Question

A 0.106-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find the time rate of change of electric flux between the plates.

Answer 1

The time rate of change of flux is
`1.34 * 10^(10)`
`(V)/(s)`

Step-by-step explanation:

Given :

Current
`I = 0.106` A

Area of plate
`A = 36 * 10^(-4)`
`m^(2)`

Plate separation
`d = 4 * 10^(-3)` m

(A)

First find the capacitance of capacitor,

`C = (\epsilon _(o) A )/(d)`

Where
`\epsilon _(o) = 8.85 * 10^(-12)`

`C = (8.85 * 10^(-12 ) * 36 * 10^(-4) )/(4 * 10^(-3) )`

`C = 7.9 * 10^(-12)` F

But
`C = (Q)/(V)`

Where
`Q = It`

`C = (It)/(V)`

`V = (It)/(C)`

Now differentiate above equation wrt. time,

`(dV)/(dt) = (I)/(C)`

`= (0.106)/(7.9 * 10^(-12) )`

`= 1.34 * 10^(10)`
`(V)/(s)`

Therefore, the time rate of change of flux is
`1.34 * 10^(10)`
`(V)/(s)`