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The capacitance of a parallel-plate capacitor is C = εA/d. Therefore, for the same dielectric and plate area, reducing the dielectric thickness d increases capacitance. Series capacitors give a lower equivalent capacitance and the combination can have a higher voltage-withstand capability. A capacitor stores energy in the electric field between its conductors. The colour-code statement is retained as the incorrect statement because the source's colour sequence is corrupted/ambiguous and cannot reliably establish 500 pF.
C = εA/d. For capacitors in series: 1/Ceq = 1/C₁ + 1/C₂ + ⋯.
📌 Remember Points
📌 C ∝ 1/d for fixed ε and A.
📌 Series capacitors → lower equivalent capacitance.
📌 Stored energy: W = ½CV².
📌 Always verify a capacitor colour code before calculating its value.
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Tolerance specifies the maximum expected deviation of the actual capacitance from its marked nominal value. Traditional paper capacitors commonly used a tolerance of about ±10%. For example, a 100 nF capacitor with ±10% tolerance can have an actual value from approximately 90 nF to 110 nF.
Tolerance (%) = |Cactual − Cnominal|/Cnominal × 100.
📌 Remember Points
📌 Tolerance describes allowable variation.
📌 100 nF ±10% → approximately 90–110 nF.
📌 Mica capacitors are traditionally associated with closer tolerances.
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Mica capacitors are traditionally preferred where close tolerance, stability and low dielectric loss are important. Mica has high insulation resistance and good temperature characteristics, making it suitable for precision and RF applications.
Tolerance is the permitted deviation from nominal capacitance. Smaller tolerance percentage → more accurate capacitance.
📌 Remember Points
📌 Mica → high stability.
📌 Mica → low dielectric loss.
📌 Mica → high insulation resistance.
📌 Useful in precision and RF circuits.
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Use Coulomb's law. The force between two point charges in free space is F = kQ₁Q₂/r², where k ≈ 9 × 10⁹ N·m²/C². With Q₁ = Q₂ = 1 C and r = 0.5 m, F = 9 × 10⁹/(0.5)² = 9 × 10⁹/0.25 = 36 × 10⁹ N.
F = kQ₁Q₂/r², where k ≈ 9 × 10⁹ N·m²/C².
📌 Remember Points
📌 F ∝ Q₁Q₂.
📌 F ∝ 1/r².
📌 Doubling distance makes force one-fourth.
📌 Like charges repel; unlike charges attract.
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Volume charge density means electric charge per unit volume. Since volume has the unit m³, the unit of volume charge density is coulomb per cubic metre. It is represented by ρᵥ. The source options were corrupted during extraction, so the dimensional forms have been restored.
ρᵥ = Q/V. Therefore [ρᵥ] = C/m³.
📌 Remember Points
📌 ρᵥ = Q/volume.
📌 Surface charge density ρₛ → C/m².
📌 Linear charge density ρₗ → C/m.
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This is Gauss's law expressed using electric displacement D. It states that the total electric flux through a closed surface equals the charge enclosed by that surface. In integral form, ∮ D·dA = Qenc. In terms of electric field intensity, ∮ E·dA = Qenc/ε.
∮D⃗·dA⃗ = Qenc. Also, D⃗ = εE⃗.
📌 Remember Points
📌 Gauss's law uses a closed surface.
📌 D → electric displacement/flux density.
📌 ∮D·dA gives enclosed charge.
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Dielectric strength is the maximum electric field that an insulating material can withstand before electrical breakdown. Mica has a relatively high dielectric strength and is therefore useful in high-voltage and high-frequency capacitor applications. The source gives approximately 50–200 kV/mm.
Dielectric strength Ebd = breakdown voltage/thickness, approximately Vbd/d for a uniform field.
📌 Remember Points
📌 Unit → V/m or kV/mm.
📌 Higher dielectric strength → better voltage withstand capability.
📌 Actual value depends on mica type, thickness and condition.
