Two isolated metallic solid spheres of radii $R$ and $2 R$ are charged such that both have same charge density $\sigma$. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is $\sigma^{\prime}$ The ratio $\frac{\sigma^{\prime}}{\sigma}$ is :

Question

A thin half ring of radius $35 \text{ cm}$ is uniformly charged with a total charge of $Q$ coulomb. If the magnitude of the electric field at centre of the half ring is $100 \text{ V/m}$, then the value of $Q$ is ______ $\text{nC}$.
($\epsilon_o = 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2$ and $\pi = 3.14$)

Answer 1

To solve this problem, we need to understand the concept of charge distribution when two metallic spheres are connected by a thin conducting wire.

Initially, we have two metallic spheres with radii $R$ and $2R$, both charged with the same charge density $\sigma$. The charges on these spheres can be calculated as follows:

The charge on the smaller sphere, $Q_1 = \sigma \cdot 4\pi R^2$.
The charge on the larger sphere, $Q_2 = \sigma \cdot 4\pi (2R)^2 = \sigma \cdot 16\pi R^2$.

When the spheres are connected by a conducting wire, charge will redistribute until both spheres have the same potential. If $q_1$ and $q_2$ are the new charges on the smaller and larger spheres respectively, then:

Total initial charge $Q = Q_1 + Q_2 = \sigma \cdot 4\pi R^2 + \sigma \cdot 16\pi R^2 = \sigma \cdot 20\pi R^2$.
Since the potential on both spheres must be the same, the potential of the spheres given by $\frac{q_i}{r_i}$ will equalize:
$\frac{q_1}{R} = \frac{q_2}{2R} \Rightarrow q_2 = 2q_1$.

Using conservation of charge:

$q_1 + q_2 = Q = \sigma \cdot 20\pi R^2$.
Substitute $q_2 = 2q_1$ into the equation above:
$q_1 + 2q_1 = \sigma \cdot 20\pi R^2 \Rightarrow 3q_1 = \sigma \cdot 20\pi R^2 \Rightarrow q_1 = \frac{\sigma \cdot 20\pi R^2}{3}$.
Thus, $q_2 = 2 \cdot \frac{\sigma \cdot 20\pi R^2}{3} = \frac{40\pi R^2 \sigma}{3}$.

The new charge density $\sigma'$ of the bigger sphere is:

$\sigma' \cdot 16\pi R^2 = \frac{40\pi R^2 \sigma}{3}$.
Solving for $\sigma'$:
$\sigma' = \frac{\frac{40\pi R^2 \sigma}{3}}{16\pi R^2} = \frac{40}{48} \sigma = \frac{5}{6} \sigma$.

Therefore, the ratio $\frac{\sigma'}{\sigma}$ is $\frac{5}{6}$.

Hence, the correct answer is

$\frac{5}{6}$

Two isolated metallic solid spheres of radii $R$ and $2 R$ are charged such that both have same charge density $\sigma$. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is $\sigma^{\prime}$ The ratio $\frac{\sigma^{\prime}}{\sigma}$ is :

The Correct Option is C

Solution and Explanation

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