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Hello, Can somebody please answer the question attached in the image below? Thank you.

Hello,

Can somebody please answer the question attached in the image below? Thank you.

Question Image
Grade:12th pass

1 Answers

Eshan
askIITians Faculty 2095 Points
3 years ago
Dear student,

The electric field due to the potential difference of5MVover a distance of0.08mis

E=\dfrac{V}{d}=6.25\times 10^7 V/m
Hence force acting on the proton=eE
Work done on proton by electric field from A to B iseE.d
This work done is equal to the energy gained by the proton.

HenceeE.d=\dfrac{1}{2}mv^2
\implies v=\sqrt{\dfrac{2eEd}{m}}

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