What is the position at 2 seconds?
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Answer:
Check attachment for better understanding
Explanation:
Given that,
Current in wire I =2.2A
Capacitor plate dimension is 2cm by 2cm
s=2cm=2/100 = 0.02m
Rate at which electric field Is changing dE/dt?
The current in the wires must also be the displacement current in the capacitor. We find the rate at which the electric field is changing from
ID = ε0•A•dE/dt
Where ε0 is a constant
ε0= 8.85×10^-12C²/Nm²
Area of the square plate is
A =s² =0.02² = 0.0004m²
Then,
Make dE/dt the subject of formula
dE/dt = ID/ε0A
dE/dt = 2.2 / (8.85×10^-12 ×4×10^-4)
dE/dt = 6.215×10^14 V/m-s
Or
dE/dt = 6.215×10^14 N/C.s
The rate at which the electric field is changing between the plates is 6.215×10^14 N/C.s
