Directions: (6-10) |
Potential Energy of the Proton |
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 qo 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 \[4\centerdot 0\times {{10}^{8}}V\,{{m}^{-1}}\] directed along the positive X-axis. The proton undergoes a displacement of \[0\centerdot 25m\] in the direction of E. |
Mass of a proton =\[1\centerdot 66\times {{10}^{-27}}\,kg\] and charge of proton \[1\centerdot 6\times {{10}^{-19}}\,C\] |
A) \[-1\times {{10}^{8}}V\]
B) \[1\times {{10}^{8}}V\]
C) \[6\centerdot 4\times {{10}^{-19}}V\]
D) \[-6\centerdot 4\times {{10}^{-19}}V\]
Correct Answer: A
Solution :
As \[\Delta V=-E\Delta x=-\left( 4\,.\,0\times {{10}^{8}}\,V/m \right)\]\[\left( 0.\,25\,m \right)=-{{10}^{8}}V\]You need to login to perform this action.
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