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AMU Medical AMU Solved Paper-2015

  • question_answer
    A proton is released from rest in a uniform electric field of magnitude \[8.0\times {{10}^{4}}\] V/m, directed along positive x-axis. The proton undergoes a displacement of 0.30m in the direction of the field. What is the change in electric potential of the proton as a result of this displacement?

    A)  \[2.4\times {{10}^{4}}\,V\]

    B)  \[4.8\times {{10}^{4}}\,V\]

    C)  \[-2.4\times {{10}^{4}}\,V\]

    D)  \[-1.2\times {{10}^{4}}\,V\]

    Correct Answer: C

    Solution :

                     Electric field of magnitude \[8.0\times {{10}^{4}}V/m\] Displacement = 0.30 m                 Therefore, change in electric potential                 \[=-Ed\]                 \[=-(8\times {{10}^{4}})\,(0.30)\]                 \[=-2.4\times {{10}^{4}}V\]


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