Chapter 16 - Electric Potential, Energy, and Capacitance - Learning Path Questions and Exercises - Conceptual Questions - Page 591: 13

a). $q=1.6\times10^{-19}C$ potential energy = $\Delta U=q\Delta V=1.6\times10^{-19}\times10^{6}=1.6\times10^{-13} J$ The gain in the Kinetic energy comes from the loss in the potential energy. Since it has no kinetic energy at start, initial velocity being zero. So, kinetic energy = $1.6\times10^{-13} J$ b). When $q^{'}=2q$, $\Delta V^{'}= 2q\Delta V$, Kinetic energy $=\Delta U^{'}= 2q\Delta V$ Also, kinetic energy = $\frac{1}{2}mv^{2}$ But, $m^{'}=4m$ So, $\Delta U^{'}= 2q\Delta V=\frac{1}{2}m'v^{2}$ $2q\Delta V=\frac{1}{2}4mv^{2}$ or, $q\Delta V=2(\frac{1}{2}mv^{2})$ So, the answer to the question part (a) is would be double.

a). $q=1.6\times10^{-19}C$ potential energy = $\Delta U=q\Delta V=1.6\times10^{-19}\times10^{6}=1.6\times10^{-13} J$ The gain in the Kinetic energy comes from the loss in the potential energy. Since it has no kinetic energy at start, initial velocity being zero. So, kinetic energy = $1.6\times10^{-13} J$ b). When $q^{'}=2q$, $\Delta V^{'}= 2q\Delta V$, Kinetic energy $=\Delta U^{'}= 2q\Delta V$ Also, kinetic energy = $\frac{1}{2}mv^{2}$ But, $m^{'}=4m$ So, $\Delta U^{'}= 2q\Delta V=\frac{1}{2}m'v^{2}$ $2q\Delta V=\frac{1}{2}4mv^{2}$ or, $q\Delta V=2(\frac{1}{2}mv^{2})$ So, the answer to the question part (a) is would be double.