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

2.7 × 10⁶ V

7.2 × 10¹¹ V

2.5 × 10¹² V

3.4 × 10⁴ V

Solution

2.7 × 10⁶ V

As shown in the figure, 0 is the centre of hexagon ABCDEF of each side 8 cm. As it is a regular hexagon OAB, OBC, etc are equilateral triangles.

OA = OB = OC = OD = OE = OF = 8 cm

= 8 x 10^-2 m

∴ V_O = V_A + V_B + V_C + V_D + V_E + V_F

V_O = $\frac{1}{4 {πε}_{0}} [\frac{q}{OA} + \frac{q}{OB} + \frac{q}{OC} + \frac{q}{OD} + \frac{q}{OE} + \frac{q}{OF}]$

$[∵ V = \frac{1}{4 π ε_{0}} \frac{q}{r}]$

V_O = $9 \times 10^{9} [\frac{4 \times 10^{- 6}}{8 \times 10^{- 2}} + \frac{4 \times 10^{- 6}}{8 \times 10^{- 2}} + \frac{4 \times 10^{- 6}}{8 \times 10^{- 2}} + \frac{4 \times 10^{- 6}}{8 \times 10^{- 2}} + \frac{4 \times 10^{- 6}}{8 \times 10^{- 2}} + \frac{4 \times 10^{- 6}}{8 \times 10^{- 2}}]$

V_O = 9 × 10⁹ × $\frac{4 \times 10^{- 6}}{8 \times 10^{- 2}} [1 + 1 + 1 + 1 + 1 + 1]$

V_O = 9 × 10⁹× $\frac{6 \times 4 \times 10^{- 6}}{8 \times 10^{- 2}}$

where $\frac{1}{4 π ε_{0}}$ = 9 × 10⁹

V_O = 2.7 × 10⁶ V

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Electrostatic Potential And Capacitance

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

2.7 × 10⁶ V

7.2 × 10¹¹ V

2.5 × 10¹² V

3.4 × 10⁴ V

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

2.7 × 10⁶ V

7.2 × 10¹¹ V

2.5 × 10¹² V

3.4 × 10⁴ V