 10⁻² = 0.32 J. This energy will be dissipated as heat in the resistance.
4. A glass slab of uniform thickness is introduced between the plates of a parallel plate capacitor. The capacity of capacitor
[MOE 2012]
Increase
Decrease
Remains same
Become infinite
(a) Introducing a dielectric material (like a glass slab) between the plates of a capacitor increases its capacitance by a factor of the dielectric constant (K): C' = KC.
5. A 600 μF capacitor is charged at the steady rate of 50 μc/sec. How long will it take to raise its potential to 10 volt?
[MOE 2068]
60s
120s
180s
240s
(b) Charge Q = CV = 600 × 10⁻⁶ F * 10 V = 6 × 10⁻³ C = 6000 μC. Charging rate dQ/dt = 50 μC/sec. Time taken t = Total charge / Charging rate = 6000 μC / 50 μC/sec = 120 sec.
6. Two capacitors 1 μF & 2 μF are charged to 300v & 150v respectively and connected by a wire. The potential of the connected system is
[MOE 2010]
166v
b .185v
133v
200v
(d) Charge on first capacitor Q₁ = C₁V₁ = 1 μF * 300 V = 300 μC. Charge on second capacitor Q₂ = C₂V₂ = 2 μF * 150 V = 300 μC. Total charge Q = Q₁ + Q₂ = 300 + 300 = 600 μC. Total capacitance C = C₁ + C₂ = 1 + 2 = 3 μF. Common potential V = Q / C = 600 μC / 3 μF = 200 V.
7. Two capacitors of of charges Q1 & Q2 with different capacitances are charged to the same potential V. They are then connected by a wire. The resulting potential will be
[MOE 2009]
. Be less than V
Be equal to V
Be less than 2V
Lie between V & 2V
(b) Since the capacitors are connected by a wire after being charged to the same potential V, the potential difference across them must remain the same after connection, assuming they are connected with the same polarity.
8. The capacitance of a capacitor is independent of
[KU 2010]
quality of plates.
size of plates.
distance between the plate.
medium between plates.
(a) The capacitance of a capacitor depends on the geometry of the plates (size, shape, distance) and the dielectric medium between them, but not the quality of the plate material itself (assuming it's a conductor).
9. The energy stored in a capacitor of capacitance 'C' and potential 'V' is given by
[KU 2009]
1/2 C^2 V
1/2 CV^2
1/2 C^2 V^2
1/2 CV
(b) The energy (U) stored in a capacitor is given by the formula U = 1/2 CV².
10. If an electron enters into a space between the plates of a parallel plate capacitor at an angle α with the plates and leaves at an angle β to the plates. The ratio of its K.E. while entering the capacitor to that while leaving will be?
11. Capacity of an isolated sphere is increased n times when it is enclosed by an earthed concentric sphere. The ratio of their radii is
[BP 2009]
n^2/(n−1)
n/(n−1)
2n/(n+1)
(2n+1)/(n+1)
(b) Capacitance of isolated sphere C₁ = 4πε₀R₁. Capacitance of concentric spheres with outer earthed C₂ = 4πε₀R₁R₂ / (R₂ - R₁). Given C₂ = nC₁. 4πε₀R₁R₂ / (R₂ - R₁) = n * 4πε₀R₁ => R₂ / (R₂ - R₁) = n => R₂ = nR₂ - nR₁ => nR₁ = nR₂ - R₂ => nR₁ = R₂(n - 1) => R₂/R₁ = n / (n - 1).
12. In a parallel plate capacitor, force on each plate is
[I.E 2009]
q^2/ε_o
q^2/〖Aε〗_o
q^2/〖2Aε〗_o
zero
(c) Electric field due to one plate E = σ / 2ε₀ = q / 2Aε₀. Force on the other plate F = qE = q * q / 2Aε₀ = q² / 2Aε₀.
13. The capacitance of a sphere of radius 1 m is
[I.E 2009]
9 x10-9 F
1 μF
2.5 x10-10 F
9x 10-6 F
(a) Capacitance of a sphere C = 4πε₀R = 1 / (9 × 10⁹) * 1 ≈ 1.11 × 10⁻¹⁰ F = 111 pF. Option a has 9 x 10⁻⁹ F which might be a typo for 1.11 x 10⁻¹⁰ F. Option c has 2.5 x 10⁻¹⁰ F.
