Finding Voltage in a Circuit with a Capacitor
I think your are stuck on our notation with the exponential.  This should help get you unstuck.
$$v(t)=10e^{j0.2t}=10cos(0.2t)$$
Since this is a steady-state, single frequency problem - you can replace that time based source with its phasor representation,
$$V=10\angle0°=10+j0 $$
BACKGROUND
If you calculate \$10e^{j\omega t}\$ as time t progresses you will see that you get complex values.  If you convert these complex values into phasor  form you will see that you have a phasor of amplitude 10 that is rotating counterclockwise on the complex plane (x-axis is real, y-axis is imaginary). Here are a few terms if we let \$\omega=0.2 \$ radians (which is \$\frac{0.2}{2\pi}=0.032\$ Hz, so the rotating phasor will make 1 complete counterclockwise revolution in \$\frac{1}{0.032}=31.4\$ seconds.  In one quarter of that time it will cover 90° etc.
$$\begin{array}{c|c|c|} 
 \text{t (sec)} &amp; \text{real} &amp; \text{imaginary} &amp; \text{angle} \\ \hline
\text{0} &amp; 10 &amp; 0 &amp; 0° \\ \hline
\text{3.93} &amp; 7.07 &amp; 7.07 &amp; 45° \\ \hline
\text{7.85} &amp; 0 &amp; 10 &amp; 90° \\ \hline
\text{11.78} &amp; -7.07 &amp; 7.07 &amp; 135° \\ \hline
\text{15.71} &amp; -10 &amp; 0 &amp; 180°  \\ \hline
\text{19.64} &amp; -7.07 &amp; -7.07 &amp; 225°  \\ \hline
\text{23.56} &amp; 0 &amp; -10 &amp; 270°  \\ \hline
\text{27.49} &amp; 7.07 &amp; -7.07 &amp; 315°  \\ \hline
\text{31.42} &amp; 10 &amp; 0 &amp; 0°=360°  \\ \hline
\end{array}$$
When plotted as time progresses it would look like this (from 2):