A parallel-plate of capacitor of capacitance `5mu F` is connected in a battery of emf `6V.` The separation between the plate is 2mm.(a) Find the charge on the positive plate.(b) find the eletric field between the plate.(c) A dielectric slab of thickness 1 mm and dielectric constant 5 is inserted into the gap to occupy the lower half of it.Find the capacitance of the new combination .(d)How much charge has flow n through the battery after the slab is inserted?

Explanation

`C = 5 mu F , v = 6V`
`d = 2 mm = 2 xx 10^(-3) m `
(a) The charge on the + ve plate q = cv
` = 5 mu F xx 6V = 30 mu C `
(b) ` E = v / d = 3 xx 10^(-3) m, `
` K = 5`
` t= 1 xx 10^(-3) m , `
`C = 5 xx 10^(-6) `
since ` C = eplison_0 A / d `
`rArr 10 xx 10^(-6) = 8.85 xx 10^(-12) xx 10^(-3) A `
`rArr 10^(4) = 8.85 xx A `
` A = 10^(4) xx 1 / 8.85`
When the dieletric placed on it .
` C_1 = 0.00000833 `
` = 8.33 mu F `
(d) C = 5 xx 10^(-6) F , v = 6 V`
So, ` Q = cv = 3 xx 10^(-5) F`
` = 30 mu F `
`C_1 = 8.3 xx 10^(-6) F`
`Q' = cv = 8.3 xx 6 xx 10^(-6) `
`= 50 mu F `
Charge flown , Q' - Q = 20 mu F `.