A homogeneously charged pyramid

The pyramid is charged with constant volume density. The word CONSTANT means 1 V is potential of whole pyramid. Hence, all that we have to calculate is the volume of remaining part of the pyramid. By assuming different dimensions I found that the part to be cutted has volume ratio 4 : 25 with the normal one. Hence, according to me , by this angle of logic the answer is \frac{21}{25} = 0.84

First let us determine what is the relation between the potential at point P and the height H of the pyramid. From dimensional analysis we have that $V_{p} \sim \frac{Q}{H} \sim \frac{\rho V}{H} \sim \frac{ H^{3}}{H}=H^{2}.$ Thus, we can write $V_{p}(H)= C_{pyr} H^{2}$ where $C_{pyr}$ is a constant that does not depend on H. This observation is sufficient to determine the potential $V'_{p}$ at point P when the pyramid is truncated. In fact, note that $V_{p}(H)= V_{p}(H-h) +V'_{p} \Rightarrow V'_{p}= C_{pyr}(H^{2}- (H-h)^{2})= (1-(1-h/H)^2) \textrm{Volts}.$ Here, we used of the fact that the potential when $h=0$ is $1V$ . Clearly, the same solution may be obtained by integration. You'll probably agree that our method is less tedious.