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Relative permittivity is the ratio of a material's permittivity to the permittivity of free space. Glass compositions vary considerably, so their dielectric constants are not represented by one universal number. For this traditional exam question, the source identifies 5–100 as the applicable broad range.
εᵣ = ε/ε₀. Relative permittivity is dimensionless.
📌 Remember Points
📌 εᵣ has no unit.
📌 Vacuum/free space → εᵣ = 1.
📌 Glass has εᵣ greater than 1.
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Variable air capacitors have traditionally been used in radio-frequency tuning circuits because their capacitance can be smoothly varied. Changing capacitance changes the resonant frequency of an LC circuit. This makes a variable air capacitor useful for selecting different radio frequencies.
f₀ = 1/(2π√LC). Increasing C decreases the resonant frequency.
📌 Remember Points
📌 Air capacitors can be variable.
📌 LC resonance depends on L and C.
📌 C ↑ → f₀ ↓.
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Conventional electrolytic capacitors are generally polarized. Therefore they are mainly used in DC applications where the applied polarity remains correct. Common applications include DC power-supply filtering, smoothing, decoupling and energy storage. Sustained reverse voltage can cause excessive leakage, heating and failure.
For a capacitor, i = C(dv/dt). An electrolytic capacitor is selected with its specified polarity and voltage rating.
📌 Remember Points
📌 Mainly used for DC filtering.
📌 Correct polarity is essential.
📌 Reverse polarity can damage the capacitor.
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Ceramic capacitors are widely used around transistor circuits because they are compact, economical, have low parasitic inductance and perform well at high frequencies. They are commonly used for bypassing, decoupling and coupling in electronic circuits.
Capacitor reactance: Xc = 1/(2πfC). At higher frequency, capacitive reactance becomes smaller.
📌 Remember Points
📌 Ceramic → small and inexpensive.
📌 Useful for bypass and decoupling.
📌 Low parasitic inductance helps high-frequency operation.
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Mica capacitors have excellent stability, low dielectric loss and high insulation resistance. These properties made them useful in coupling, tuned and oscillator circuits, especially where stable capacitance is important. The source identifies mica as the intended answer.
For a capacitor in AC operation, Xc = 1/(2πfC). Stable C helps maintain predictable circuit behavior.
📌 Remember Points
📌 Mica → high stability.
📌 Mica → low dielectric loss.
📌 Useful in precision and RF circuits.
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Capacitance measures the ability to store charge for a given voltage. Its reciprocal is called elastance. Since C = Q/V, taking the reciprocal gives 1/C = V/Q. Elastance is therefore the opposition to electric charge storage in the same conceptual sense that resistance opposes current flow.
C = Q/V, therefore elastance S = 1/C = V/Q.
📌 Remember Points
📌 Elastance = 1/C.
📌 Capacitance unit → farad.
📌 Elastance is the reciprocal quantity.
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For a uniform electric field, the potential changes at a constant rate with distance. The relationship is E = −dV/dx. Therefore, if E is constant, the magnitude of the potential gradient dV/dx is also constant. The negative sign shows that electric field points toward decreasing potential.
E⃗ = −∇V; in one dimension E = −dV/dx.
📌 Remember Points
📌 Potential gradient unit → V/m.
📌 Electric field points toward lower potential.
📌 The original source wording has been clarified to specify a uniform field.
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Electric displacement and electric field are related by D = εE. Therefore E = D/ε. For the same D, a smaller permittivity produces a larger electric field. Since the magnitude of the potential gradient equals the electric field magnitude, the potential gradient is greater in the lower-permittivity material.
D = εE, so E = D/ε = |dV/dx| for a one-dimensional field.
📌 Remember Points
📌 D constant → E ∝ 1/ε.
📌 Low permittivity → high electric stress.
📌 High permittivity → lower electric field for the same D.
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A capacitor stores energy primarily in the electric field between its plates. The field exists because opposite charges are established on the two conductors. The energy density in a linear dielectric is u = ½εE², while total capacitor energy is W = ½CV².
u = ½εE² and W = ½CV².
