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F 10 m E. Find the area of the hexagonal room witheach side 10 m.10 mWOTD22 m10 m10 mB 10 mc С. |
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Answer» 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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