Math, asked by faisalma263, 6 months ago

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.​

Answers

Answered by manjumeena80369
2

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

Attachments:
Similar questions