Final answer:
The rate at which the electric field changes between the plates of a capacitor with a separation of 1.2 mm when the voltage across them is changing at a rate of 110V/s is 91666.67 V/m ext{s}.
Step-by-step explanation:
To determine the rate at which the electric field changes in a parallel-plate capacitor, we can use the relationship between the electric field (E), the voltage (V), and the separation (d) of the plates. The electric field in a parallel-plate capacitor is given by the equation E = V/d. If the voltage changes at a rate of dV/dt, the rate of change of the electric field will be directly proportional to this rate, and is given by dE/dt = dV/dt / d.
The given diameter of the plates is 8.0 cm, but for the electric field calculation, we only need the separation between the plates, which is 1.2 mm, or 0.0012 meters. Since the rate of change of the voltage is 110 V/s, we can calculate the rate of change of the electric field (dE/dt) as follows:
dE/dt = 110 V/s / 0.0012 m = 91666.67 V/m ext{s}