at the corners A,B,C of square ABCD (each side 0.2m) positive charges of 2×10^-9C, 4×10^-9C and 6×10^-9C are placed. what is the intensity of electric field at D?
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Answer:
The charge is placed as shown in the figure.
2×10
−9
at positionA,4×10
−9
at position B and6×10
−9
at C
E
A
,E
B
andE
C
is produce due to charge present at position A ,B and C
⇒
E
A
=
d
2
−Kq
A
j
^
=−
(0.2)
2
9×10
9
×2×10
−9
j
^
=−50
j
^
N/C.
⇒
E
C
=
d
2
−Kq
c
i
^
=
(0.2)
2
9×10
9
×6×10
−9
i
^
=−125
i
^
N/C.
⇒
E
B
=
2
2
d
2
−Kq
B
i
^
−
2
2
d
2
Kq
B
j
^
=−
2×
2
×(0.2)
2
9×10
9
×6×10
−9
i
^
=−
2×
2
×(0.2)
2
9×10
9
×6×10
−9
i
^
−
2×
2
×(0.2)
2
9×10
9
×6×10
−9
j
^
=−
2
50
i
^
−
2
50
j
^
resultant electric field intensity
⇒
E
=
E
A
+
E
B
+
E
C
=(−125−
2
50
)
i
^
+(−50−
2
50
)
j
^
=−160.3
i
^
−85.3
j
^
⇒E=181.5N/C
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