$\therefore \int \frac{1}{ dr }=\int_{ R _1}^{ R _2} \frac{ tdx }{\rho \pi x }$

$\frac{1}{ r }=\frac{ t }{\pi \rho} \ln \left(\frac{ R _2}{ R _1}\right)$

$\text { Resistance }=\frac{\pi \rho}{ t \ln \left(\frac{ R _2}{ R _1}\right)}$

$i=\frac{ V _0 t \ln \left(\frac{ R _2}{ R _1}\right)}{\pi \rho}$

$(-e $$\overrightarrow{ E }$ ) will be inward direction in order to provide centripetal acceleration. Therefore electric field will be radially outward

$V_{\text {outer }} < V_{\text {inmer }}$       ($C$)

$E =\frac{ m V_d^2}{ qr } \quad\left( I = meAV _{ d } \Rightarrow V _{ d } \propto i \right)$

$\Delta V =\int \overrightarrow{ E } \cdot \overrightarrow{ dr }$

$\Delta V \propto V _{ d }^2$

$\Delta V \propto I ^2$