At a point A, there is an electric field of 400V/m
and potential difference of 2000 V. Then the
distance between the point charge and A is
Answers
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Explanation:
Here, the electric field,
E
=Ecos45
i
^
+Esin45
j
^
=
2
E
(
i
^
+
j
^
)
now, V
A
−V
B
=∫
A
B
E
.
dr
=∫
A
B
2
E
(
i
^
+
j
^
).(dx
i
^
+dy
j
^
)=
2
E
[∫
0
0.03
dx+∫
0.02
0
dy]
or V
A
−V
B
=
2
400
[0.03−0.02]=2.8V
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