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Two parallel plate air capacitors have their plate areas `100 and 500 cm^(2)` respectively. If they have the same charge and potential and the distance between the plates of the first capacitor of 0.5 mm, what is the distance between the plates of second capacitor ?A. 0.25 cmB. 0.52 cmC. 0.75 cmD. 1 cm |
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Answer» Correct Answer - A Let `A_(1) and d_(1)` be the area of plates and distance between the plates of first capacitor, `A_(2) and d_(2)` be corresponding values in case of second capacitor. If `C_(1) and C_(2)` are the capacitances of two capacitors, then `C_(1)=(epsilon_(0)A_(1))/(d_(1)) and C_(2)=(epsilon_(0)A_(2))/(d_(2))` We know,`" "C=(q)/(V)` Since, the two capacitors have same charge and potential, their capacitance must be equal i.e. `C_(1)=C_(2)` `"or "(epsilon_(0)A_(1))/(d_(1))=(epsilon_(0)A_(2))/(d_(2)) or d_(2)=(A_(2))/(A_(1))d_(1)` Here, `A_(1)=100cm^(2), a+(2)=500cm^(2) and d_(1)=0.5mm = 0.05 cm` `therefore" "d_(2)=(500xx0.05)/(100)=0.25cm` |
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