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A regular hexagon of side $10 \mathrm{~cm}$ has a charge $5 \mu \mathrm{C}$ at each of its vertices. Calculate the potential at the centre of the hexagon.
Solution:
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Verified Answer
Geometrically each vertex is at a distance of $10 \mathrm{~cm}$ (length of side) from centre of the hexagon.
So, for all charges,
$$
\begin{aligned}
&\mathrm{r}=10 \mathrm{~cm}=0.1 \mathrm{~m}, \\
&\mathrm{q}=5 \mu \mathrm{C}=5 \times 10^{-6} \mathrm{C}, \mathrm{n}=6, \mathrm{~V}=?
\end{aligned}
$$
By formula, $\mathrm{V}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{nq}}{\mathrm{r}}$
$$
\mathrm{V}=\frac{9 \times 10^9 \times 6 \times 5 \times 10^{-6}}{0.1}=2.7 \times 10^6 \mathrm{~V}
$$
So, for all charges,
$$
\begin{aligned}
&\mathrm{r}=10 \mathrm{~cm}=0.1 \mathrm{~m}, \\
&\mathrm{q}=5 \mu \mathrm{C}=5 \times 10^{-6} \mathrm{C}, \mathrm{n}=6, \mathrm{~V}=?
\end{aligned}
$$
By formula, $\mathrm{V}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{nq}}{\mathrm{r}}$
$$
\mathrm{V}=\frac{9 \times 10^9 \times 6 \times 5 \times 10^{-6}}{0.1}=2.7 \times 10^6 \mathrm{~V}
$$
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