Question

A
A regular hexagon of
side
10cm
has
a charge of
5mc
at
each
of its vertices.
calculate the potential at the
centre
OF
hexagon​

Answer 1

Explanation:

$\Large{\red{\underline{\underline{\sf{\blue{Solution:}}}}}}$

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

$\mapsto$ The given figure shows six equal amout of charges, q, at the vertices of regular hexagon.

 ‎ ‎ ‎ ‎ ‎ ‎

Where,...

$\hookrightarrow$ $\sf Charge,\,q\:=\:5\mu C\:=\:5\times 10^{-6}C$

$\hookrightarrow$ 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,

$\longrightarrow$ $\underline{\boxed{\sf{\orange{V\:=\: \dfrac{6\times q}{4\pi \epsilon_od}}}}}$

 ‎ ‎ ‎ ‎ ‎ ‎

Where,

$\hookrightarrow$ $\sf \epsilon_o$ = Permittivity of free space

$\hookrightarrow$ $\sf \dfrac{1}{4\pi \epsilon_o}\:=\:9\times 10^9C^{-2}m^{-2}$

 ‎ ‎ ‎ ‎ ‎ ‎

$\leadsto$ $\sf \therefore\:V\:=\: \dfrac{6\times 9\times 10^9\times 5\times 10^{-6}}{0.1}$

 ‎ ‎ ‎ ‎ ‎ ‎

$\sf \implies\:V\:=\:2.7\times 10^6\,V$

 ‎ ‎ ‎ ‎ ‎ ‎

$\longrightarrow$ Hence, the potential at centre of hexagon is $\sf{\red{2.7\times 10^6\,V}}$

Answer 2

Answer:

• Side of Regular Hexagon - 10 cm
• Charge = 5${ \mu}$c

To Find:

• Potential at the centre of Hexagon?

There are two key elements on which the electric potential energy of an object depends:

• It’s own electric charge.
• It’s relative position with other electrically charged objects.

$\\ {\boxed{\sf{ V = \dfrac{6q}{4π\epsilon_{0} d} }}}$

Solution:

$\\ \longrightarrow\purple{\sf{ V = \dfrac{6q}{4π\epsilon_{0} d} }} \\ \\ \\ \implies{\sf{ V = 2.7 \times 10^6 V}}$

Hence,

The potential at the centre of Hexagon is $\blue{\sf{ V = 2.7 \times 10^6 V}}.$

Extra Dose:

The electric potential at any point around a point charge q is given by:

$\\ {\underline{\underline\red{\boxed{\sf{ V = k \times \dfrac{q}{r} }}}}}$

Where,

• V = electric potential energy,
• q = point charge
• r = distance between any point around the charge to the point charge
• k = Coulomb constant; k = 9.0 × ${10^9}$ N.