📌 Remember Points
📌 Capacitor energy → ½CV².
📌 Inductor energy → ½LI².
📌 Electric field is responsible for capacitor energy storage.
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Use W = ½CV². Convert 5 μF to farads: 5 × 10⁻⁶ F. The voltage is 10,000 V, so V² = 10⁸. Therefore W = ½ × 5 × 10⁻⁶ × 10⁸ = 250 J.
W = ½CV² = ½(5 × 10⁻⁶)(10,000)² = 250 J.
📌 Remember Points
📌 Always convert μF to F when using SI units.
📌 W ∝ V².
📌 Doubling voltage makes stored energy four times at constant C.
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For a single-core coaxial cable, maximum electric stress occurs at the conductor surface. The maximum stress is gmax = V/[r ln(R/r)]. The conductor radius is 5 mm = 0.005 m and sheath radius is 12.5 mm = 0.0125 m. Thus R/r = 2.5 and ln(2.5) ≈ 0.9163. Hence gmax ≈ 33000/[0.005 × 0.9163] ≈ 7.21 × 10⁶ V/m.
gmax = V/[r ln(R/r)].
📌 Remember Points
📌 Convert diameter to radius.
📌 Electric stress decreases from conductor toward sheath.
📌 Use natural logarithm ln(R/r).
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First find the uniform electric field: E = V/d = 100/0.01 = 10,000 V/m. The electric force magnitude on an electron is F = eE = 1.6 × 10⁻¹⁹ × 10⁴ = 1.6 × 10⁻¹⁵ N. Using F = ma and me = 9.11 × 10⁻³¹ kg, a = 1.6 × 10⁻¹⁵/(9.11 × 10⁻³¹) ≈ 1.76 × 10¹⁵ m/s².
E = V/d; F = qE; a = qE/m.
📌 Remember Points
📌 e = 1.6 × 10⁻¹⁹ C.
📌 me = 9.11 × 10⁻³¹ kg.
📌 Uniform plate field → E = V/d.
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Electric charge describes the net excess or deficiency of electrons. If a body gains electrons, it becomes negatively charged; if it loses electrons, it becomes positively charged. The charge magnitude is related to the number of elementary charges by Q = ne.
Q = ne, where e = 1.6 × 10⁻¹⁹ C.
📌 Remember Points
📌 Charge unit → coulomb.
📌 Current is rate of charge flow: I = dQ/dt.
📌 One electron has charge magnitude 1.6 × 10⁻¹⁹ C.
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Relative permittivity is a ratio: εr = ε/ε0. Both absolute permittivity ε and free-space permittivity ε0 have units of F/m, so their ratio is dimensionless. Therefore relative permittivity has no unit. The source answer shown elsewhere as Wb/m is technically incorrect.
εr = ε/ε0.
📌 Remember Points
📌 ε0 ≈ 8.854 × 10⁻¹² F/m.
📌 εr has no unit.
📌 Vacuum/free space → εr = 1.
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When a charged body is brought near an uncharged conductor, the free charges inside the conductor redistribute. For a positive rod, electrons move toward the rod, making the near side negative and the far side positive. This redistribution is called electrostatic induction. With suitable grounding, the conductor can acquire a net charge without direct contact.
Electrostatic induction is a charge-redistribution phenomenon caused by an external electric field.
📌 Remember Points
📌 No direct contact is required.
📌 Charges redistribute inside the conductor.
📌 Grounding can be used to leave a net induced charge.
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In classical electrostatic terminology, a Faraday tube represents a tube of electric flux containing one unit of electric flux. The terminology is associated with Faraday's lines or tubes of electric flux. The source question was incomplete because of extraction, so the intended wording has been restored.
Electric flux ΦE = ∫E⃗·dA⃗.
📌 Remember Points
📌 Electric flux is associated with an electric field.
📌 Do not confuse electric flux with magnetic flux.
