Answer
(a) Proton will move from lower potential to higher potential against the electric field.
(b) $3340$ volt
Work Step by Step
Given:
Proton initial speed = 8,00,000 m/s
Proton final speed =0
(a) Proton will move from lower potential to higher potential against the electric field.
(b) By conservation of energy
Loss in kinetic energy = Gain in potential energy
$e\Delta V=\frac{1}{2}m_{p}v_{p}^{2}-0$
$e\Delta V=\frac{1}{2}m_{p}v_{p}^{2}$
$\Delta V=\frac{{\frac{1}{2}m_{p}v_{p}^{2}}}{1.6 \times 10^{-19}}$
$=\frac{1}{2}\times \frac{{ 1.67 \times 10^{-27} \times(8\times 10^{5})^{2}}}{1.6 \times 10^{-19}}$
$=3340$ volt
