5 because we have nothing new to say about the physics governing quantum fluctuations, the above argument does not, by itself, solve the cosmological constant problem. However, the interpretation of cosmic acceleration as due to entropic force helps to understand why the accelerating component is expected to be small today (of order H as in Eq. (7)), in contrast to the embarasingly large value predicted by quantum field theory combined with general relativity. We shall next amplify on the distinction, and study more the entropy and surface screen considerations, showing that even the present fraction of the critical energy associated with acceleration can thereby be understood. The next portion derives an expression for the pressure, which is negative and thus a tension in the direction of the screen. The following results rely on simple principles of entropy and thermodynamics and are not dependent on any specific model. The entropy on the Hubble Horizon, e.g. the Hubble radius R$_{H}$ = c/H , is Increasing the radius R$_{H}$ , by Δ r , increases the entropy by Δ S$_{H}$ according to The entropic force is simply where the minus sign indicates pointing in the direction of increasing entropy or the screen, which in this case is the horizon. The pressure from entropic force exerted is where ρ$_{critical}$ is the critical energy density ρ$_{critical}$ ≡ 3 H $^{2}$/ 8 πG . This is close to the value of the currently measured dark energy/cosmological constant S$_{H}$ = k$_{B}$c 3 G$_{ℏ}$ A 4 = k$_{B}$c 3 G$_{ℏ}$ πR 2 H = k$_{B}$c 3 G$_{ℏ}$ π ( c H$^{)}$$^{2}$ ∼ (2 . 6 ± 0 . 3) × 10 $^{122}$k$_{B}$ (8) Δ S$_{H}$ = k$_{B}$c 3 G$_{ℏ}$ 2 πR$_{H}$ Δ r = k$_{B}$c 3 G$_{ℏ}$ 2 π ( c H$^{)}$ Δ r ∼ (2 . 6 ± 0 . 3) × 10 $^{122}$k$_{B}$ Δ r/R$_{H}$ (9) F$_{r}$ = − dE dr = − T dS dr = − T$_{β}$ dS$_{H}$ dr = − ℏ k$_{B}$ H 2 π k$_{B}$c 3 G$_{ℏ}$ 2 π ( c H$^{)}$ = − c 4 G (10) P = F$_{r}$ A = − 1 A T dS dr = − 1 A c 4 G = − 1 4 πc $^{2}$/H2 c 4 G = − c $^{2}$H2 4 πG = − 2 3 ρ$_{critical}$c 2 (11)