📌 Flux is obtained by surface integration of E.
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Electric flux represents the total electric field flux passing through a surface. Mathematically, ΦE = ∫E·dA. For a uniform electric field perpendicular to a flat surface, ΦE = EA. Its unit is N·m²/C, equivalently V·m.
ΦE = ∫E⃗·dA⃗; for uniform perpendicular field, ΦE = EA.
📌 Remember Points
📌 ΦE → electric flux.
📌 D → electric flux density.
📌 Uniform normal field → Φ = EA.
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Electric field intensity is force per unit positive test charge. Therefore E = F/Q, giving the SI unit N/C. Electric field can also be expressed as V/m, and these two units are equivalent: 1 N/C = 1 V/m.
E = F/Q. Also E = −dV/dx, so the unit can be V/m.
📌 Remember Points
📌 E is a vector quantity.
📌 N/C = V/m.
📌 Direction is the force direction on a positive test charge.
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Coulomb's law gives the electric field produced by a point charge. In free space, E = (1/4πε0)(Q/r²) = kQ/r². Thus field magnitude is directly proportional to charge and inversely proportional to the square of distance.
E = kQ/r², where k = 1/(4πε0) ≈ 9 × 10⁹ N·m²/C².
📌 Remember Points
📌 E ∝ Q.
📌 E ∝ 1/r².
📌 Doubling distance → E becomes one-fourth.
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Electric flux density D describes electric flux per unit area when the field is normal to the surface. In a linear isotropic dielectric, D = εE. Therefore, the quantity associated with the number of electric flux lines per unit area is electric flux density.
D = dΨ/dA; for a uniform normal field D = Ψ/A, and D = εE.
📌 Remember Points
📌 D unit → C/m².
📌 E → electric field intensity.
📌 D = εE.
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Electric field intensity is related to electric potential by E⃗ = −∇V. In one dimension, E = −dV/dx. Therefore, the magnitude of electric field intensity equals the magnitude of the potential gradient. The negative sign indicates that the field points toward decreasing potential.
E⃗ = −∇V; in one dimension E = −dV/dx.
📌 Remember Points
📌 Potential gradient unit → V/m.
📌 Electric field unit → V/m or N/C.
📌 Negative sign indicates direction.
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Electric displacement D, also called electric flux density in this context, has magnitude and direction and is therefore a vector quantity. In a linear isotropic dielectric, D = εE. Its SI unit is C/m².
D⃗ = εE⃗.
📌 Remember Points
📌 D unit → C/m².
📌 D and E have the same direction in an isotropic dielectric.
📌 D = εE.
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The electric field is the negative spatial gradient of electric potential. In one dimension, E = −dV/dx. The negative sign is essential because the electric field points from higher potential toward lower potential.
E⃗ = −∇V; one-dimensional form: E = −dV/dx.
📌 Remember Points
📌 E → electric field intensity.
📌 V → electric potential.
📌 Negative sign → direction of decreasing potential.
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Dielectric strength is the maximum electric field that an insulating material can withstand before breakdown. Because it is an electric-field quantity, its SI unit is V/m. In high-voltage engineering it is also commonly expressed in kV/mm.
Ebd = Vbd/d for a uniform electric field.
📌 Remember Points
📌 Higher dielectric strength → greater voltage withstand capability.
📌 Breakdown occurs when electric stress exceeds the material limit.
📌 kV/mm is common in insulation specifications.
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This question requires a distinction between breakdown voltage and dielectric strength. In practical insulation systems, the measured dielectric strength can decrease as thickness increases because of defects, voids and statistical breakdown effects. However, the total breakdown voltage generally increases with thickness. The source gives B.
Dielectric strength ≈ Vbd/d. Breakdown voltage and dielectric strength are not identical quantities.
📌 Remember Points
📌 Breakdown voltage generally increases with insulation thickness.
📌 Practical dielectric strength may decrease with thickness.
📌 Material quality and defects strongly influence breakdown.
