Question

Find the resistivity of conductor which carries a current density of 2.5x106Am-2 and electric field of is 15 V/m

Answer 1

$\underline{ \boxed{ \mathfrak{ \huge{ \purple{Answer}}}}} \\ \\ \star \rm \: \pink{Given} \\ \\ \implies \rm \: current \: density \: (J) = 2.5 \times {10}^{6} \: \frac{a}{ {m}^{2} } \\ \\ \implies \rm \: electric \: field \: (E) = 15 \: \frac{v}{m} \\ \\ \star \rm \: \pink{To \: Find} \\ \\ \implies \rm \: resistivity \: ( \rho) \\ \\ \star \rm \: \pink{Formula} \\ \\ \implies \rm \: relation \: between \: J ,\: E \: and \: \rho \: is \: given \: as \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{ \huge{ \rm{ \red{E = J\times \rho}}}}} \\ \\ \star \rm \: \pink{Calculation} \\ \\ \implies \rm \: \rho = \frac{E}{J} = \frac{15}{2.5 \times {10}^{6} } \\ \\ \implies \underline{\boxed{ \rm{ \red{ \rho = 6 \times {10}^{ - 6} \: \mho \: {m}^{ - 1} }}}}$