Question

A hexagon of side 8 cm has a charge 4 $\mu C$ at each of its vertices. The potential at the centre of the hexagon is

• A. $2.7\times {10}^6$ V
• B. $7.2\times {10}^{11}$ V
• C. $2.5\times {10}^{12}$ V
• D. $3.4\times {10}^4$ V

Answer 1

Answer: (A)

$\text{Step 1: Calculating Distances}$

As shown in the figure, $O$ is the centre of hexagon $ABCDEF$ of each side $8cm$. As it is a regular hexagon $OAB$, $OBC$ etc are equilateral triangles.
$\therefore$ $OA=OB=OC=OD=OD=OF=8cm=8\times {10}^{-2}m$

$\text{Step 2 : Calculating Potential}$

Since, Distances of each charge from O is same, therefore Potential due to to all the charges at O would be same.

& Potential being a scalar quantity will be added at O due to each charge

$\therefore V_O=V_A+V_B+V_C+V_D+V_E+V_F=6V_A$

$\Rightarrow V_O=6\times \frac{q}{4\pi {\epsilon }_{0}r}=\frac{6\times 9\times {10}^9\times (4\times {10}^{-6}C)}{8\times {10}^{-2}m}=2.7\times {10}^{6}V$

Hence , Option(A) is correct.