
Show the derivation to find the dimensional formula of electric potential and choose the correct answer.
A. $[ML^2 T^{-3}A^{-1}]$
B. $[ML^2 T^{-2}A^{-1}]$
C. $[ML^2 T^{-1}A^{-1}]$
D. $[ML^2 T^{-2}]$
Answer
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Hint: First, we will use the potential energy formula to find its dimensional formula and then we will find out the dimensional formula for the charge of a particle. And at last by dividing the dimensional formula of potential energy by that of charge, we will get the dimensional formula for the electric potential.
Complete step-by-step answer:
The dimensional notations used here for mass, length, time and current are M, L, T and A respectively.
We know that, Potential energy = Particle’s charge $\times$ Electric potential
Thus, we can write, electric potential = Potential energy $\times$ [Particle’s charge]$^{-1}$ …… (i)
Since, electric current is the electric charge per unit time,
i.e. charge = current $\times$ time.
Therefore, the dimensional formula for the charge = $[A^1 T^1]$ …… (ii)
Also, we know that potential energy = M $\times$ g $\times$ h …… (iii)
where the dimensions of height, h = $[M^0 L^1 T^0]$, and mass, m = $[M^1 L^0 T^0]$ …… (iv)
Similarly, the dimensional formula for the acceleration due to gravity, g = $[M^0 L^1 T^{-2}]$ …… (v)
Now, let us substitute equations (iv) and (v) in equation (iii), that will give the dimensional formula for the potential energy = $[M^1 L^0 T^0] \times [M^0 L^1 T^{-2}] \times [M^0 L^1 T^0] = [M^1 L^2 T^{-2}]$ …… (vi)
Now, we will substitute the equations (vi) and (ii) in equation (i),
Thus, dimensional formula for the electric charge = $[M^1 L^2 T^{-2}] \times [A^1 T^1]^{-1}= [M^1 L^2 T^{-3} A^{-1}]$
So, the correct answer is “Option A”.
Note: The derivation of dimensional formulas of any quantity can be done in various ways but the main aim should be to break the formula into base units. Like in this question, the dimensional formula of potential energy can also be found out from the formula of energy produced in a spring block system.
Complete step-by-step answer:
The dimensional notations used here for mass, length, time and current are M, L, T and A respectively.
We know that, Potential energy = Particle’s charge $\times$ Electric potential
Thus, we can write, electric potential = Potential energy $\times$ [Particle’s charge]$^{-1}$ …… (i)
Since, electric current is the electric charge per unit time,
i.e. charge = current $\times$ time.
Therefore, the dimensional formula for the charge = $[A^1 T^1]$ …… (ii)
Also, we know that potential energy = M $\times$ g $\times$ h …… (iii)
where the dimensions of height, h = $[M^0 L^1 T^0]$, and mass, m = $[M^1 L^0 T^0]$ …… (iv)
Similarly, the dimensional formula for the acceleration due to gravity, g = $[M^0 L^1 T^{-2}]$ …… (v)
Now, let us substitute equations (iv) and (v) in equation (iii), that will give the dimensional formula for the potential energy = $[M^1 L^0 T^0] \times [M^0 L^1 T^{-2}] \times [M^0 L^1 T^0] = [M^1 L^2 T^{-2}]$ …… (vi)
Now, we will substitute the equations (vi) and (ii) in equation (i),
Thus, dimensional formula for the electric charge = $[M^1 L^2 T^{-2}] \times [A^1 T^1]^{-1}= [M^1 L^2 T^{-3} A^{-1}]$
So, the correct answer is “Option A”.
Note: The derivation of dimensional formulas of any quantity can be done in various ways but the main aim should be to break the formula into base units. Like in this question, the dimensional formula of potential energy can also be found out from the formula of energy produced in a spring block system.
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