Question 46: Electrostats & Capacitance

Question

Two metal spheres of radius R and 3R have same surface charge density $\sigma$. If they are brought in contact and then separated, the surface charge density on smaller and bigger sphere becomes $\sigma_1$ and $\sigma_2$, 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 $V = \frac{\sigma r}{\epsilon_0}$, where $r$ is the radius of the sphere and $\epsilon_0$ is the permittivity of free space.When the two spheres are brought into contact and then separated, they equalize their potentials:$V1 = V2$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

A. $\frac{1}{3}$

Answer

B. $\frac{1}{9}$

Answer

C. $9$

Answer

D. $3$

Electrostats & Capacitance

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

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