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

Question:

A regular hexagon of side 10 cm has a charge 5 micro C at each of its vertices. calculate the potential at the center of the hexagon?

Answer:

• The Potential (V) will be 2.7 × 10⁶ V

Given:

1. Side of Hexagon (a) = 10 cm = 0.1 m
2. Each vertices has charge (q) = 5 μ C

Explanation:

$\rule{300}{1.5}$

We know, that the hexagon is made of 6 equilateral triangles. Therefore, OA = OB = AB

Hence,

⇒ OA = OB = 10 cm = 0.1 m

And, as we know that potential is a scalar quantity, we will be adding all the potential acting on the center by scalar addition.

Now,

$\bigstar\;\underline{\boxed{\sf V=V_{1}+V_{2}+V_{3}+V_{4}+V_{5}+V_{6}}}$

Solving,

$\displaystyle\longrightarrow\sf V=\dfrac{1}{4\;\pi \epsilon_o}\Bigg\lgroup \dfrac{q_1}{r_1} + \dfrac{q_2}{r_2}+\dfrac{q_3}{r_3}+\dfrac{q_4}{r_4}+\dfrac{q_5}{r_5}+\dfrac{q_6}{r_6}\Bigg\rgroup$

Here, the Charges are same i.e. 5 μC = 5 × 10⁻⁶ C and distance of separation is same i.e. r = 0.1 m.

$\displaystyle\longrightarrow\sf V=\dfrac{1}{4\;\pi \epsilon_o}\Bigg\lgroup \dfrac{q+q+q+q+q+q}{r}\Bigg\rgroup\\\\\\\longrightarrow\sf V=\dfrac{1}{4\;\pi \epsilon_o}\Bigg\lgroup \dfrac{6\times q}{r}\Bigg\rgroup\\\\\\\longrightarrow\sf V=\dfrac{6}{4\;\pi \epsilon_o}\Bigg\lgroup \dfrac{q}{r}\Bigg\rgroup$

Substituting the values,

$\longrightarrow\sf V=\dfrac{6}{4\;\pi \epsilon_o}\Bigg\lgroup \dfrac{5\times 10^{-6}}{10^{-1}}\Bigg\rgroup\\\\\\\longrightarrow\sf V=\dfrac{6}{4\;\pi \epsilon_o}\Bigg\lgroup 5\times 10^{-6+1}\Bigg\rgroup\\\\\\\longrightarrow\sf V=\dfrac{6}{4\;\pi \epsilon_o}\Bigg\lgroup 5\times 10^{-5}\Bigg\rgroup\\\\\\\longrightarrow\sf V=\dfrac{30\times 10^{-5}}{4\;\pi\epsilon_o}\\\\\\\longrightarrow\sf V=30\times 10^{-5} \times 9\times 10^{9}$

$\displaystyle\longrightarrow\sf V=270\times 10^{9-5}\\\\\\\longrightarrow\sf V = 270\times 10^{4}\\\\\\\longrightarrow\sf V = 2.7\times 10^{6}\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf V = 2.7\times 10^{6}\;J/C}}}}$

∴ The Potential (V) will be 2.7 × 10⁶ V.

$\rule{300}{1.5}$

Answer 2

$\huge\underline\mathtt\red{Answer:-}$

•The Potential (V) will be 2.7 × 10⁶ V