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Capacitance is the ability of a capacitor to store electric charge for a given potential difference. It is defined as the ratio of stored charge to voltage. A larger capacitance stores more charge at the same voltage.
C = Q/V.
📌 Remember Points
📌 C = Q/V.
📌 Q = CV.
📌 Higher C → more charge for the same V.
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A capacitor resists sudden changes in its terminal voltage. Its current is i = C(dv/dt). For a finite current, an instantaneous change in voltage would require an extremely large rate of voltage change. Therefore capacitor voltage cannot change instantaneously under ordinary finite-current conditions.
i = C(dv/dt).
📌 Remember Points
📌 Capacitor voltage is continuous in ordinary finite-current circuits.
📌 i = C dv/dt.
📌 Inductor relation: v = L di/dt.
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The RC time constant is τ = RC. Here R = 8000 Ω and C = 100 pF = 100 × 10⁻¹² F. Therefore τ = 8000 × 100 × 10⁻¹² = 8 × 10⁻⁷ s = 0.8 μs.
τ = RC = 8000 × 100 × 10⁻¹² = 0.8 μs.
📌 Remember Points
📌 τ = RC.
📌 100 pF = 100 × 10⁻¹² F.
📌 At t = τ, charging voltage reaches 63.2% of final value.
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The electric stress in a single-core coaxial cable varies inversely with radial distance from the conductor. The stress is therefore greatest at the smallest radius, which is the conductor surface, and decreases toward the sheath.
g(x) = V/[x ln(R/r)].
📌 Remember Points
📌 Maximum stress → conductor surface.
📌 Stress decreases toward the sheath.
📌 The field in cable insulation is non-uniform.
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For capacitor charging, Vc = V(1 − e^(−t/RC)). At one time constant, t = τ = RC. Thus Vc = V(1 − e⁻¹) ≈ 0.632V, or 63.2% of the final value.
Vc(t) = V[1 − e^(−t/RC)]; at t = τ, Vc ≈ 63.2% V.
📌 Remember Points
📌 At t = τ → Vc = 63.2% of final.
📌 Charging current → 36.8% of initial.
📌 Discharging voltage → 36.8% of initial.
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During charging, current is i = I0e^(−t/RC). At t = τ = RC, i = I0e⁻¹ ≈ 0.368I0. Therefore the charging current falls to about 36.8%, usually rounded to 37%, of its initial value.
i(t) = I0e^(−t/RC); at t = τ, i/I0 = e⁻¹ ≈ 0.368.
📌 Remember Points
📌 τ = RC.
📌 Current decays exponentially.
📌 36.8% ≈ 37%.
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For a parallel-plate capacitor, C = εA/d. Thus capacitance depends directly on plate area A, inversely on plate separation d, and directly on dielectric permittivity ε. Since ε = ε0εr, the dielectric material also affects capacitance.
C = εA/d = ε0εrA/d.
📌 Remember Points
📌 A ↑ → C ↑.
📌 d ↑ → C ↓.
📌 εr ↑ → C ↑.
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A basic capacitor consists of two conducting electrodes separated by an insulating dielectric. The conductors may be metal plates, foils or conductive coatings. The dielectric may be air, paper, mica, ceramic, plastic film or an oxide layer in an electrolytic capacitor.
C = εA/d.
📌 Remember Points
📌 Two conductors store equal and opposite charge.
📌 Dielectric provides insulation.
📌 Energy is stored in the electric field.
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Permittivity ε is a property describing the electric response of a material. It is related to relative permittivity by ε = ε0εr. Since ε0 has units of F/m and εr is dimensionless, the unit of absolute permittivity is farad per metre.
ε = ε0εr; [ε] = F/m.
📌 Remember Points
📌 ε0 ≈ 8.854 × 10⁻¹² F/m.
📌 εr is dimensionless.
📌 Higher ε generally increases capacitance for the same geometry.
