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Exercise:
A double capacitor consists of two fixed outer plates and two connected inner plates. The inner plates can be moved from left to right thereby changing the effective overlapping plate area of the two parallel plate capacitors with the same plate distance. When the area of the upper capacitor increases the area of the lower one decreases and vice versa see figure. abcliste abc Find an expression for the series capacitance of the two capacitors as a function of the position x. abc Determine the position x_ for the maximum capacitance. abcliste center includegraphicswidth.textwidth#image_path:advanced-doublcapacitor# center

Solution:
abcliste abc With a maximum plate area A and a distance d between the pairs of plates we can write for the two capacitances: C_ varepsilon_fracx Ad C_ varepsilon_frac-x Ad with xin . For the series capacitance we find sscCs leftfracC_+fracC_right^- leftfracdvarepsilon_ x A+fracdvarepsilon_ -x Aright^- varepsilon_fracAdleftfracx+frac-xright^- varepsilon_fracAdleftfrac-x+xx-xright^- varepsilon_fracAd x-x resultvarepsilon_fracAd x-x^ abc In order to find the maximum of the capacitance we can derive the expression from a with respect to the position x: fracmathrmdsscCsmathrmdx varepsilon_fracAd - x At an extremal po xx_ the derivative is equal to zero. This leads to - x_ x_ resultfrac The result can easily be verified with a graph of fxx-x^ in the range : center includegraphicswidth.textwidth#image_path:graph-advanced-doublcapacitor# center abcliste
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Exercise:
A double capacitor consists of two fixed outer plates and two connected inner plates. The inner plates can be moved from left to right thereby changing the effective overlapping plate area of the two parallel plate capacitors with the same plate distance. When the area of the upper capacitor increases the area of the lower one decreases and vice versa see figure. abcliste abc Find an expression for the series capacitance of the two capacitors as a function of the position x. abc Determine the position x_ for the maximum capacitance. abcliste center includegraphicswidth.textwidth#image_path:advanced-doublcapacitor# center

Solution:
abcliste abc With a maximum plate area A and a distance d between the pairs of plates we can write for the two capacitances: C_ varepsilon_fracx Ad C_ varepsilon_frac-x Ad with xin . For the series capacitance we find sscCs leftfracC_+fracC_right^- leftfracdvarepsilon_ x A+fracdvarepsilon_ -x Aright^- varepsilon_fracAdleftfracx+frac-xright^- varepsilon_fracAdleftfrac-x+xx-xright^- varepsilon_fracAd x-x resultvarepsilon_fracAd x-x^ abc In order to find the maximum of the capacitance we can derive the expression from a with respect to the position x: fracmathrmdsscCsmathrmdx varepsilon_fracAd - x At an extremal po xx_ the derivative is equal to zero. This leads to - x_ x_ resultfrac The result can easily be verified with a graph of fxx-x^ in the range : center includegraphicswidth.textwidth#image_path:graph-advanced-doublcapacitor# center abcliste
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Attributes & Decorations
Branches
Electrostatics
Tags
capacitance, capacitor, electrostatic, parallel plate capacitor, series circuit
Content image
Difficulty
(4, default)
Points
0 (default)
Language
ENG (English)
Type
Calculative / Quantity
Creator by
Decoration