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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Solution Steps
1
Condition for Equilibrium
When two conducting spheres are brought in contact, charge flows between them until they reach a common electric potential $V$.
Potential of a sphere of radius $r$ and charge $q$ is given by $V = \frac{kq}{r}$.
2
Equating Potentials
After separation, let the charges be $q_1$ and $q_2$ for spheres with radii $r_1=R$ and $r_2=3R$. Since they were in contact:
$$V_1 = V_2 \implies \frac{kq_1}{r_1} = \frac{kq_2}{r_2}$$
$$\frac{q_1}{q_2} = \frac{r_1}{r_2} = \frac{R}{3R} = \frac{1}{3}$$
3
Relating Charge to Density
Surface charge density $\sigma$ is defined as charge per unit area: $\sigma = \frac{q}{4\pi r^2}$.
Therefore, the ratio of final densities $\sigma_1$ and $\sigma_2$ is:
$$\frac{\sigma_1}{\sigma_2} = \frac{q_1 / (4\pi r_1^2)}{q_2 / (4\pi r_2^2)} = \frac{q_1}{q_2} \times \frac{r_2^2}{r_1^2}$$
4
Simplify the Ratio
Substitute the charge ratio $\frac{q_1}{q_2} = \frac{r_1}{r_2}$ into the density ratio equation:
$$\frac{\sigma_1}{\sigma_2} = \left(\frac{r_1}{r_2}\right) \times \left(\frac{r_2^2}{r_1^2}\right) = \frac{r_2}{r_1}$$
5
Final Calculation
Using the given radii $r_1 = R$ and $r_2 = 3R$:
$$\frac{\sigma_1}{\sigma_2} = \frac{3R}{R} = 3$$
Final Answer: 3
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Theory
1. Distribution of Charges on Conductors
When two conductors are connected, charge redistributes to minimize potential energy. In equilibrium, the entire connected system forms an equipotential surface. For spherical conductors, this leads to the charge being proportional to the radius ($q \propto r$). This is why larger spheres hold more charge than smaller ones when brought to the same potential.
2. Surface Charge Density and Curvature
Surface charge density $\sigma$ is higher on regions with smaller radii of curvature. Since $V = \frac{\sigma r}{\epsilon_0}$, for a constant potential $V$, we see that $\sigma \propto \frac{1}{r}$. This physical principle explains why sharp points on conductors have very high charge densities, leading to effects like corona discharge and the functionality of lightning rods.
3. Capacitance of Isolated Spheres
The ability of a sphere to store charge is its capacitance, defined as $C = q/V$. For a sphere in a vacuum, $C = 4\pi\epsilon_0 r$. When two spheres are in contact, they act as capacitors in parallel, and the total charge is distributed such that $q_1/C_1 = q_2/C_2$. This confirms the $q \propto r$ relationship derived from the potential formula.
4. Conservation of Charge
In any isolated system, total charge remains constant. If spheres initially have density $\sigma$, their initial charges are $4\pi R^2\sigma$ and $4\pi(3R)^2\sigma$. Upon contact, the sum of these charges is preserved, but they are redistributed. While the total charge helps find the absolute values of $\sigma_1$ and $\sigma_2$, the ratio depends solely on the geometric constraint of equal potential.
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FAQs
1
Does the initial charge density $\sigma$ affect the final ratio?
No, the final ratio $\sigma_1/\sigma_2 = r_2/r_1$ is independent of the initial charges, provided the spheres reach equilibrium.
2
What is the common potential $V$ after contact?
It is the total initial charge divided by the total capacitance: $V = (q_{total}) / (4\pi\epsilon_0(R + 3R))$.
3
Why does charge flow between the spheres?
Charge flows because of a potential difference. It stops when the potential becomes equal across both spheres.
4
How is $\sigma \propto 1/r$ derived?
From $V = kq/r$ and $q = \sigma(4\pi r^2)$, we get $V = k(4\pi r^2 \sigma)/r = 4\pi kr\sigma$. Thus $\sigma = V/(4\pi kr)$.
5
What if the spheres are very far apart after separation?
The problem assumes they are separated enough so their fields don’t interfere, but the charges they acquired during contact remain.
6
Is energy conserved during the process?
No, some energy is usually lost as heat in the spark or the connecting wire during the redistribution of charge.
7
Can this apply to non-metal spheres?
No, this logic requires the spheres to be conductors so that charge can move freely to reach an equipotential state.
8
What is the ratio of final charges $q_1/q_2$?
The ratio of charges is $R/3R = 1/3$.
9
Does the medium affect the ratio?
No, the dielectric constant would cancel out in the ratio of potentials.
10
Why is $\sigma_1$ higher than $\sigma_2$?
Because the smaller sphere has a tighter curvature, it requires a higher charge density to reach the same potential as the larger sphere.
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