Looks a bit like an integrator, although there is some detail missing from the circuit. If the base is held 10V below the +10V rail, then the emitter will be held at ~9.4V below the 10V rail (provided enough current can flow out of the base from the 10V source). So, there will be a constant ~9.6mA current flowing into the emitter, and so about the same from the collector. Assuming the capacitor is discharged, the voltage will rise linearly from -10V towards 0V.