Polarity of input signal and capacitor in voltage clamper
The Diode as a current rectifier will have resistance &lt; 10 Ohms or 0 compared to 10k with a 0.6V drop (@ 1mA).

So the cap charges up for negative swings with the diode conducting thus R1 ends up with a + Voltage as it slowly decays the cap and ends up with a POSITIVE steady-state half-wave ripple voltage.  You know this is true as the Diode blocks negative peak waves.

Since RC = 1ms is the 63% response time and f = 1kHz you end up with R1 sees a half-wave rectified sine wave but the 1st pulse height is only a little over half of the final peak voltage as the cap charges up.

To add more clarity using Falstad.com sim
 
The Cap. is shown in red as diode conducts with the
Polarity chosen so that Vcap + Vdiode = Vsource 
The diode voltage is ~ 500mV due to I=C dV/dt with a power diode.