Consider a parallel-plate capacitor with a K-dielectric that is only half filled. Although it occupies a little portion of the capacitor's surface area, the dielectric is the same height as the distance between its plates. When the dielectric is all gone, the capacitor has a capacitance of&nbsp;C0. How does this capacitor's capacitance C(f) change with f?In terms of&nbsp;C0&nbsp;and K, express C(f).

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Consider a parallel-plate capacitor with a K-dielectric that is only half filled. Although it occupies a little portion of the capacitor&#039;s surface area, the dielectric is the same height as the distance between its plates. When the dielectric is all gone, the capacitor has a capacitance of&amp;nbsp;C0. How does this capacitor&#039;s capacitance C(f) change with f?In terms of&amp;nbsp;C0&amp;nbsp;and K, express C(f).

Clelioo · Accepted Answer

AS THE CAPACITOR IS PARTIALLY FILLED WITH DIELECTRIC IN THEABOVE FIGURE  Then there will be combination of two capacitors which are in series capacitance of 1st capacitor filled with directric C1=k△0fAd capacitance of 2nd capacitor filled with air (remainingportion in the above figure)  C2=△0(1−f)Ad  C0=capacitance of total capacitor when directric is removed=Ad  therefore C1=kfC0  C2=(1−f)C0  effective capacitance =C1C2C1+C2  PLUG THE VALUES OF C1 AND C2 IN  1 WE EFFECTIVE CAPACITANCE IN TERMS OF C0,f,k

FSP02510402311.jpgFSZConsidera parallel-plate capacitor that is partially filled with adielectric of dielectric constant K.

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