find b part
fast....
Answers
Explanation:
Solution:
\mapsto↦ The given figure shows six equal amout of charges, q, at the vertices of regular hexagon.
Where,...
\hookrightarrow↪ \sf Charge,\,q\:=\:5\mu C\:=\:5\times 10^{-6}CCharge,q=5μC=5×10
−6
C
\hookrightarrow↪ Side of the hexagon, l = AB = BC = CD = DE = EF = FA = 10cm
\hookrightarrow↪ Distance of each vertex from centre O, d = 10cm
Electric potential at point O is given by,
\longrightarrow⟶ \underline{\boxed{\sf{\orange{V\:=\: \dfrac{6\times q}{4\pi \epsilon_od}}}}}
V=
4πϵ
o
d
6×q
Where,
\hookrightarrow↪ \sf \epsilon_oϵ
o
= Permittivity of free space
\hookrightarrow↪ \sf \dfrac{1}{4\pi \epsilon_o}\:=\:9\times 10^9C^{-2}m^{-2}
4πϵ
o
1
=9×10
9
C
−2
m
−2
\leadsto⇝ \sf \therefore\:V\:=\: \dfrac{6\times 9\times 10^9\times 5\times 10^{-6}}{0.1}∴V=
0.1
6×9×10
9
×5×10
−6
\sf \implies\:V\:=\:2.7\times 10^6\,V⟹V=2.7×10
6
V
\longrightarrow⟶ Hence, the potential at centre of hexagon is \sf{\red{2.7\times 10^6\,V}}2.7×10
6
V