The electric intensity at a point 3 metre away from a charge of 50 coulomb kept in air is *
50
500
5000
50000
Answers
Answered by
0
Answer:
Electric field are vector quantities in this case they are acting in opposite direction.
E due to A= kq/r2=
(0⋅3) 2
(9×10 )
9
)(50μC)
(
x
^
)
E due to B=
(0⋅3)
2
(9×10
9
)(100μC)
(−
x
^
)
∴Net E=
(0⋅3)
2
9×10
9
(50−100)×10
−6
x
^
E=5×10
6
(−
x
^
) V/m
Answered by
1
Answer:
50
Explanation:
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