A regular hexagon of side $5 \mathrm{~cm}$ has a charge $10 \mu \mathrm{C}$ at each of its vertices. The potential at the centre of hexagon is

Question

Aicharya · Accepted Answer

Given, length of sides of hexagon, $r=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$ charge on each vertices of hexagon, $q=10 \mu \mathrm{C}$ $=10 \times 10^{-6} \mathrm{C}$ Now, according to the question, Due to one charge, potential on hexagon's centre $P$ is given as, potential, $V=\frac{1}{4 \pi \varepsilon_0}, \frac{q}{r}$ Potential, $V=\frac{k q}{r}$ $\left[\because k=\frac{1}{4 \pi \varepsilon_0}, k=9 \times 10^9 \mathrm{C}^2 / \mathrm{N}-\mathrm{m}^2\right]$ $V=\frac{9 \times 10^9 \times 10^{-1} \times 10^{-6}}{5 \times 10^{-2}}=9 \times 2 \times 10^3 \times 10^2$ $V=18 \times 10^5 \mathrm{~N}-\mathrm{m}^2 / \mathrm{C}^2$ Due to all six charges, total potential at the centre of hexagonal is $V_{\text {total }}=6 \times\left(V_1\right)$ Now, from Eq. (i), we get $V_{\text {lotal }}=6 \times 18 \times 10^5 \Rightarrow V_{\text {lotal }}=1.08 \times 10^7 \mathrm{~V}$

A regular hexagon of side \(5 \mathrm{~cm}\) has a charge \(10 \mu \mathrm{C}\) at each of its vertices. The…

Solution