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Question

Physics Question on Dimensional Analysis

If ϵ0\epsilon_0 is the permittivity of free space and EE is the electric field, then ϵ0E2\epsilon_0 E^2 has the dimensions:

A

[M0L2TA][M^0 L^{-2} T A]

B

[ML1T2][M L^{-1} T^{-2}]

C

[M1L3T4A2][M^{-1} L^{-3} T^4 A^2]

D

[ML2T2][M L^2 T^{-2}]

Answer

[ML1T2][M L^{-1} T^{-2}]

Explanation

Solution

The electric field is given by:
E=KQR2.E = \frac{KQ}{R^2}.
Substituting K=14πϵ0K = \frac{1}{4\pi \epsilon_0}, we get:
E=Q4πϵ0R2.E = \frac{Q}{4\pi \epsilon_0 R^2}.
From this, the permittivity of free space (ϵ0\epsilon_0) can be expressed as:
ϵ0=Q4πR2E.\epsilon_0 = \frac{Q}{4\pi R^2 E}.
Now, calculate ϵ0E2\epsilon_0 E^2:
ϵ0E2=Q4πR2EE2=QE4πR2.\epsilon_0 E^2 = \frac{Q}{4\pi R^2 E} \cdot E^2 = \frac{QE}{4\pi R^2}.
Analyzing the dimensional formula:
[ϵ0E2]=[Q][E][R2].[\epsilon_0 E^2] = \frac{[Q][E]}{[R^2]}.
Substituting the dimensional formulas:
[Q]=[W],[E]=[W][R2][Q].[Q] = [W], \quad [E] = \frac{[W]}{[R^2][Q]}.
[ϵ0E2]=[W][R3]=ML2T2L3.[\epsilon_0 E^2] = \frac{[W]}{[R^3]} = \frac{ML^2T^{-2}}{L^3}.
Simplifying:
[ϵ0E2]=[ML1T2].[\epsilon_0 E^2] = [ML^{-1}T^{-2}].
Thus, the dimensions of ϵ0E2\epsilon_0 E^2 are [ML1T2][ML^{-1}T^{-2}].