Potential is an essential concept of mechanics. It can be defined for any force field like a gravitational field, electrostatic field, etc. In this note of the Dimension formula for Electric Potential, we will discuss Potential for the electrostatic field.
Electric Potential is nothing but the work required to bring a charged particle from infinity to the point of reference. We must keep in mind that the acceleration of that point charge should be zero. When a positive direction is brought into a field, a repulsion occurs. The work done against that repulsive force gets stored as potential energy.
A point continually defines Electric Potential. To that point, the amount of work is required to bring a positive unit charge from infinity (where the electric field is considered to have zero influence). This is called the Potential of that point. The Potential of the point w.r.t, which is measured, is taken as zero.
For a point charge, the potential is defined as follows-
V = k. q/r
Electric Potential = Work Done / Unit Charge
Electric Potential in SI units:
V = W/q = Joules/ coulomb = Volts
Electric Potential is thus measured in Volts or Voltage.
1 Volt = 1 Joule/ 1 coulomb
One volt means, for one coulomb of electricity to move, one joule of work must be performed.
Electric Potential is a scalar quantity as we only measure the amount of work done. No directional property is associated with it.
The dimensional formula of any bodily amount is defined as the expression that represents how and which of the bottom portions are protected in that amount. It is denoted through enclosing the symbols for base portions with suitable strength in rectangular brackets, i.e. [].
An example is the dimension formula of mass which is given as [M].
According to the above definition, we can define electric Potential as:
V= W/q
q = electric charge
W = work done
So, dimension of electric potential= dimension of work done / dimension of electric charge.
W= F.S
So, the dimension of W = [MLT-2]x[L] = [ML2T-2]
So, the dimension of V = [ML2T-2]/ [IT] = [MI-1L2T-3]
So, the dimension of Electric potential is [MI-1L2T-3]
By the superposition principle, the Electric Potential for point charges can be defined as:
V = Kq/ r
Where: V is the electric potential produced by a point charge with a charge of magnitude Q; r is the distance from the point charge, and k is a constant with a value of 8.99 x 109 N m2/C2.
If three charges q1, q2, and q3 are arranged at the vertices of a triangle, the expected energy of the framework is:
U =U12 + U23 + U31 = (1/4πε0) × [q1q2/d1 + q2q3/d2 + q3q1/d3]
If four charges q1, q2, q3, and q4 are arranged at the sides of a square, the likely electric energy of the framework is:
U = (1/4πε0) × [(q1q2/d) + (q2q3/d) + (q3q4/d) + (q4q1/d) + (q4q2/√2d) + (q3q1/√2d)]
In the field of a charge Q, assuming a charge q is moved against the electric field from a distance ‘a’ to a distance ‘b’ from Q, the work done is given by:
W = (Vb – Va) × q
= [(1/4πε0) × (Qq/b)] – [(1/4πε0) × (Qq/a)]
= (Qq/4πε0)[1/b – 1/a]
= (Qq/4πε0)[(a-b)/ab]
In a physical relation, the dimensions are examined through dimensional analysis. These analyses can be used in conversion, correction, and derivation.
It determines the dimensional consistency, homogeneity, and accuracy of the mathematical expressions.
In these notes of the Dimension formula for Electric Potential, we learned how to deduce a dimensional formula for Electric Potential and some basic concepts regarding this.
V = kQ/r is the electric potential of a point charge. The difference between a scalar and a vector is the electric Potential. The total electric potential is obtained by adding the voltages as numbers, whereas the entire electric field is obtained by adding the individual areas as vectors. I hope now you have all the necessary information related to this topic. For better understanding, you must go through this topic thoroughly.
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