Animated Solution for Physics - Physics and Measurement: The dimensions of $\frac{1}{2}{\epsilon }_0{E}^2$ (${\epsilon }_0$ : permittivity of free space; $E$ : electric field) is

Have you ever looked at a complex physics expression and felt a sudden urge to calculate everything from scratch? It happens to the best of us. When faced with finding the dimensions of a composite term like $\frac{1}{2}{\epsilon }_0{E}^2$, the instinct is often to break it down into its absolute fundamental parts. But what if I told you there is a much more elegant, almost magical shortcut?

Let's embark on a journey to decode this expression, and along the way, we'll uncover a fascinating truth about competitive exams that every JEE aspirant must know.

In physics, expressions are rarely just random collections of symbols. They tell a story about the universe. The term $\frac{1}{2}{\epsilon }_0{E}^2$ is a prime example.

Imagine a parallel plate capacitor, fully charged, with an electric field $\vec{E}$ established between its plates. The space between those plates isn't just empty void; it is brimming with stored electrical energy. The expression $\frac{1}{2}{\epsilon }_0{E}^2$ represents exactly this: the Energy Density of the electric field.

Energy Density

By definition, energy density is the amount of energy stored per unit volume.

This simple realization is our master key. Instead of painstakingly deriving the dimensional formula for the permittivity of free space (${\epsilon }_0$) and the electric field ($E$), squaring one, multiplying them, and hoping we didn't drop a minus sign, we can just use the dimensions of energy and volume!

Let's recall the fundamental dimensions we need. Energy is equivalent to work done (Force $\times$ Displacement), so its dimensional formula is:

Volume is simply length cubed:

Now, we substitute these into our energy density relation. Remember, the constant $\frac{1}{2}$ is a pure number and is completely dimensionless.

Bringing the ${\text{L}}^3$ to the numerator, the powers of length subtract ($2-3=-1$):

And there we have it! A clean, elegant derivation in just a few seconds.

Trusting Yourself

Now, look back at the options provided in the original question: (a) $[{\text{MLT}}^{-1}]$ (b) $[{\text{ML}}^2{\text{T}}^{-2}]$ (c) $[{\text{MLT}}^{-2}]$ (d) $[{\text{ML}}^2{\text{T}}^{-1}]$

Do you see our answer? No. None of the options match $[{\text{ML}}^{-1}{\text{T}}^{-2}]$.

This brings us to a crucial lesson for competitive exams like JEE. Sometimes, the question itself is flawed. There will be times when all the given options are incorrect. In such moments, panic is your worst enemy.

If you have solved a problem using a solid conceptual foundation and double-checked your math, trust yourself. Do not waste precious exam minutes trying to force your correct answer to match an incorrect option. Mark it for review, move on, and know that such questions are typically awarded as bonus marks to everyone.

Physics is about truth, not just ticking boxes. Keep your concepts strong, and you'll navigate any exam with confidence!