The energy stored in a capacitor of capacitance \( C \) charged to voltage \( V \) is
Answer: A
The factor of one half arises because the voltage rises from zero to its final value as the charge is delivered, so the mean voltage during charging is half the final one. The same energy may be written as \( Q^2/2C \) or \( \tfrac{1}{2}QV \).
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Charging a capacitor from a constant voltage source dissipates in the circuit resistance
Answer: B
The source delivers \( QV \) while the capacitor retains \( \tfrac{1}{2}QV \), so exactly half is lost whatever the resistance. Reducing the resistance makes the loss happen faster but does not reduce it, which is a result worth knowing.
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The dielectric strength of a material is
Answer: C
It is quoted in kilovolts per millimetre or megavolts per metre, and it sets the minimum thickness of insulation for a given working voltage. Air manages about three megavolts per metre and good solid insulants ten times that or more.
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In a series arrangement of two dielectrics between capacitor plates, the larger field appears in the layer of
Answer: D
The normal flux density is common to both layers, so the field intensity is inversely proportional to the permittivity. A thin air gap inside solid insulation therefore carries a much higher field than the solid, which is why voids are the usual starting point of failure.
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A parallel plate capacitor of area 0.01 m² and gap 1 mm in air has a capacitance of about
Answer: A
Correct answer: 88.5 pF
\[ C = \frac{\epsilon_0 A}{d} = \frac{8.854\times10^{-12} \times 0.01}{0.001} = 8.85\times10^{-11}\ \mathrm{F} = 88.5\ \mathrm{pF}. \] The figure is worth remembering as a scale: a hundred square centimetres a millimetre apart gives well under a nanofarad, which is why practical capacitors use very thin dielectrics and very large rolled or stacked areas.
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Filling that same capacitor with a dielectric of relative permittivity 4 raises its capacitance to
Answer: B
The capacitance is multiplied by the relative permittivity, so \( 4 \times 88.5 = 354 \) pF. The stored energy at a given voltage rises by the same factor, which is why high permittivity ceramics are used where compactness matters.
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Two capacitors of 3 \( \mu \)F and 6 \( \mu \)F in series give a combined capacitance of
Answer: C
Correct answer: 2 \( \mu \)F
In series the reciprocals add: \[ \frac{1}{C} = \frac{1}{3}+\frac{1}{6} = \frac{1}{2}, \qquad C = 2\ \mu\mathrm{F}. \] The result is smaller than either capacitor, which is the check to apply - series capacitance behaves like parallel resistance, and adding them directly gives the 9 offered.
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Those same two capacitors in series carry 12 V across the pair. The voltage across the 3 \( \mu \)F capacitor is
Answer: D
The same charge sits on both, so the voltages divide in inverse proportion to the capacitances and the smaller capacitor takes the larger share, two thirds of twelve volts. This is the opposite of how a series resistive divider behaves with respect to resistance.
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A coaxial cable with inner radius 1 mm and outer radius 3 mm, filled with a dielectric of relative permittivity 2.5, has a capacitance per metre of about
Answer: A
Correct answer: 127 pF
For a coaxial line, \[ C = \frac{2\pi\epsilon_0\epsilon_r}{\ln(b/a)} = \frac{2\pi \times 8.854\times10^{-12} \times 2.5}{\ln 3} = \frac{1.391\times10^{-10}}{1.0986} = 127\ \mathrm{pF/m}. \] The logarithm makes the capacitance insensitive to the exact radii - tripling the ratio again would change it by only a third - while the permittivity multiplies it directly.
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A capacitor of 10 \( \mu \)F is charged to 100 V. The energy stored is
Answer: B
Correct answer: 50 mJ
\[ W = \tfrac{1}{2}CV^{2} = \tfrac{1}{2} \times 10\times10^{-6} \times 100^{2} = \tfrac{1}{2} \times 10^{-5} \times 10^{4} = 0.05\ \mathrm{J} = 50\ \mathrm{mJ}. \] Omitting the factor of one half gives the 100 mJ offered. Note that the energy goes as the square of the voltage, so doubling the voltage quadruples the stored energy.
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