A hexagon of side 8 ~cm has a charge 4 μ C at each of its vertices....

A hexagon of side

8 \mathrm{~cm}

has a charge

4 \mu \mathrm{C}

at each of its vertices. The potential at the centre of the hexagon is A hexagon of side 8 ~cm has a charge 4 μ C at each of its vertices. The potential at the centre of the hexagon is

A hexagon of side 8 ~cm has a charge 4 μ C at each of its vertices. The potential at the centre of the hexagon is

Options

A $2.7 \times 10^6 \mathrm{~V}$
B $7.2 \times 10^{11} \mathrm{~V}$
C $2.5 \times 10^{12} \mathrm{~V}$
D $3.4 \times 10^4 \mathrm{~V}$

Correct answer

A. $2.7 \times 10^6 \mathrm{~V}$

Step-by-step solution

As shown in the figure,

O

is the centre of hexagon

A B C D E F

of each side

8 \mathrm{~cm}

. As it is a regular hexagon

O A B, O B C

, etc are equilateral triangles.

\begin{aligned} \therefore \quadO A=O B=O C= & O D=O E=O F=8 \mathrm{~cm}\\ & =8 \times10^{-2} \mathrm{~m} \end{aligned}

The potential at

O

\begin{aligned}V & =6 \times \frac{q}{4 \pi \varepsilon_0 r}\\ & =\frac{6 \times 9 \times 10^9 \times 4 \times 10^{-6}}{8 \times 10^{-2}}=2.7 \times10^6 \mathrm{~V} \end{aligned}

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