To find the rate at which the potential difference ($V$) must be changed to produce a displacement current ($I_d$) of 2.2 A in a parallel-plate capacitor with a capacitance ($C$) of 2.3 μF, we can use the relationship between displacement current, capacitance, and the rate of change of voltage:

$I_d = C \frac{dV}{dt}$

Where:

• $I_d$ is the displacement current,
• $C$ is the capacitance,
• $\frac{dV}{dt}$ is the rate of change of voltage.

Rearranging this equation to solve for $\frac{dV}{dt}$:

$\frac{dV}{dt} = \frac{I_d}{C}$

Now substituting the values:

• $I_d = 2.2 \, \text{A}$
• $C = 2.3 \, \mu\text{F} = 2.3 \times 10^{-6} \, \text{F}$

Now, calculate $\frac{dV}{dt}$:

$\frac{dV}{dt} = \frac{2.2 \, \text{A}}{2.3 \times 10^{-6} \, \text{F}} \approx 956521.74 \, \text{V/s}$

Thus, the potential difference must be changed at a rate of approximately 956,521.74 V/s to produce a displacement current of 2.2 A.

If you have any questions or need further details, feel free to ask! Here are some related questions you might consider:

1. What factors affect the displacement current in a capacitor?
2. How does changing the area of the plates impact the capacitance?
3. What happens to the displacement current if the capacitance is doubled?
4. How does the potential difference affect the energy stored in the capacitor?
5. What are the practical applications of displacement current in electrical engineering?

Tip: When working with capacitors, always keep track of the units, especially when dealing with microfarads (μF) and converting them to farads (F).