Answer
See the graph below.
Work Step by Step
We know that the energy stored in a capacitor is given by
$$U_C=\dfrac{Q^2}{2C} $$
Plug the known;
$$U_C=\dfrac{Q^2}{2(2\times 10^{-6})}$$
$$U_C=\dfrac{Q^2}{4\times 10^{-6}}\tag 1 $$
We have here two stages:
- The first stage is from $t=0$ s to $t=3$ s, where $Q=\dfrac{200}{3}t$
Hence, from (1),
$$U_{C1}=\dfrac{\left(\dfrac{200\times 10^{-6}}{3}t\right)^2}{4\times 10^{-6}} $$
$$U_{C1}=\dfrac{1}{900}t^2 $$
- The second stage is from $t=3$ s to $t=4$ s, where $Q=200\;\rm \mu C$.
Hence, from (1),
$$U_{C2}=\dfrac{(200\times 10^{-6})^2}{4\times 10^{-6}}=\bf 0.01\;\rm J$$
See the graph below.
