
$q_{2}=Q-q_{1}=\frac{Q}{2}$
$q_{3}=-q_{2}=\left(-\frac{Q}{2}\right)$
$E_{A}=E_{1}+E_{4}(\text { towards left })$
$=\frac{Q}{4 A \varepsilon_{0}}+\frac{Q}{4 A \varepsilon_{0}}=\frac{Q}{2 A \varepsilon_{0}}$
$E_{C}=-E_{A}(\text { towards right })$
$=\frac{Q}{2 A \varepsilon_{0}}$
$E_{B}=E_{2}+E_{3}(\text { towards right })$
$=\frac{Q}{4 A \varepsilon_{0}}+\frac{Q}{4 A \varepsilon_{0}}=\frac{Q}{2 A \varepsilon_{0}}$


$(A)$ $I =\frac{V_0 t}{\pi \rho} \ln \left(\frac{R_2}{R_1}\right)$
$(B)$ the outer surface is at a higher voltage than the inner surface
$(C)$ the outer surface is at a lower voltage than the inner surface
$(D)$ $\Delta V \propto I ^2$


