Question

is the electric flux due to these charges through a sphere of radius 4a with its centre at the origin?

ii. A Carnot refrigerator operates between 150 K and 200 K. Calculate its coefficient of performance.

Answer 1

i. The electric flux due to these charges through the sphere is $\frac{-5Q}{\epsilon_0}$. ii. The coefficient of performance of the Carnot refrigerator is $3$.

Answer 2

The given problem consists of two independent parts.

Part i: Electric Flux

Explanation: This part of the question is incomplete as it refers to "these charges" without defining them. Assuming, based on the provided similar question, that the charges are $-5Q$ located at $(2a, 0)$ and $+3Q$ located at $(5a, 0)$. According to Gauss's Law, the total electric flux ($\Phi$) through any closed surface is equal to the total electric charge ($Q_{enclosed}$) enclosed within that surface divided by the permittivity of free space ($\epsilon_0$). The given closed surface is a sphere of radius $4a$ with its center at the origin.

1. The charge $-5Q$ is located at $(2a, 0)$. Its distance from the origin is $2a$. Since $2a < 4a$ (radius of the sphere), this charge is inside the sphere.
2. The charge $+3Q$ is located at $(5a, 0)$. Its distance from the origin is $5a$. Since $5a > 4a$ (radius of the sphere), this charge is outside the sphere.

Only the charge enclosed within the sphere contributes to the electric flux. Therefore, the total enclosed charge $Q_{enclosed} = -5Q$. Applying Gauss's Law: $\Phi = \frac{Q_{enclosed}}{\epsilon_0} = \frac{-5Q}{\epsilon_0}$

Part ii: Carnot Refrigerator

Explanation: The coefficient of performance (COP) of a Carnot refrigerator operating between a low temperature reservoir ($T_L$) and a high temperature reservoir ($T_H$) is given by the formula: $\text{COP} = \frac{T_L}{T_H - T_L}$ Given: $T_L = 150 \, \text{K}$ $T_H = 200 \, \text{K}$ Substitute these values into the formula: $\text{COP} = \frac{150 \, \text{K}}{200 \, \text{K} - 150 \, \text{K}} = \frac{150 \, \text{K}}{50 \, \text{K}} = 3$