Question

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

Answer 1

Explanation:

A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon. The given figure shows six equal amount of charges, q, at the vertices of a regular hexagon. Therefore, the potential at the centre of the hexagon is 2.7 × 106 V

Answer 2

$\mathfrak{\huge{\green{\underline{Question:-}}}}$

A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the center of the hexagon.

$\mathfrak{\huge{\green{\underline{Given:-}}}}$

➤ Side of the hexagon = 10 cm

➤ Charge of the hexagon at each of it's vertices = 5 µC

$\mathfrak{\huge{\green{\underline{To \: Find:-}}}}$

➤ The potential at the center of the hexagon.

$\mathfrak{\huge{\green{\underline{Solution:-}}}}$

The regular hexagon containing charges q, at each of its vertices.

Here,

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

Length of each side of hexagon,

$\sf AB =BC = CD = DE = EF = FA = 10 \: cm.$

Distance of the vertices from the center O, d = 10 cm

Electric potential at point O,

$\sf V=\dfrac{1}{4 \pi \epsilon_0} .\dfrac{6xq}{d}$

Here,

$\sf \epsilon_0=Permittivity \: of \: free \: space \: and \: \dfrac{1}{4 \pi \epsilon_0} = 9 \times 10^{9} \: Nm^{2} \: C^{-2}$

$\sf V=\dfrac{9x \ 10^{9}\ x6x5x10_-_6}{0.1} = \underline{\underline{2.7 \times 10^{6}\ V}}$

$\mathfrak{\huge{\green{\underline{Note:-}}}}$

➤ Electrostatic potential is a state dependent function as electrostatic forces are conservative forces.

➤ The potential at a point due to a positive charge is positive while due to negative charge, it is negative.

➤ Electrostatic potential due to a point charge q at any point P lying at a distance r from it is given by $\sf V= \dfrac{1}{4 \pi \epsilon_0} . \dfrac{q}{r}$