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Question

There exists a region of electric field with potential difference of 25 V as shown in figure. An electron enters this region at speed of v = 3 ×106 m/s at angle of α and emerges out of region at angle β. The sin αsin β is [Electric field exist in given region only]
e=1.6×1019,m=9.1×1031 (2 decimal places)

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Solution

u sin α=vsin αsin β=vu
12mu2+qV=12mv2 vu=1+2qVmu2=1.406

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