Question:medium
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
- \(0\)
- \(1\)
- \(-1\)
- \(2\)
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The Correct Option is A
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