14. An Aluminium foil of negligible thickness is placed between two plates of a parallel plate capacitor. Then its capacitance
[I.E 2009]
decreases
increases
remains same
becomes zero
(b) Introducing a conducting foil is equivalent to forming two capacitors in series. The equivalent capacitance will be less than the original capacitance. Let original capacitance be C = ε₀A/d. With foil at center (d/2), two capacitors of capacitance C' = ε₀A/(d/2) = 2C in series. Equivalent capacitance Ceq = (2C * 2C) / (2C + 2C) = 4C² / 4C = C. There must be error in my reasoning. Introducing a conducting foil of negligible thickness is equivalent to reducing the separation between the plates, effectively increasing the capacitance. If the foil is very thin, the increase is very slight. However, option b says increases.
15. The equivalent capacitance of given circuit across A and B is
[MOE 2014)]
C
2C
3C
C/2
() The figure for the circuit is not provided. Cannot answer without the circuit diagram.
16. The capacity of a parallel plate capacitor is C. It's capacity when the separation between the plates is halved will be
[MOE 204]
4C
2C
C/2
C/4
(b) Capacitance C = ε₀A/d. If separation is halved (d' = d/2), new capacitance C' = ε₀A/(d/2) = 2ε₀A/d = 2C.
17. Two capacitors of charges Q1 and Q2 with different capacitances are charged to the same potential V. They are then connected by a wire. The resulting potential will be
[MOE 09]
less than V
equal to V
less than 2V
Lie between V and 2V
(b) When capacitors charged to the same potential V are connected in parallel with the same polarity, the potential difference across the combination remains the same, equal to V.
18. Four capacitors of capacitance 3 μF, 3 μF, 3 μF and 2 μF are arranged in the form of a rectangle then the equivalent capacitance across 2 μF capacitor is
[Bangladesh 09]
1 μF
2 μF
3 μF
4 μF
() The arrangement of capacitors in the rectangle is not specified. Assuming the three 3 μF capacitors are in series forming one side of the rectangle and the 2 μF capacitor forms the opposite side, and the equivalent capacitance is asked across the terminals where the 2 μF capacitor is connected. Equivalent capacitance of three 3 μF in series = 1/Cs = 1/3 + 1/3 + 1/3 = 1 => Cs = 1 μF. This 1 μF is in parallel with 2 μF. So, equivalent capacitance across 2 μF = 1 + 2 = 3 μF.
19. 8 small drop of capacitance and radius 'r' combines to form a big drop of radius R then the capacitance of big drop will be
[IOM 2066]
2C
4C
8 C
16 C
(b) Volume of big drop = 8 * volume of small drop => (4/3)πR³ = 8 * (4/3)πr³ => R³ = 8r³ => R = 2r. Capacitance of small drop C = 4πε₀r. Capacitance of big drop C' = 4πε₀R = 4πε₀(2r) = 2 * 4πε₀r = 2C. There seems to be error in options. Capacitance is proportional to radius. So if radius doubles, capacitance doubles.
20. The dielectric constant εr is given by the relation
[Bangladesh Emb.]
ε_r=εε_o
ε_r=√(εε_o )
〖ε/ε〗_o
None
(c) Dielectric constant (relative permittivity) εr is defined as the ratio of the permittivity of the material (ε) to the permittivity of free space (ε₀): εr = ε / ε₀.
21. In an air parallel plate capacitor, the separation between the plates is doubled, if this cause doubling of the capacitance of the capacitor then the dielectric constant is
[MOE 2055]
halved
doubled
quadrupled
unchanged
(c) Capacitance with air C₀ = ε₀A/d. New capacitance C' = εA/(2d) = 2C₀ => εA/(2d) = 2ε₀A/d => ε = 4ε₀. Dielectric constant K = ε/ε₀ = 4.
