The capacitance of a capacitor can beaffected by dielectric material that,although not inside the capacitor, is nearenough to the capacitor to be polarized bythe fringing electric field that exists near acharged capacitor. This effect is usually ofthe order of picofarads (pF), but it can be used with appropriateelectronic circuitry to detect a change in the dielectric material sur-rounding the capacitor. Such a dielectric material might be thehuman body, and the effect described above might be used in thedesign of a burglar alarm. Consider the simplified circuit shown inFigure. . The voltage source has anf €= 1000 V, and tecapacitor has capacitance C = 10.0 pF. The electronic circuitryfor detecting the current, represented as an ammeter in the dia-gram, has negligible resistance and is capable of detecting a cur-rent that persists at a level of at least 1.00 μA for at least 200 μsafter the capacitance has changed abruptly from C to C’. The bur-glar alarm is designed to be activated if the capacitance changes by10%. (a) Determine the charge on the 10.0-pF capacitor when it isfully charged. (b) If the capacitor is fully charged before theintruder is detected, assuming that the time taken for the capaci-tance to change by 10% is short enough to be ignored, derive anequation that expresses the current through the resistor R as a func-tion of the time f since the capacitance has changed. (c) Determinethe range of values of the resistance R that will meet the designspecifications of the burglar alarm. What happens if R is too small?Too large? (Hint: You will not be able to solve this part analyticallybut must use numerical methods. Express R as a logarithmic func-tion of R plus known quantities. Use a trial value of R and calculatefrom the expression a new value. Continue to do this until the inputand output values of R agree to within three significant figures.)
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