To establish an instantaneous current of 2 A through a 1$$\mu$$F capacitor; the potential difference across the capacitor plates should be changed at the rate of:

The relationship between the current through a capacitor and the rate of change of the potential difference across its plates is given by the formula:

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

where:

• $$i$$ is the instantaneous current,
• $$C$$ is the capacitance,
• $$\frac{dV}{dt}$$ is the rate of change of potential difference with respect to time.

Given in the question:

• Instantaneous current, $$i = 2$$ A,
• Capacitance, $$C = 1 \mu F = 1 \times 10^{-6}$$ F (since 1 μF = 10^{-6} F).

Substitute these values into the formula:

$$2 = (1 \times 10^{-6}) \times \frac{dV}{dt}$$

To solve for $$\frac{dV}{dt}$$, rearrange the equation:

$$\frac{dV}{dt} = \frac{2}{1 \times 10^{-6}}$$

Dividing by $$10^{-6}$$ is equivalent to multiplying by $$10^6$$:

$$\frac{dV}{dt} = 2 \times 10^6$$

Therefore, the potential difference across the capacitor plates must be changed at the rate of $$2 \times 10^6$$ V/s.

Comparing with the options:

• A. $$2 \times 10^4$$ V/s
• B. $$4 \times 10^6$$ V/s
• C. $$2 \times 10^6$$ V/s
• D. $$4 \times 10^4$$ V/s

Hence, the correct answer is Option C.