A particle of charge 2 uC and mass 1.6 g
is moving with a velocity 4ims-1. At
t = 0 the particle enters in a region having
an electric field Ē (in N C-1) = 80 î +60 j.
Find the velocity of the particle at t = 5 s.
(2/3, 2020)
Answers
Answered by
4
Answer:
Given that :-
q = charge =2μC=2×10
−6
C
mass =m=1.6g=1.6×10
−3
kg
V= velocity =4
i
^
m/s
E
=(80
i
^
+60j)N/C
Since particle has initial velocity only in x-direction
u
x
=v=4m/s;u
y
=0m/s
Force =F=q
E
⇒
F
=2×10
−6
(80
i
^
+60j)=(160
i
^
+120j)×10
−6
a
= acceleration =
a
=
m
F
=
1.6×10
−3
(160
i
^
+120j)×10
−6
a
=(100
i
^
+75
j
^
)×10
−3
=(a
x
i
^
+a
y
j
^
)
time =t=5sec
v
x
=u
x
+a
x
t=(4+100×10
−3
×5)
v
x
=4+0.5=4.5m/s
v
y
=u
y
+a
y
t=0+75×10
−3
×5=0.375m/s
Where u
x
,v
x
are initial and final velocity in x-direction and u
y
and v
y
are initial and final velocity in y-direction.
∴ Final velocity =v=v
x
i
^
+v
y
j
^
=4.5
i
^
+0.375j
v=(4.5
i
^
+0.375
j
^
)m/s
Answered by
0
Explanation:
answer- 4.5i + 0.375 j
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