In steady state heat conduction,the equations | vedclass

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(B) In electrostatics,the electric field $E$ at a distance $R$ from a point charge $q$ is given by $E = \frac{1}{4\pi\epsilon_0} \frac{q}{R^2}$.The po

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$1$

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(B) In electrostatics,the electric field $E$ at a distance $R$ from a point charge $q$ is given by $E = \frac{1}{4\pi\epsilon_0} \frac{q}{R^2}$.The potential $V$ is related to the electric field by $E = -\frac{dV}{dR}$.By the given analogy,the heat current density $j$ is equivalent to the electric field $E$,and the temperature $T$ is equivalent to the potential $V$.Thus,the heat current density $j$ at a distance $R$ from a source is proportional to $\frac{1}{R^2}$.The total rate of heat flow $\dot{Q}$ through a spherical surface of radius $R$ is given by the product of the heat current density $j$ and the surface area $A = 4\pi R^2$.Therefore,$\dot{Q} = j \cdot A \propto \left( \frac{1}{R^2} \right) \cdot R^2 = R^0$.However,for a sphere maintained at a constant temperature difference relative to infinity,the heat flow rate $\dot{Q}$ is proportional to the capacitance of the sphere,which is proportional to $R$.Specifically,$\dot{Q} = \frac{\Delta T}{R_{th}}$,where $R_{th} = \frac{1}{4\pi k R}$.Thus,$\dot{Q} = 4\pi k R \Delta T \propto R^1$.Therefore,$n = 1$.

Heat flow	Electrostatics
$T(r)$	$V(r)$
$j(r)$	$E(r)$

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