Consider two conductors of some shape. Use them in two alternative ways, as a capacitor and as a resistor as shown in (a) and (b) of figure respectively. In case (a), the space between the conductors is filled with a homogeneous material of permittivity $ε$ while in case (b), it is filled with a homogeneous material of finite conductivity $σ$ .

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Value of RC is given by $\frac{ε}{σ}$ (where R is resistance in case (b) and C is capacitance in case (a)) when conductors are square in shape.

Value of RC is given by $\frac{2 ε}{σ}$ (where R is resistance in case (b) and C is capacitance in case (a)) when conductors are square in shape.

Time constant of the system is independent of shape of the conductors when a medium having permittivity e and conductivity $σ$ is inserted between the conductors.

Time constant of the system depends on shape of the conductors when a medium having permittivity $ε$ and conductivity $σ$ is inserted between the conductors.

answer is A, C.

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Detailed Solution

From Gauss Law,
$\begin{array}{l} \int \vec{E} \cdot d \vec{A} = \frac{Q}{ε} = \frac{CV}{ε} \\ \Rightarrow \frac{1}{V} \int \vec{E} \cdot d \vec{A} = \frac{C}{ε} . . . . . (i) \end{array} Also, \vec{J} = σ \vec{E} and I = \vec{J} \cdot d \vec{A} \begin{array}{l} ∴ I = \frac{V}{R} \Rightarrow \int \vec{J} \cdot d \vec{A} = \frac{V}{R} \\ \Rightarrow \frac{σ}{V} \int \vec{E} \cdot d \vec{A} = \frac{1}{R} \\ \Rightarrow \frac{1}{V} \int \vec{E} \cdot d \vec{A} = \frac{1}{σR} . . . . (ii) \end{array} ∴ from (i) \& (ii), 1 = \frac{CσR}{ε} \Rightarrow RC = \frac{ε}{σ}$