If I understand correctly that is not correct; the N-E leakage will add to the total leakage.Hence, for, say, 50m of cable that would be 5,000 pF (5 nF) - so, about 640 kΩ at 50 Hz, resulting in an L-E 'leakage' current of about 0.36mA at 230V. Since the N-E potential difference will usually be fairly negligible, the N-E leakage current (which will reduce the L-N imbalance) is also likely to be fairly negligible but, in any event, the maximum L-N imbalance would be about 0.36 mA for 50m of T+E.