Three charges 4nc and 8nc are placed at three corners of square of side 2cm. Find the electric field at the 4th corner.
Answers
Answer:ANSWER
AD=2 cm=2×10
−2
m
CD=2 cm=2×10
−2
m
BD=2
2
cm=2
2
×10
−2
m
E
A
=
4πε
0
1
r
2
Q
=9×10
9
×
(2×10
−2
)
2
4×10
−9
=9×10
4
NC
−1
along AD
E
C
=
4πε
0
1
=9×10
9
×
(2×10
−2
)
2
4×10
−9
=9×10
4
NC
−1
along CD
E
B
=
4πε
0
1
r
2
Q
B
=9×10
9
×
(2
2
×10
−2
)
2
4×10
−9
=4.5×10
4
NC
−1
along BD
X−component of the field E
x
=E
A
+E
B
cos45=9×10
4
+4.5×10
4
×0.707
E
x
=12.1815×10
4
NC
−1
Y−component of the field E
y
=E
C
+E
B
cos45=9×10
4
+4.5×10
4
×0.707
E
x
=12.1815×10
4
NC
−1
Magnitude E=
E
x
2
+E
y
2
=
(12.1815×10
4
)
2
+(12.1815×10
4
)
2
E=
2×(12.1815×10
4
)
2
=17.2272×10
4
NC
−1
OR
E
AC
=
E
A
2
+E
C
2
=
(9×10
4
)
2
+(9×10
4
)
2
E
AC
=
2(9×10
4
)
2
=12.726×10
4
NC
−1
along BD
E=E
B
+E
AC
=4.5×10
4
+12.724×10
4
=17.226×10
4
NC
−1
along B
Explanation: PLEASE MARK AS BRAINLEST
Answer:
=2 cm=2×10
−2
m
CD=2 cm=2×10
−2
m
BD=2
2
cm=2
2
×10
−2
m
E
A
=
4πε
0
1
r
2
Q
=9×10
9
×
(2×10
−2
)
2
4×10
−9
=9×10
4
NC
−1
along AD
E
C
=
4πε
0
1
=9×10
9
×
(2×10
−2
)
2
4×10
−9
=9×10
4
NC
−1
along CD
E
B
=
4πε
0
1
r
2
Q
B
=9×10
9
×
(2
2
×10
−2
)
2
4×10
−9
=4.5×10
4
NC
−1
along BD
X−component of the field E
x
=E
A
+E
B
cos45=9×10
4
+4.5×10
4
×0.707
E
x
=12.1815×10
4
NC
−1
Y−component of the field E
y
=E
C
+E
B
cos45=9×10
4
+4.5×10
4
×0.707
E
x
=12.1815×10
4
NC
−1
Magnitude E=
E
x
2
+E
y
2
=
(12.1815×10
4
)
2
+(12.1815×10
4
)
2
E=
2×(12.1815×10
4
)
2
=17.2272×10
4
NC
−1
OR
E
AC
=
E
A
2
+E
C
2
=
(9×10
4
)
2
+(9×10
4
)
2
E
AC
=
2(9×10
4
)
2
=12.726×10
4
NC
−1
along BD
E=E
B
+E
AC
=4.5×10
4
+12.724×10
4
=17.226×10
4
NC
−1
along BD
solution
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