= Solution
Let $v_C$ denote the <capacitor> <voltage> with the same polarity as the applied <voltage>. By <Kirchhoff's voltage law>, <Ohm's law> and $I=C\,dv_C/dt$,
$$
RC\frac{dv_C}{dt}+v_C=V,\qquad v_C(0)=0.
$$
Solving this first-order <differential equation> gives the <RC low-pass step response>
$$
\boxed{v_C(t)=V\left(1-e^{-t/\tau}\right),\qquad \tau=RC=0.050\,\mathrm s.}
$$
The <capacitor> <voltage> is continuous at switching because an instantaneous finite jump would require an impulsive <electric current>.
For the <square-wave response of an RC low-pass filter>, take equal positive and negative half-periods. Let $A=10\,\mathrm V$, $T=1/f$, $h=T/2$ and $\alpha=e^{-h/\tau}$. After the transient, symmetry makes the <capacitor> <voltage> alternate between $-v_p$ and $+v_p$. Over the positive half-period,
$$
v_C(t)=A-(A+v_p)e^{-t/\tau},\quad 0\leq t<h.
$$
Requiring $v_C(h)=v_p$ gives
$$
\boxed{v_p=A\frac{1-\alpha}{1+\alpha}=A\tanh\!\left(\frac{T}{4\tau}\right).}
$$
The next half-period is the sign-reversed copy. These exact expressions explain the requested sketches:
* At $1\,\mathrm{kHz}$, $h/\tau=0.01$. The <voltage> barely changes in each half-period, so the output is nearly triangular about zero, with extrema about $\pm0.0500\,\mathrm V$. Its small-signal slope is approximately $\pm A/\tau=\pm200\,\mathrm{V\,s^{-1}}$.
* At $10\,\mathrm{Hz}$, $h=\tau$. Clearly curved exponential rises and falls reach only $\pm4.621\,\mathrm V$; the output never settles to either input level.
* At $0.1\,\mathrm{Hz}$, $h=100\tau$. The output closely follows the input plateaux at $\pm10\,\mathrm V$, joined by continuous exponential transitions with <time constant> $0.050\,\mathrm s$.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2019-resit-rc-square-wave-response.png]
{title=Steady periodic capacitor voltages for three square-wave frequencies}
{description=The blue curves use the exact periodic <RC low-pass filter> solution with $RC=0.050\,\mathrm s$. Dashed grey curves show the applied <square wave> in the lower panels; the first panel uses an expanded output scale. Time is measured in input periods.}
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