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

Answer:

The time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

Explanation:

Given :

Current $I = 0.106$ A

Area of plate $A = 36 \times 10^{-4}$ $m^{2}$

Plate separation $d = 4 \times 10^{-3}$ m

(A)

First find the capacitance of capacitor,

   $C = \frac{\epsilon _{o} A }{d}$

Where $\epsilon _{o} = 8.85 \times 10^{-12}$

   $C = \frac{8.85 \times 10^{-12 } \times 36 \times 10^{-4} }{4 \times 10^{-3} }$

   $C = 7.9 \times 10^{-12}$ F

But   $C = \frac{Q}{V}$

Where $Q = It$

  $C = \frac{It}{V}$

  $V = \frac{It}{C}$

Now differentiate above equation wrt. time,

  $\frac{dV}{dt} = \frac{I}{C}$

       $= \frac{0.106}{7.9 \times 10^{-12} }$

       $= 1.34 \times 10^{10}$ $\frac{V}{s}$

Therefore, the time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$