Answer:
The time rate of change of flux is [tex]1.34 \times 10^{10}[/tex] [tex]\frac{V}{s}[/tex]
Explanation:
Given :
Current [tex]I = 0.106[/tex] A
Area of plate [tex]A = 36 \times 10^{-4}[/tex] [tex]m^{2}[/tex]
Plate separation [tex]d = 4 \times 10^{-3}[/tex] m
(A)
First find the capacitance of capacitor,
[tex]C = \frac{\epsilon _{o} A }{d}[/tex]
Where [tex]\epsilon _{o} = 8.85 \times 10^{-12}[/tex]
[tex]C = \frac{8.85 \times 10^{-12 } \times 36 \times 10^{-4} }{4 \times 10^{-3} }[/tex]
[tex]C = 7.9 \times 10^{-12}[/tex] F
But [tex]C = \frac{Q}{V}[/tex]
Where [tex]Q = It[/tex]
[tex]C = \frac{It}{V}[/tex]
[tex]V = \frac{It}{C}[/tex]
Now differentiate above equation wrt. time,
[tex]\frac{dV}{dt} = \frac{I}{C}[/tex]
[tex]= \frac{0.106}{7.9 \times 10^{-12} }[/tex]
[tex]= 1.34 \times 10^{10}[/tex] [tex]\frac{V}{s}[/tex]
Therefore, the time rate of change of flux is [tex]1.34 \times 10^{10}[/tex] [tex]\frac{V}{s}[/tex]
