Question
Physics Question on Dimensional Analysis
If ϵ0 is the permittivity of free space and E is the electric field, then ϵ0E2 has the dimensions:
A
[M0L−2TA]
B
[ML−1T−2]
C
[M−1L−3T4A2]
D
[ML2T−2]
Answer
[ML−1T−2]
Explanation
Solution
The electric field is given by:
E=R2KQ.
Substituting K=4πϵ01, we get:
E=4πϵ0R2Q.
From this, the permittivity of free space (ϵ0) can be expressed as:
ϵ0=4πR2EQ.
Now, calculate ϵ0E2:
ϵ0E2=4πR2EQ⋅E2=4πR2QE.
Analyzing the dimensional formula:
[ϵ0E2]=[R2][Q][E].
Substituting the dimensional formulas:
[Q]=[W],[E]=[R2][Q][W].
[ϵ0E2]=[R3][W]=L3ML2T−2.
Simplifying:
[ϵ0E2]=[ML−1T−2].
Thus, the dimensions of ϵ0E2 are [ML−1T−2].