Question

At a given instant, a 2.2 A current flows in the wires connected to a parallel-plate capacitor. What is the rate at which the electric field is changing between the plates if the square plates are 2.0 cm on a side? Express your answer using two significant figures

Answer 1

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