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Dielectric strength is the maximum electric field an insulating material can withstand before breakdown. In practical insulation systems it is affected by moisture, temperature, thickness, defects, material composition and other operating conditions. Moisture is particularly harmful because it can increase leakage and weaken insulation.
Dielectric strength ≈ breakdown voltage/thickness, expressed in V/m or kV/mm.
📌 Remember Points
📌 Dielectric strength → maximum electric stress before breakdown.
📌 Moisture generally reduces insulation performance.
📌 Temperature affects many insulating materials.
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When the applied electric field exceeds the dielectric strength, electrical breakdown occurs. In solid insulation this can cause puncture, cracking, carbonization and permanent damage. Therefore, exceeding breakdown voltage can physically damage the insulating medium.
E = V/d; breakdown occurs when E exceeds the dielectric strength.
📌 Remember Points
📌 Breakdown → loss of insulation strength.
📌 Solid insulation may puncture or crack.
📌 Repeated overvoltage can permanently damage insulation.
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Air has a comparatively low dielectric strength compared with solid insulating materials such as glass, quartz and paraffin wax. Under ordinary atmospheric conditions, the breakdown field of air is approximately 3 kV/mm, although the actual value depends on pressure, temperature, humidity, electrode shape and gap distance.
Air breakdown field under ordinary conditions ≈ 3 kV/mm.
📌 Remember Points
📌 Dielectric strength depends on physical conditions.
📌 Air is a gaseous insulator.
📌 Electrode geometry affects air breakdown.
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Electric field intensity can be written as E = F/Q and also as E = V/d. Therefore V/m and N/C are equivalent units. Hence 1 V/m = 1 N/C. The source options contained an OCR error, so the intended dimensional relationship is restored here.
1 V/m = 1 N/C.
📌 Remember Points
📌 E = V/d.
📌 E = F/Q.
📌 1 V/m and 1 N/C are equivalent.
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Potential difference is defined as work done per unit charge. Therefore V = W/Q. Since work is measured in joules and charge in coulombs, one volt equals one joule per coulomb.
V = W/Q, so 1 V = 1 J/C.
📌 Remember Points
📌 1 V = 1 J/C.
📌 1 A = 1 C/s.
📌 1 W = 1 J/s.
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For an ideal parallel-plate capacitor, C = εA/d. Increasing the plate area increases capacitance, while decreasing the plate separation also increases capacitance. The applied voltage does not directly determine capacitance for an ideal linear capacitor.
C = εA/d.
📌 Remember Points
📌 A ↑ → C ↑.
📌 d ↓ → C ↑.
📌 Voltage does not normally determine C for an ideal linear capacitor.
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The time constant of a first-order RC circuit is the product of resistance and capacitance. It indicates the speed of charging or discharging. At one time constant, the charging capacitor reaches 63.2% of its final voltage and the charging current falls to 36.8% of its initial value.
τ = RC.
📌 Remember Points
📌 τ has units of seconds.
📌 At t = τ → 63.2% charged.
📌 Charging current at τ → 36.8% of initial.
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The Schering bridge is an AC bridge used for measuring capacitance and dielectric loss. It is particularly useful for testing practical capacitors and insulating materials because it can also determine dissipation factor and insulation characteristics.
Schering bridge is used for capacitance and dielectric-loss measurements.
📌 Remember Points
📌 Wheatstone bridge → resistance.
📌 Maxwell/Hay bridges → inductance.
📌 Schering bridge → capacitance/dielectric loss.
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A healthy capacitor initially draws current from the ohmmeter and therefore may show a low resistance reading. As it charges, current decreases and the indicated resistance rises toward a very high value. If the reading immediately goes practically to zero and remains there, the capacitor has a continuous low-resistance path and is therefore short-circuited.
Healthy capacitor: initial low resistance → high resistance after charging. Shorted capacitor: R ≈ 0 Ω continuously.
📌 Remember Points
📌 Good capacitor → low initially, then very high.
📌 Shorted capacitor → approximately zero.
📌 Leaky capacitor → resistance rises but remains abnormally low.