4. Positive charges of 9 nC and 4 nC are placed at the points A and C respectively of a right angled Triangle ABC in which <B -90°. AB = 3 cm and BC =2cm. Find the electric intensity at B.ans is...... (12.73 10' N/C making a=45° with E, and E.)
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Explanation:
Let F
1
and F
2
exerted by the charge placed at A and B point of the equilateral triangle on C point as shown in the figure
F
1
=
r
2
Kq
A
q
c
=
(.3)
2
9×10
9
×3×10
−9
×2×10
−9
=6×10
−7
F
2
=
r
2
Kq
B
q
c
=
(.3)
2
9×10
9
×(−3)×10
−9
×2×10
−9
=−6×10
−7
both F
1
and F
2
make angle of 120^0 between each other
resultant of F
1
and F
2
is F
⇒F
2
=F
1
2
+F
2
2
+F
1
F
2
cosθ
⇒(6×10
−7
)
2
+(−6×10
−7
)
2
+2×6×10
−7
×(−6)×10
−7
cos120
0
⇒(36+36−36)×10
−14
=36×10
−14
⇒F=6×10
−7
N
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