A parallel plate air capacitor is made using two plates 0.2 m square , spaced 1 cm apart . It is connected to a 50V battery.
(a) what is the capacitance ?
(b) what is the charge one each plane ?
(c) what is energy stored in the capacitor ?
(d) what is the electric field between the plates ?
(e) If the battery is disconnected and then the plates are pulled apart to a separation of 2 cm , what are the answer to the above parts ?

Explanation

(a)`C_(0) = (e_(0)A)/(d_(0)) = (8.85 xx 10^(-12) xx 0.2 0.2)/(0.01)`
`C_(0) = 3.5 xx 10^(-11)muF`
(b) `Q_(0) = C_(0) V_(0) = (3.5 xx 10^(-11) xx 50) mu C`
`=1.77 xx 10^(-3) mu C`
(c) `U_(0) = 1//2C_(0)V_(0)^(2) = 1//2(3.54 xx 10^(-11))(50)^(2)`
` U_(0) = 4.42 xx 10^(-8)J`.
(d) `E_(0) = (V_(0))/(d_(0)) = (50)/(0.01) = 5000 V//m`
(e) If the battery is disconneted, the charge on the capacitor plates remains constant while the potential difference between plates can change .
`C =(A in_(0))/(2d) = 1.77 xx 10^(-5) mu F` ,
` Q = Q_(0) =1.77 xx 10^(-3) mu C`
`V = (Q)/(C) = (Q_(0))/(C_(0)//2) = 2V_(0) = 100` volts .
`U = (1)/(2)(Q^(2))/(C) = (1)/(2) (Q_(0)^(2))/((C_(0)//2)) = 2U_(0) = 8.84 xx 10^(-8)J`
`E = (V)/(d) = 5000V //m`
work has to be done against the attraction of plates when they are separated . This gets stored in the energy of the capacitor .