Q.A metallic sphere A isolated from ground is charged to +50 $\mu C$. This sphere is brought in contact with other isolated metallic sphere B of half the radius of sphere A. The charge on the two spheres will be now in the ratio (A) 1:2 (B) 2:1 (C) 4:1 (D) 1:1

[!TLDR] Contact between isolated conductors of the same material equalises their potential, so charge splits in the ratio of the …

When two isolated conducting spheres are brought into contact, the combined system settles at electrostatic equilibrium at a single common potential $V$ shared by both spheres (since they are now, in effect, one connected conductor). For an isolated charged sphere, $V=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R}$, so equality of potential requires $\dfrac{Q_A}{R_A}=\dfrac{Q_B}{R_B}$. Sphere B has HALF the radius of sphere A, i.e. $R_B=R_A/2$, so $\dfrac{Q_A}{R_A}=\dfrac{Q_B}{R_A/2}=\dfrac{2Q_B}{R_A}$, giving $Q_A=2Q_B$, i.e. $Q_A:Q_B=2:1$. Combined with total-charge conservation, $Q_A+Q_B=50\,\mu C$, this gi …

Equate the final potentials of the two spheres after contact ($V=Q/(4\pi\epsilon_0R)$ for each), which shows charge spli …

Assuming the charge simply splits equally (1:1) between the two spheres regardless of their sizes, ignoring that a large …