Acceleration of the particles due to electric field, a=mF=mqE=(2×1011)(1.8×103)
=3.6×1014ms−2
Time to cross plates =vd
⇒t=3×1070.10
Now deflection,
y=21at2=21(3.6×1014)(3×1070.10)2
=0.002m
=2mm
A stream of a positively charged particles having mq=2×1011Ckg−1 and velocity v0=3×107i^ms−1 is deflected by an electric field 1.8j^kVm−1. The electric field exists in a region of 10cm along x direction. Due to the electric field, the deflection of the charge particles in the y direction is _____ mm.
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