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The dimensional formula of \(\frac{1}{2}{\epsilon }_{0}{E}^{2}\left({\epsilon }_{0}=\right.\) permittivity of vacuum and…

Q1

The dimensional formula of \(\frac{1}{2}{\epsilon }_{0}{E}^{2}\left({\epsilon }_{0}=\right.\) permittivity of vacuum and E = electric field) is \({M}^{a}{L}^{b}{T}^{c}.\) The value of 2 a - b + c =_________

[JEE Main 2026, 2 Apr (Shift 1)]

a

0

b

1

c

-1

d

2

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