A parallel-plate air capacitor has rectangular plates each of size `40 cm xx 10cm`, separated by 1mm, if a charge of `10^(-10)` C is given to the capacitor calcualte the potential difference and electric field between the plates.

Explanation

Data: `Q=10^(-10) C, l=40 cm, b=10cm, d=1 mm = 10^(-3), epsilon_(0) = 8.85 xx 10^(-12) C^(2)//N.m^(2),k=1` (air)
Area of the plates, `A=lb = 40 xx 10 = 400 cm^(2) = 0.04 m^(2)`
(1) The capacitance of the capacitor,
`C=(Akepsilon_(0))/d = (0.04 xx 1 xx 8.85 xx 10^(-12))/(10^(-3))`
`=3.54 xx 10^(-10)F`
The potential difference between the plates,
`V=Q/C = 10^(-10)/(3.54 xx 10^(-10)) = 1/(3.54) = 0.2825` V
(ii) The electric field between the plates,
`E=V/d = 0.2825/10^(-3) = 282.5` N/C