Question 183: Electrostatics & 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
σ
1
and
σ
2
, 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

Electrostatics & 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 σ 1 and σ 2 , respectively. The ratio \frac{\sigma_1}{\sigma_2} is

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