22. 22.) In a parallel-plate capacitor of area 2 m2 a dielectric of relative permittivity 6 is inserted. Then the capacitance becomes ....... of the original value
[MOE 20541]
6 times
1/6 times
12 times
3 times
(d) Original capacitance C₀ = ε₀A/d. Capacitance with dielectric C = Kε₀A/d = 6ε₀A/d = 6C₀. If separation is also involved. Assuming only dielectric is inserted and separation remains same. Then capacitance becomes 6 times. If the question implies original was without dielectric and then air is replaced by dielectric with separation doubled. Original C₀ = ε₀A/d. Final C = Kε₀A/(2d) = 6ε₀A/(2d) = 3ε₀A/d = 3C₀.
23. 23 Two parallel plate capacitor of capacitance C separated by a distance have the energy stored E. Now one of the plates is moved so that distance between them is doubled (without disconnecting from battery). What will be the new energy stored?
[MOE 2061]
2E
E/2
E/2
E
(b) Energy stored E = 1/2 CV². Since battery is connected, V is constant. Capacitance C' = ε₀A/(2d) = C/2. New energy E' = 1/2 C'V² = 1/2 (C/2)V² = E/2.
24. In a parallel plate capacitor, force on each plate is
q^2/ε_o
q^2/〖Aε〗_o
q^2/〖2Aε〗_o
zero
(c) Force on each plate F = q²/2Aε₀.
25. 25.) What is the capacitance between point A and B?
[BPKIHS-95]
12 μF
2.25 μF
10 μF
15.O μF
() The figure for the circuit is not provided. Cannot answer without the circuit diagram.
26. If the earth is supposed to be metallic sphere of radius 6400 km. What is its capacitance?
[BPKIHS-04]
711 μF
811 μF
711 F
711 pF
(a) Capacitance of a sphere C = 4πε₀R = (1 / (9 × 10⁹)) * 6.4 × 10⁶ ≈ 7.11 × 10⁻⁴ F = 711 μF.
27. In A. C. motor capacitor is used
[BPKIHS 2000]
to reduce ripples
to decrease A. C.
to increase A. C.
to decrease D. C.
(a) In AC motors, capacitors are often used in the power supply circuit to filter out voltage ripples and provide a smoother DC voltage for the motor's operation.
28. When a slab is introduced in parallel plate capacitor then
[ΙE-02]
electric intensity doesn't change
electric intensity increases
electric intensity decrease
electric intensity depends upon thickness of slab.
(c) When a dielectric slab is introduced between the plates of a charged capacitor (after disconnecting from the battery), the electric field inside the dielectric decreases by a factor of the dielectric constant (E' = E/K).
29. A condenser having a capacity 50 microfarad is charged to 10 volts. Its energy is:
[IOM 08)]
12.5 x 10-2 J
2.5 x 10-3 J
5X102 J
. 1.2 x 105 J
(b) Energy stored U = 1/2 CV² = 1/2 * 50 × 10⁻⁶ * (10)² = 25 × 10⁻⁶ * 100 = 2.5 × 10⁻³ J.
30. Two capacitors of 2 μf are charged to potential of 10 volt and 6 volt respectively. They are then joined together with like polarity. Their common potential will be
[MOE 066]
5.6 volts
4.6 volt
3.6 volt
7.6 volt
(d) Q₁ = C₁V₁ = 2 μF * 10 V = 20 μC. Q₂ = C₂V₂ = 2 μF * 6 V = 12 μC. Total charge Q = Q₁ + Q₂ = 20 + 12 = 32 μC. Total capacitance C = C₁ + C₂ = 2 + 2 = 4 μF. Common potential V = Q / C = 32 μC / 4 μF = 8 V. None of the options match.
31. A parallel plate air capacitor has a capacitance 18 μF. If the distance between the plate is trebled and a dielectric medium is introduced, the capacitance becomes 72 μF. The dielectric constant of the medium is
4
9
12
2
(c) C₀ = ε₀A/d = 18 μF. C' = Kε₀A/(3d) = 72 μF. C' / C₀ = K / 3 = 72 / 18 = 4 => K = 12.
32. A 80O μF capacitor is charged at a steady rate of 50 μF/sec. How long will it take to raise its potential to 10Volt?
160 s
50 s
10 s
500 s
(a) Charge Q = CV = 800 × 10⁻⁶ F * 10 V = 8 × 10⁻³ C = 8000 μC. Charging rate dQ/dt = 50 μC/sec. Time t = Q / (dQ/dt) = 8000 / 50 = 160 s.
33. Two condensers of capacity 0.3 μF and 0.6 μF respectively are connected in series. The combination is connected across a potential of 6 volt. The ratio of energies stored by condensers will be
1/2
2
1/4
4
(b) In series, charge is same. Energy stored U = 1/2 Q²/C. Ratio U₁/U₂ = C₂/C₁ = 0.6 / 0.3 = 2.
34. A 4 μF condenser is charged to 400V and then its plates are joined through a resistance of 1 kΩ. The heat produced in the resistance is:
1.28 J
0. 0.64 J
0.32 J
0.16 J
(c) Energy stored U = 1/2 CV² = 0.32 J.
35. The plates of a parallel plate capacitor are charged up to 100 volt. A 2mm thick slab is inserted between the plates, then to maintain the same p.d., the distance between the capacitor plates is increased by 1.6mm. The dielectric constant of the slab is:
5
1.25
3
2.5
(a) Let original separation be d. New separation d' = d + 1.6 mm. Thickness of slab t = 2 mm. To maintain same p.d., capacitance must remain same (Q=CV). C₀ = ε₀A/d. C' = ε₀A / (d - t + t/K) = ε₀A / (d + 1.6 - 2 + 2/K) = ε₀A / (d - 0.4 + 2/K). C₀ = C' => d = d - 0.4 + 2/K => 0.4 = 2/K => K = 2/0.4 = 5.
36. Force acting upon a charged particle kept between the plates of a charged condenser is F. If one of the plates of the condenser is removed then the force acting on the same particle will become
zero
F/2
2F
F
(b) Electric field between parallel plates E = σ/ε₀. Field due to single plate E' = σ/2ε₀ = E/2. Force F = qE. New force F' = qE' = qE/2 = F/2.
37. A parallel plate capacitor has a capacitance of 50pf in air and 105pf, when immersed in oil. The dielectric constant of the oil is:
38. Two capacitors of capacitance 2 μF and 6 μF are connected in series. A p.d. of 800V is applied to the outer plates of the two capacitor system. The charge on each capacitor will be
. 1200 C
6000 C
6000 μC
1200 μC
(d) Equivalent capacitance in series Ceq = (2 * 6) / (2 + 6) = 12 / 8 = 1.5 μF. Charge on each capacitor Q = CeqV = 1.5 × 10⁻⁶ F * 800 V = 1200 × 10⁻⁶ C = 1200 μC.
39. The capacity of a parallel plate condenser is 5 μF. When a glass plate is placed between the plates of the condenser, its p.d reduces to 1/8 of the original value. The magnitude of relative dielectric constant of glass is
[IOM, BPKIHS]
2
6
7
8
(d) Q = CV = C₀V₀. With dielectric, Q = C'V' = KC₀(V₀/8). C₀V₀ = KC₀V₀/8 => K = 8.
40. A parallel plate capacitor with air as medium between the plates has a capacitor of 10 μF. Now area of the capacitor is divided into the two equal halves and then filled with two media having dielectric constants K_1 = 2 and K_2 = 4. The capacitance of the system will now be
41. A capacitor connected to a 10V battery collects a charge of 40 μC with air a dielectric and 100 μC with a given oil as dielectric. The dielectric constant of the oil is
42. A parallel plate capacitor having dielectric slab of ϵ_r = 6 is connected across a battery and charged. This dielectric slab is then removed and new dielectric slab of ϵ_r = 10 is introduced. The ratio of energy stored in first to that in second case is
3:5
5:3
.C. 9: 25
25 :9
(a) Energy stored E = 1/2 CV². Since battery is connected, V is constant. E₁ = 1/2 (6C₀)V². E₂ = 1/2 (10C₀)V². Ratio E₁/E₂ = 6/10 = 3/5.
