= Solution
Use an unloaded <RC low-pass filter>: a series <resistor> between the source and the output node, with a <capacitor> from that node to ground, and measure the output across the <capacitor>. The components are ideal in this first-order model; “ideal” here does not mean a brick-wall <frequency> response.
The <time constant> is $\tau=RC$, hence
$$
\boxed{C=\frac{10^{-6}}{10^4}=10^{-10}\,\mathrm F=100\,\mathrm{pF}.}
$$
The capacitive <electrical impedance> gives the <transfer function> $H(i\omega)=1/(1+i\omega RC)$, so
$$
|H|=\frac1{\sqrt{1+(f/f_c)^2}},\qquad \boxed{f_c=\frac1{2\pi RC}=1.59\times10^5\,\mathrm{Hz}.}
$$
On the magnitude <Bode plot>, $20\log_{10}|H|=-10\log_{10}[1+(f/f_c)^2]$. The gain approaches $0\,\mathrm{dB}$ for $f\ll f_c$, is $-3.01\,\mathrm{dB}$ at $f_c$, and falls with slope $-20\,\mathrm{dB}$ per decade for $f\gg f_c$. These results assume negligible source resistance beyond the specified $R$ and a sufficiently large load impedance.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2018/paper-1-rc-low-pass.png]
{title=Unloaded RC low-pass circuit and its first-order Bode magnitude}
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