F 10 m E
. Find the area of the hexagonal room with
each side 10 m.
10 m
WOT
D
22 m
10 m
10 m
B 10 m
c С.
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Answer:
ABCDEF is a regular hexagon of side 10 cm each. At each corner, the charge q=5μC is placed. O is the center of the hexagon
Given AB=BC=CD=DE=EF=FA=10cm. As the hexagon has six equilateral triangles so the distance of center O from every vertex is 10 cm.
i.e. OA=OB=OC=OD=OE=OF=10cm.
Potential at point O = Sum of potential at center O due to individual point charge
∴V
O
=V
A
+V
B
+V
C
+V
D
+V
E
+V
F
As V=
4πε
0
1
r
q
then
V
O
=
4πε
0
1
.[
OA
q
+
OB
q
+
OC
q
+
OD
q
+
OE
q
+
OF
q
]
Putting the values, we get
V
O
=9×10
9
[
10×10
−2
5×10
−6
+
10×10
−2
5×10
−6
+
10×10
−2
5×10
−6
+
10
−2
5×10
−6
+
10×10
−2
5×10
−6
+
10×10
−2
5×10
−6
]
V
O
=9×10
9
×
10×10
−2
6×10
−6
×5
V
O
=2.7×10
6
V
solution
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