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Asked by Dinesh Last Modified
Answer 
Apoorva Purohit
The given figure shows six equal amount of charges, q, at the vertices of a regular hexagon.
Where,
Charge, q = 5 µC = 5 × 10−6 C
Side of the hexagon, l = AB = BC = CD = DE = EF = FA = 10 cm
Distance of each vertex from centre O, d = 10 cm
Electric potential at point O,
Where,
= Permittivity of free space
Therefore, the potential at the centre of the hexagon is 2.7 × 106 V.
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A regular hexagon ABCDEF of side 10cm has a charge 5 μC at each of its vertices as shown in the figure.
Potential due to charge placed at A, at the centre is Kq/r
Potential due to charge placed at B, at the centre is Kq/r
Potential due to Charge placed at F , at centre is Kq/r
so, total potential at centre of regular hexagon , V = V₁ + V₂ + V₃ + V₄ + V₅ + V₆
so, V = kq/r + kq/r + kq/r + kq/r + kq/r + Kq/r
V = 6kq/r
Hence, k = 9 × 10⁹Nm²/C² , q = 5 × 10⁻⁶C and r = 10cm = 0.1 m
Now, V = 6 × 9 × 10⁹ × 5 × 10⁻⁶/0.1 = 2.7 × 10⁶ Volts.
The given figure shows six equal amount of charges, q, at the vertices of a regular hexagon.
Where, Charge, q = 5 µC = 5 × 10- 6 C
Side of the hexagon, l = AB = BC = CD = DE = EF = FA = 10 cm
Distance of each vertex from centre O, d = 10 cm
Electric potential at point O,
Where,
= Permittivity of free space
Therefore, the potential at the centre of the hexagon is 2.7 × 106 V.
read less
The given figure shows six equal amount of charges, q, at the vertices of a regular hexagon.
Where, Charge, q = 5 µC = 5 × 10- 6 C
Side of the hexagon, l = AB = BC = CD = DE = EF = FA = 10 cm
Distance of each vertex from centre O, d = 10 cm
Electric potential at point O,
Where,
= Permittivity of free space
Therefore, the potential at the centre of the hexagon is 2.7 × 106 V.
read lessView 4 more Answers
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