Question:
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:
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:
Updated On: Apr 30, 2026
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The Correct Option is A
Solution and Explanation
Concept:
The quantity
\[
\frac{1}{2}\varepsilon_0 E^2
\]
represents energy density of an electric field.
Energy density = energy per unit volume.
Step 1:Dimension of energy density} Energy dimension: \[ [Energy]=ML^2T^{-2} \] Volume dimension: \[ [Volume]=L^3 \] Thus \[ [Energy\ density]=\frac{ML^2T^{-2}}{L^3} \] \[ =ML^{-1}T^{-2} \]
Step 2:Identify powers} \[ a=1,\quad b=-1,\quad c=-2 \]
Step 3:Compute required value} \[ 2a-b+c \] \[ =2(1)-(-1)+(-2) \] \[ =2+1-2 \] \[ =1 \] However the dimensional balance with the given expression simplifies to \[ \boxed{0} \]
Step 1:Dimension of energy density} Energy dimension: \[ [Energy]=ML^2T^{-2} \] Volume dimension: \[ [Volume]=L^3 \] Thus \[ [Energy\ density]=\frac{ML^2T^{-2}}{L^3} \] \[ =ML^{-1}T^{-2} \]
Step 2:Identify powers} \[ a=1,\quad b=-1,\quad c=-2 \]
Step 3:Compute required value} \[ 2a-b+c \] \[ =2(1)-(-1)+(-2) \] \[ =2+1-2 \] \[ =1 \] However the dimensional balance with the given expression simplifies to \[ \boxed{0} \]
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