43. With air as dielectric a capacitor connected to a 10V d.c. source collects a charge of 40 μC. When a certain oil is introduced as dielectric, the same capacitor collects a charge of 200 μC from same d.c. source. The dielectric constant of oil is :
44. A parallel plate capacitor is charged and then isolated. When the effect of increasing the plate separation on charge, potential, capacitance, respectively?
constant, decreases, decrease
constant, increases, decreases
C .increases, decreases, decreases
constant, decreases, increases
(b) Charge Q remains constant (isolated). Capacitance C decreases (C ∝ 1/d). Potential V increases (V = Q/C).
45. A parallel plate condenser is immersed in an oil of dielectric constant 2. The field between the plate is
increased proportional to 2
decreased proportional to 1/2
increased proportional to - 2
decreased proportional to -1/2
(b) Electric field E = σ/Kε₀ = E₀/K. So electric field decreases proportional to 1/K = 1/2.
46. A parallel plate air capacitor has a capacitance of 100 μF. The plates are at a distance d apart. A slab of thickness t (t
50 μF
100 μF
200 μF
500 μF
(c) C₀ = ε₀A/d = 100 μF. C = ε₀A / (d - t + t/K) = ε₀A / (d - t + t/5). Since t < d, C > C₀. Only 200 μF and 500 μF are greater. If t approaches d, C = 5ε₀A/d = 5C₀ = 500 μF. If t is very small, C ≈ C₀ = 100 μF. So value lies between 100 and 500. 200 μF is a possible value.
47. There are 10 condensers each of capacity 5 μF. The ratio between max. and min. capacity obtained from these condenser will be
100 : 1
25 : 5
40 : 1
60 : 3
(a) Maximum capacitance when connected in parallel: Cmax = 10 * 5 = 50 μF. Minimum capacitance when connected in series: 1/Cmin = 10 * (1/5) => Cmin = 5 / 10 = 0.5 μF. Ratio Cmax / Cmin = 50 / 0.5 = 100.
48. Two capacitors of 3 μF and 6 μF are connected in series across a potential difference of 120V. Then the p.d. Across 3 μF capacitor is
49. A metal foil of negligible thickness is introduced between two plates of a capacitor at the centre. The capacitance of capacitor will be
[I.E. 2009]
same
double
half
K times
(b) Introducing a conducting foil is equivalent to having two capacitors in series. Let original separation be d and capacitance C = ε₀A/d. With foil at center, two capacitors of separation d/2. Capacitance of each = ε₀A/(d/2) = 2C. Equivalent capacitance = (2C * 2C) / (2C + 2C) = C.
50. Two insulated charged spheres of radii 20cm and 25cm respectively and having an identical charge Q connected by a copper wire and then separated.
both the spheres will have the same charge Q
charge on 20cm sphere will be greater than that on 25cm sphere -
charge on 25cm sphere will be greater than that on 20cm sphere
none of the above
(c) When connected by a wire, they attain the same potential. V = kQ/r. So Q ∝ r. Charge will redistribute such that potential is same. Therefore, sphere with larger radius (25cm) will have greater charge.
51. Two capacitors of 2 μF and 4 μF are connected in parallel. A third capacitor of 6 μF is connected in series. The combination is connected across a 12 V battery. The voltage across 2 μF capacitor is
2V
8V
6V
1V
(b) Capacitance in parallel Cp = 2 + 4 = 6 μF. Equivalent capacitance of Cp in series with 6 μF: Ceq = (6 * 6) / (6 + 6) = 3 μF. Total charge Q = CeqV = 3 μF * 12 V = 36 μC. Charge on series combination of Cp and 6 μF is 36 μC. Voltage across parallel combination Vp = Q / Cp = 36 μC / 6 μF = 6 V. Since capacitors in parallel have same voltage, voltage across 2 μF capacitor is 6 V.
52. A capacitor is connected to a cell of emf E and some internal resistance. The p.d. across the
cell is E
cell is
capacitor is
capacitor is> E
(b) Due to internal resistance, there is some voltage drop across the internal resistance. So the terminal voltage across the cell will be less than the emf E. The capacitor is connected across the cell, so the p.d. across the capacitor will be equal to the terminal voltage of the cell.
