An electron is taken from a point A to point B along the path AB in a uniform electric field of intensity E = 10 Vm™
1 Side AB = 5 mand side BC = 3 m Then, the amount of work done is
a) 50 eV
b) 40 eV
c) - 50 eV
d) - 40 eV
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Answered by
0
Answer:
W
AB
=W
AC
+W
CB
W
CB
should be zero, because in moving from C to B, we always move perpendicular to field. Hence, force applied by field and displacement will be at 90
∘
.
so work done in BC will be 0
W
AC
=−e(V
C
−V
A
)
V
C
−V
A
=−E×AC=−10×4=−40
∴W
AB
=40eJ=40eV
Explanation:
Hope it's help you...
Answered by
1
Answer:
-40ev
Explanation:
work done = force.displacement ( vectors)
from A to C
w = delta(U) = - 10×AC
= -40
work done in moving electron along bc is zero as work done in moving charge perpendicular to field is zero(w = F.s.cos(x) and x = 90°)
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