True or false: two capacitors in series always have a combined value smaller than either of them.
Answer: A
The reciprocals add, so the reciprocal of the total exceeds each individual reciprocal and the total is smaller than the smallest member. When one is much larger than the other the total tends to the smaller one but never exceeds it.
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Assertion: a thin void in solid insulation is where breakdown begins. Reason: the field in the void is higher than in the surrounding solid.
Answer: B
The flux density is continuous across the void, so the field there is higher by the ratio of the permittivities, and air is weaker than the solid besides. Both effects act the same way, which is why voids are the chief enemy of high voltage insulation.
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Assertion: the field within a dielectric is less than the applied field. Reason: the bound surface charge produces a field opposing the applied one.
Answer: C
The polarisation charge on the faces of the slab acts like a pair of oppositely charged sheets whose field opposes the applied one. The reduction is by the factor of the relative permittivity, and it is exactly why a dielectric raises the capacitance.
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Assertion: half the energy drawn from a source is lost when charging a capacitor through a resistor. Reason: the loss is proportional to the resistance in the circuit.
Answer: D
The loss is exactly half whatever the resistance, because a smaller resistance simply makes a larger current flow for a shorter time. The reason given is therefore false, though the assertion it accompanies is a standard and correct result.
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Match the arrangement with its capacitance.
1. Parallel plates 2. Coaxial cylinders 3. Isolated sphere
P. \( 4\pi\epsilon a \) Q. \( \epsilon A/d \) R. \( 2\pi\epsilon/\ln(b/a) \) per unit length
Answer: A
The three formulae correspond to the three symmetries and to the three ways the field falls off with distance. Recognising which geometry a problem presents settles which expression to reach for, and each may be derived in a line from the appropriate Gaussian surface.
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Match the change with its effect on a capacitor's capacitance.
1. Doubling the plate area 2. Doubling the separation 3. Doubling the relative permittivity
P. Halves it Q. Doubles it R. Doubles it also
Answer: B
Area and permittivity appear in the numerator and separation in the denominator, so the first and third double the capacitance and the second halves it. All three effects are exploited in practical capacitor design, thin dielectrics and large rolled areas most of all.
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The steps of computing a composite dielectric capacitor are listed below. Put them in order.
1. Combine the parts as series or parallel 2. Compute the capacitance of each part alone 3. Decide whether the voltage or the flux is shared 4. Check the field in each layer against its dielectric strength
Answer: C
Deciding whether the arrangement is series or parallel comes first because it determines what a part even is; then each part is computed and the two combined; and the stress check comes last, since it needs the field in each layer, which follows from the completed solution.
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Consider the following statements about dielectrics.
1. The relative permittivity exceeds unity 2. The field inside is reduced 3. The bound charge can be conducted away
Answer: D
The first two are correct and closely related, the reduction of the interior field being precisely what a permittivity above one expresses. Bound charge is not free to move and cannot be drained away, which is what distinguishes it from the free charge on the electrodes.
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Consider the following statements about capacitors.
1. Series capacitors share the same charge 2. Parallel capacitors share the same voltage 3. The stored energy goes as the first power of the voltage
Answer: A
The first two are the defining conditions from which both combination rules follow. The energy goes as the square of the voltage, which is why doubling the working voltage of a capacitor bank quadruples the energy it can deliver.
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Which of the following would NOT increase the capacitance of a parallel plate capacitor?
Answer: B
A larger separation reduces the field for a given voltage and therefore the charge held, so the capacitance falls. The other three all increase either the effective area or the permittivity and so raise the capacitance.
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