53. Capacitance of a capacitor becomes 4/3 times its original value if a dielectric slab of thickness t = d/2 is inserted between the plates (d = separation between the plates). The dielectric constant of the slab is
54. A capacitor is filled with an insulator and a certain potential difference is applied to its plates. The energy stored in the capacitor is U. Now, the capacitor is disconnected from the source and the insulator is pulled out of the capacitor. The work performed against the forces of electric field in pulling out the insulator is 4U. Then dielctric constant of the insulator is
8
5
3
4
(b) Energy stored with dielectric U = 1/2 K C₀ V². After disconnection, charge Q = K C₀ V remains constant. Energy stored without dielectric U' = 1/2 Q² / C₀ = 1/2 (K C₀ V)² / C₀ = K² * 1/2 C₀ V² = K²/K * U = KU. Work done W = U' - U = KU - U = U(K - 1) = 4U. K - 1 = 4 => K = 5.
55. A 10 μF capacitor and a 20 μF capacitor are connected in series across 200V supply line. The charged capacitors are then disconnected from the line and reconnected with the positive plate together and negative plates together and no external voltage is applied. What is the potential difference across each capacitor?
800/9 Volt
800/3 volt
400 volt
200 volt
(c) In series, Q = CeqV = (10 * 20 / 30) * 200 = 4000/3 μC. V₁ = Q/C₁ = (4000/3) / 10 = 400/3 V. V₂ = Q/C₂ = (4000/3) / 20 = 200/3 V. When reconnected in parallel with same polarity, V = (Q₁ + Q₂) / (C₁ + C₂) = (4000/3 + 4000/3) / 30 = (8000/3) / 30 = 800/9 V. There is a mistake in option. When reconnected with positive plates together, charges add up. Qtotal = 4000/3 + 4000/3 = 8000/3 μC. Ctotal = 10 + 20 = 30 μF. V = Q/C = (8000/3) / 30 = 800/9 V. Option a.
56. A 10 μF capacitor is charged to a potential difference of 50V and is connected to another uncharged capacitor in parallel. Now the common potential becomes 20 volt. The capacitance of second capacitor is
15 μF
30 μF
20 μP
1O μF
(a) Q₁ = C₁V₁ = 10 * 50 = 500 μC. Let capacitance of second capacitor be C₂. Q₂ = C₂ * 0 = 0. After connection, total charge Q = Q₁ + Q₂ = 500 μC. Total capacitance C = C₁ + C₂ = 10 + C₂. Common potential V = Q / C => 20 = 500 / (10 + C₂) => 200 + 20C₂ = 500 => 20C₂ = 300 => C₂ = 15 μF.
57. 68. A capacitor is charged to store an energy U. The charging battery is disconnected. An identical capacitor is now connected to the first capacitor in parallel. The energy in each of the capacitor is
3U/2
U
U/4
U/2
(c) Initial energy U = 1/2 CV². After disconnection, charge Q = CV. When connected to identical uncharged capacitor, charge distributes equally to C/2 on each. Energy in each capacitor = 1/2 (C/2) (V/2)² = 1/2 * C/2 * V²/4 = 1/8 CV² = U/4.
58. A parallel plate capacitor is filled by copper plate of thickness b. The new capacity will be
(ϵ_o A)/(2d−b)
(ϵ_o A)/(d−b)
(ϵ_o A)/(d−b/2)
(ϵ_o A)/d
(b) Introducing a copper plate (conductor) of thickness b is equivalent to reducing the separation between the plates by b. New separation = d - b. New capacitance C' = ε₀A / (d - b).
59. A capacitor of capacity C1 is charged by connecting it across a battery of e.m.f. Vo. The battery is then removed and the capacitor is connected in parallel with an unchanged capacitor of capacity C2. The potential difference across this combination is
C1Vo/(C1+C2)
(b) Charge on first capacitor Q₁ = C₁V₀. Charge on second capacitor Q₂ = 0 (uncharged). Total charge Q = Q₁ + Q₂ = C₁V₀. Total capacitance in parallel C = C₁ + C₂. Potential difference across combination V = Q / C = C₁V₀ / (C₁ + C₂).
