Point charges of +2nC and - 4nC are placed at A and C respectively of a triangle ABC in which angle ABC=90degree, AB=BC=0.5m. Find the resultant electric intensity at B.
Answers
Answered by
0
Answer:
Let 4C charge which is placed at C has an electric field intensity $$-E_{1}\hat{i} at position B
and electric field intensity due to 9C charge which is placed at position A is $$-E_{2}\hat{j} at position B
⇒E
1
=
r
2
Kq
1
=
2
2
×10
−4
9×10
9
×4×10
−9
=9×10
4
N/C.
⇒E
2
=
r
2
Kq
2
=
3
2
×10
−4
9×10
9
×9×10
−9
=9×10
4
N/C.
resultant E=−E
1
i
^
−E
2
j
^
=−9×10
4
i
^
−9×10
4
j
^
=
9
2
+9
2
×10
4
.
⇒3
2
×10
4
N/C
Similar questions