The charge on a capacitor plate in a circuit, as a function of time, is shown in the figure. What is the value of current at $$t = 4$$ s?

First, recall the fundamental relation between current and charge. The electric current $$I$$ through any conductor (or here, through the leads of the capacitor) is defined as the time‐rate of change of charge $$q$$ flowing past a given point. Mathematically, the definition is

$$I \;=\;\dfrac{dq}{dt}$$

We are given a graph of charge on the capacitor plate as a function of time. The question asks for the value of the current at the particular instant $$t = 4\;\text{s}$$. From the definition above, this current is nothing but the slope of the $$q$$-versus-$$t$$ curve at that instant.

Now, observe the portion of the graph that covers the interval from $$t = 3\;\text{s}$$ to $$t = 5\;\text{s}$$. In this stretch, the plotted charge remains perfectly horizontal, indicating that the value of charge does not change with time. Expressed algebraically, during this interval we have

$$q(t) = \text{constant}$$

Since the function is constant, its derivative with respect to time is zero:

$$\dfrac{dq}{dt} = 0$$

Substituting this derivative back into the current definition, we obtain

$$I = \dfrac{dq}{dt} = 0$$

Hence, at $$t = 4\;\text{s}$$—which lies squarely in the mentioned flat region—the current through the circuit is zero.

Hence, the correct answer is Option C.