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.
Solution and Explanation
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, \(v=\frac{1}{4πe_0}.\frac{6xq}{d}\)
Where,
Where, = Permittivity of free space and \(\frac{1}{4pi_0}\)= 9 ×10-9 Nm2 c-2
\(v=\frac{9x10^9x6x5x10^-6}{0.1}=2.7x10^6V\)
Therefore, the potential at the centre of the hexagon is 2.7 × 106 V
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- Find the inverse of each of the matrices,if it exists \(\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}\)
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What is the Planning Process?
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Concepts Used:
Electrostatic Potential and Capacitance
Electrostatic Potential
The potential of a point is defined as the work done per unit charge that results in bringing a charge from infinity to a certain point.
Some major things that we should know about electric potential:
- They are denoted by V and are a scalar quantity.
- It is measured in volts.
Capacitance
The ability of a capacitor of holding the energy in form of an electric charge is defined as capacitance. Similarly, we can also say that capacitance is the storing ability of capacitors, and the unit in which they are measured is “farads”.
Read More: Electrostatic Potential and Capacitance
The capacitor is in Series and in Parallel as defined below;
In Series
Both the Capacitors C1 and C2 can easily get connected in series. When the capacitors are connected in series then the total capacitance that is Ctotal is less than any one of the capacitor’s capacitance.
In Parallel
Both Capacitor C1 and C2 are connected in parallel. When the capacitors are connected parallelly then the total capacitance that is Ctotal is any one of the capacitor’s capacitance.
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