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प्रश्न
A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon.
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उत्तर
The given figure shows six equal amounts 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,
`V = (6 xx q)/(4pi ε_0 d)`
Where,
`ε_0` = Permittivity of free space
`1/(4pi ε_0) = 9 xx 10^9` N C−2 m−2
∴ `V = (6 xx 9 xx 10^9 xx 5 xx 10^-6)/0.1`
= 270 × 103
= 2.7 × 106 V
Therefore, the potential at the centre of the hexagon is 2.7 × 106 V.
संबंधित प्रश्न
Define an equipotential surface.
Two charges 2 μC and −2 µC are placed at points A and B 6 cm apart.
- Identify an equipotential surface of the system.
- What is the direction of the electric field at every point on this surface?
What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning?
(Hint: The earth has an electric field of about 100 Vm−1 at its surface in the downward direction, corresponding to a surface charge density = −10−9 C m−2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)
What is the geometrical shape of equipotential surfaces due to a single isolated charge?
Why is there no work done in moving a charge from one point to another on an equipotential surface?
Define equipotential surface.
Two identical point charges, q each, are kept 2m apart in the air. A third point charge Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q.
Statement - 1: For practical purpose, the earth is used as a reference at zero potential in electrical circuits.
Statement - 2: The electrical potential of a sphere of radius R with charge Q uniformly distributed on the surface is given by `Q/(4piepsilon_0R)`.
A particle of mass 'm' having charge 'q' is held at rest in uniform electric field of intensity 'E'. When it is released, the kinetic energy attained by it after covering a distance 'y' will be ______.
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Equipotential surfaces ______.
Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.
Find the equation of the equipotentials for an infinite cylinder of radius r0, carrying charge of linear density λ.
Equipotential surfaces are shown in figure. Then the electric field strength will be ______.
What is an equipotential surface?
A curve connecting points that all have equal electric potential is best described as:
Equipotential surfaces are denser (closer together) in regions where:
The direction in which electric potential decreases is:
The surface of any conductor in electrostatic equilibrium is an equipotential surface because:
