The dimensional formula of \( \frac{1}{2}\varepsilon_0 E^2 \) (\(\varepsilon_0\) = permittivity of vacuum and \(E\) = electric field) is \(M^aL^bT^c\). The value of \(2a-b+c\) is:
Step 1: Understanding the Concept:
The expression \(\frac{1}{2} \epsilon_0 E^2\) represents the energy density (energy per unit volume) of an electric field in a vacuum.
By determining the dimensions of energy and volume, we can find the dimensional formula for energy density and extract the exponents \(a, b, \text{ and } c\). Step 2: Key Formula or Approach:
1. Energy Density (\(u\)) = \(\frac{\text{Energy}}{\text{Volume}}\).
2. Dimensional formula of Energy (\(E_{energy}\)) = \([ML^2 T^{-2}]\).
3. Dimensional formula of Volume (\(V\)) = \([L^3]\). Step 4: Detailed Explanation:
Calculate the dimensions of energy density:
\[ [u] = \frac{[ML^2 T^{-2}]}{[L^3]} = [M^1 L^{-1} T^{-2}] \]
Comparing this with the given form \(M^a L^b T^c\), we identify:
\(a = 1, b = -1, c = -2\).
Now, calculate the required value:
\[ 2a - b + c = 2(1) - (-1) + (-2) \]
\[ = 2 + 1 - 2 = 1 \] Step 4: Final Answer:
The value of \(2a - b + c\) is 1.
