Two metal spheres of radius R and 3R have same surface charge density σ. If they are brought in contact and then separated, the surface charge density on smaller and bigger sphere becomes σ₁ and σ₂, respectively. The ratio $ \frac{\sigma_1}{\sigma_2} $ is

<p>When two conducting spheres are in contact, they achieve the same electric potential. Consider two metal spheres with radii $ R $ and $ 3R $ that have the same surface charge density $\sigma$.</p> <p>For a conductive sphere, the potential $ V $ of a sphere can be expressed as $ V = \frac{\sigma r}{\varepsilon_0} $, where $ r $ is the radius of the sphere and $\varepsilon_0$ is the permittivity of free space.</p> <p>When the two spheres are brought into contact and then separated, they equalize their potentials: </p> <p>$ V_1 = V_2 $</p> <p>Thus, the equations become:</p> <p>$ \sigma_1 r_1 = \sigma_2 r_2 $</p> <p>Substituting for the radii:</p> <p>$ \sigma_1 R = \sigma_2 (3R) $</p> <p>From this, we can solve for the ratio of the surface charge densities:</p> <p>$ \frac{\sigma_1}{\sigma_2} = \frac{3R}{R} = 3 $</p> <p>Therefore, the ratio of the surface charge densities on the smaller sphere to the larger sphere after contact is 3.</p>