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 ?

Options

A. 0.25 cm
B. 0.52 cm
C. 0.75 cm
D. 1 cm

Answer

A

Explanation

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`