Question

a regular hexagon of side 10cm has a charge 5 MC at each of its verticle. calculate the potential at the center of the hexagon​

Answer 1

$\small{\underline{\sf{\green{Given:}}}}$

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

$\small{\underline{\sf{\orange{To\:Find:}}}}$

• 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.

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

$\small{\underline{\sf{\pink{Solution:}}}}$

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

Hence,

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

$\small{\underline{\sf{\red{More\:To\:know:}}}}$

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

$\begin{gathered}\\ {\underline{\underline\pink{\boxed{\sf{ V = k \times \dfrac{q}{r} }}}}} \\\end{gathered}$

Where,

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

Answer 2

$\tt{In \: the \: attached \: figure ,O \: is \: centre \: of \: hexagon \: ABCDEF \: of \: each \: side \: 10cm \: as \: is \: clear}$

$\tt{from \: the \: figure ,OAB , OBC \: etc \: are \: equilateral \: triangle}$

$\tt{Therefore, OA = OB = OC = OD = OE = OF = r = 10cm}$

$\tt{ = {10}^{ - 1} m. \: As \: potential \: is \: scalarquantity,}$

$\tt therefore \: potential \: at \: O \: is \: V$

$= 6 \times \frac{q}{4 \pi ∈ _{0}r }$

$\tt = \frac{9 \times {10}^{9} \times 6 \times 5 \times {10}^{ - 6} }{ {10}^{ - 1} } = 2.7 \times {10}^{6} V.$