The dimensional formula of `(1)/(2) epsilon_(0) E^2` is `(E = "electric field")`

Options

A. `[ M L T^(-1) ]`
B. `[ M L^(2) T^(-2) ]`
C. `[ M L^(-1) T^(-2) ]`
D. `[ M L^(2) T^(-1) ]`

Answer

To find the dimensional formula of \(\frac{1}{2} \epsilon_0 E^2\), we will break down the components involved: \(\epsilon_0\) (the permittivity of free space) and \(E\) (the electric field).

### Step 1: Dimensional Formula of Electric Field \(E\)

The electric field \(E\) is defined as the force \(F\) experienced by a charge \(q\) per unit charge:

\[ E = \frac{F}{q} \]

**Dimensional Formula of Force \(F\)**:
\[ F = m \cdot a \quad \text{(where \(m\) is mass and \(a\) is acceleration)} \]
The dimensional formula for mass \(m\) is \([M]\) and for acceleration \(a\) is \([L T^{-2}]\). Thus, the dimensional formula for force \(F\) is:
\[ [F] = [M][L T^{-2}] = [M L T^{-2}] \]

**Dimensional Formula of Charge \(q\)**:
The dimensional formula for charge is not directly defined, but we can express it in terms of electric field. However, for our purpose, we will assume the dimensional formula for charge is \([Q]\).

Now substituting back into the formula for electric field:

\[ [E] = \frac{[F]}{[q]} = \frac{[M L T^{-2}]}{[Q]} = [M L T^{-2} Q^{-1}] \]

### Step 2: Dimensional Formula of \(\epsilon_0\)

The permittivity of free space \(\epsilon_0\) can be derived from Coulomb's law, which states:

\[ F = \frac{q_1 q_2}{4 \pi \epsilon_0 r^2} \]

Rearranging gives:

\[ \epsilon_0 = \frac{q_1 q_2}{4 \pi F r^2} \]

Using the dimensional formulas we have:

- Dimensional formula of \(q^2\) is \([Q^2]\)
- Dimensional formula of \(r^2\) is \([L^2]\)
- Dimensional formula of \(F\) is \([M L T^{-2}]\)

Thus, we can write:

\[ [\epsilon_0] = \frac{[Q^2]}{[M L T^{-2}] [L^2]} = \frac{[Q^2]}{[M L^3 T^{-2}]} \]

This simplifies to:

\[ [\epsilon_0] = [M^{-1} L^{-3} T^{2} Q^{2}] \]

### Step 3: Dimensional Formula of \(\epsilon_0 E^2\)

Now we can find the dimensional formula of \(\epsilon_0 E^2\):

\[ [E^2] = \left([M L T^{-2} Q^{-1}]\right)^2 = [M^2 L^2 T^{-4} Q^{-2}] \]

Now substituting back into \(\epsilon_0 E^2\):

\[ [\epsilon_0 E^2] = [M^{-1} L^{-3} T^{2} Q^{2}] \cdot [M^2 L^2 T^{-4} Q^{-2}] \]

Multiplying these gives:

\[ [\epsilon_0 E^2] = [M^{1} L^{-1} T^{-2}] \]

### Step 4: Dimensional Formula of \(\frac{1}{2} \epsilon_0 E^2\)

Since the numerical coefficient \(\frac{1}{2}\) is dimensionless, the dimensional formula of \(\frac{1}{2} \epsilon_0 E^2\) remains the same:

\[ \left[\frac{1}{2} \epsilon_0 E^2\right] = [M L^{-1} T^{-2}] \]

### Final Answer:
The dimensional formula of \(\frac{1}{2} \epsilon_0 E^2\) is:

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