Question 299: Electrostats & Capacitance

Question

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

Accepted Answer

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 $\sigma$.
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.
When the two spheres are brought into contact and then separated, they equalize their potentials:
$ V_1 = V_2 $
Thus, the equations become:
$ \sigma_1 r_1 = \sigma_2 r_2 $
Substituting for the radii:
$ \sigma_1 R = \sigma_2 (3R) $
From this, we can solve for the ratio of the surface charge densities:
$ \frac{\sigma_1}{\sigma_2} = \frac{3R}{R} = 3 $
Therefore, the ratio of the surface charge densities on the smaller sphere to the larger sphere after contact is 3.

Answer

$ \frac{1}{3} $

Answer

$ \frac{1}{9} $

Answer

9

Answer

3

Electrostats & Capacitance

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

Choose the correct answer: