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 electric 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 flown 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 `.