question_answer

Direction: Q.6 to Q.10

Potential difference \[\left( \Delta V \right)\]between two points A and B separated by a distance x, in a uniform electric field E is given by \[\Delta V=-Ex\]where x is measured parallel to the field lines. If a charge \[{{q}_{0}}\]moves from P to Q, the change in potential energy \[\left( \Delta U \right)\]is given as \[\Delta U={{q}_{0}}\Delta V\]. A proton is released from rest in uniform electric field of magnitude \[4.0\times {{10}^{8}}V{{m}^{-1}}\]directed along the positive X-axis. The proton undergoes a displacement of 0.25 m in the direction of E.

Mass of a proton \[=1.66\times {{10}^{-27}}kg\]and charge of proton\[=1.6\times {{10}^{-19}}C\]

Read the given passage carefully and give the answer of the following questions.

The change in electric potential of the proton between the points A and B is:

A) \[-1\times {{10}^{8}}V\]

B)   \[1\times {{10}^{8}}V\]

C) \[6.4\times {{10}^{-19}}V\]     

D) \[-6.4\times {{10}^{-19}}V\]