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)

$2.7 \times 10^{6}V$

Answer 2

:

As shown in the figure, O is the centre of hexagon ABCDEF of each side 8 cm. As it a regular hexagon OAB, OBC, etc are equilateral triangles.

$\therefore$OA = OB = OC = OD = OE = OF = 8 cm = $8 \times 10^{- 2}m$

The potential at O is

$V = 6 \times \frac{q}{4\pi\varepsilon_{0}r} = \frac{6 \times 9 \times 10^{9} \times 4 \times 10^{- 6}}{8 \times 10^{- 2}} = 2.7 \times 10^{6}V$