Physics - Electrostatics Question with Solution | TestHub

$\left|Q_B\right|>\left|Q_A\right|$ (given). Inside the shell $B$, the potential is a constant and negative. But $V_A$ inside $r_A$ is positive and greater than that of $B$ because $r_A<r_B$. Upto $r_B$, one as hence positive potential. Beyond $r_A,V_A\propto (1/r)$, this is $>V_B$, one gets the course $\propto 1/r$. At $B$, the negative potential is more therefore the potential line is still negative. (c) explains all these factors. Note: However the question asked is the field and the diagrams given are for the potential. This is an error. $$\begin{array}{rl}& \text{ Resistance }=\frac{\text{ potential difference }}{\text{ current }}\\ & =\frac{{\mathrm{ML}}^2{\mathrm{T}}^{-3}{\mathrm{A}}^{-1}}{\mathrm{A}}=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-3}{\mathrm{A}}^{-2}\right]\text{. }\\ & \end{array}$$