Question

The dimension of $\frac{1}{2} \epsilon_0 E^2$ is $M^a L^b T^c$ then value of $a - 2b + c = ?$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is A

To find the dimensional formula of the given expression \(\frac{1}{2} \epsilon_0 E^2\), we need to analyze each component based on their fundamental dimensions. This will help us deduce the expression's overall dimensional formula and find the value of \(a - 2b + c\).

1. The electric field \(E\) is defined as force per unit charge, i.e., \(E = \frac{F}{q}\).
   • The dimension of force \(F\) is \([M L T^{-2}]\).
   • The dimension of charge \(q\) is \([I T]\) (current \(I\) times time \(T\)).
2. The permittivity of free space \(\epsilon_0\) has the dimension \([M^{-1} L^{-3} T^4 I^2]\).
3. Thus, the dimension of \(\epsilon_0 E^2\) is:
   • \(\epsilon_0: [M^{-1} L^{-3} T^4 I^2]\)
   • \(E^2: ([M L I^{-1} T^{-3}])^2 = [M^2 L^2 I^{-2} T^{-6}]\)
4. Now, comparing \([M L^{-1} T^{-2}]\) with \(M^a L^b T^c\) gives:
   • \(a = 1\)
   • \(b = -1\)
   • \(c = -2\)
5. Finally, calculate \(a - 2b + c\):
   • \(a - 2b + c = 1 - 2(-1) - 2 = 1 + 2 - 2 = 1\)

Therefore, the value of \(a - 2b + c\) is 1. The correct answer is option 1.