Question

At a given instant, a 3.5. A current flows in the wires connected to a parallel-plate capacitor. What is the rate at which the electric field is charging between the plates if the square plates are 1.40 cm on a side? Express your answer using two significant figures.

Answer 1

I= C dV/dt

Capacitance of the capacitor is

C = ε_oA/d = 8.85x10^-12 (0.014²) /(0.014) =1.239x10^-14F (Square plates area is a² = 0.014² and d = 0.014)

dV/dt = I/C = 3.5/1.239x10^-14 =2.83x10¹⁴V/s

the electric field in a parallel plate capacitor is

dE/dt = (dV/dt )/ d = 2.83x10¹⁴ / 0.014 = 2.02x10¹⁶ V/ ms