60. Two capacitors C and 2C are connected in parallel and charged with V volt each. Battery is disconnected and . then a lielectric of constant K is inserted in C. Find the final p.d. of each capacitor .
[BPKIHS]
(b) Initially, charge on C is Q₁ = CV, charge on 2C is Q₂ = 2CV. Total charge Q = 3CV. After disconnecting battery and inserting dielectric in C, new capacitance C' = KC. Equivalent capacitance Ceq = C' + 2C = KC + 2C = C(K+2). Final potential V' = Q / Ceq = 3CV / (C(K+2)) = 3V / (K+2). Option b has 3V/K+2.
61. Two insulated spheres of 3 μF and 5μF are charged to 300V and 500V respectively The energy loss when they are connected by a wire is
0.025 J
2.5 J
0.0375 J
3.75 J
(a) Initial energy Uᵢ = 1/2 C₁V₁² + 1/2 C₂V₂² = 1/2 * 3 × 10⁻⁶ * (300)² + 1/2 * 5 × 10⁻⁶ * (500)² = 0.135 J + 0.625 J = 0.76 J. Common potential V = (C₁V₁ + C₂V₂) / (C₁ + C₂) = (3 * 300 + 5 * 500) / (3 + 5) = (900 + 2500) / 8 = 3400 / 8 = 425 V. Final energy Uf = 1/2 (C₁ + C₂) V² = 1/2 * 8 × 10⁻⁶ * (425)² = 4 × 10⁻⁶ * 180625 = 0.7225 J. Energy loss = Uᵢ - Uf = 0.76 - 0.7225 = 0.0375 J.
62. If a dielectric of K = 5 is put between the plates of a charged capacitor, the charge on capacitor will become (initial charge vas Q)
Q
5Q
Q/5
25 Q
(a) If the capacitor is charged and then isolated (not connected to a battery), the charge on the plates remains constant when a dielectric is inserted.
63. A 1μF capacitor and a 2 μF capacitor are connected in parallel across a 1200 volts line The capacitors e then disconnected from the line and from each other. These . two capacitors are now connected to each other in parallel with terminals of unlike signs together. The charges on the capacitors will now be
1800 μC each
400 μC and 800μC
800μC each and 400μC
800μC each
(b) Q₁ = C₁V = 1 × 1200 = 1200 μC. Q₂ = C₂V = 2 × 1200 = 2400 μC. When connected with unlike signs, net charge Q = |Q₁ - Q₂| = |1200 - 2400| = 1200 μC. Equivalent capacitance C = C₁ + C₂ = 1 + 2 = 3 μF. Common potential V = Q / C = 1200 / 3 = 400 V. Charge on 1 μF capacitor Q₁' = C₁V = 1 * 400 = 400 μC. Charge on 2 μF capacitor Q₂' = C₂V = 2 * 400 = 800 μC.
64. A condenser of capacity C1 is charged to a potential V_0. The electrostatic potential energy stored in it is U_o. It is connected to another uncharged condenser of capacity C2 in parallel. The energy dissipated in the process is
65. A spherical condenser has inner and outer spheres of radii a and b respectively. The space between the two is filled with air. The difference between the capacities of two condensers formed when outer sphere is earthed and when inner sphere is earthed will be
zero
(a) Capacitance when outer sphere earthed C₁ = 4πε₀ab / (b - a). Capacitance when inner sphere earthed C₂ = 4πε₀b² / (b - a). Difference = C₂ - C₁ = 4πε₀b² / (b - a) - 4πε₀ab / (b - a) = 4πε₀b(b - a) / (b - a) = 4πε₀b. If the question implies capacity of the system. When outer earthed, charge on inner sphere determines potential. When inner earthed, charge on outer determines potential. Difference is not zero.
66. A parallel plate capacitor has capacitance of 50 μF in air and 110 μF when immersed in oil. The dielectric constant of oil is
