JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement
If \(\varepsilon_0\) is the permittivity of free space and \(E\) is the electric field, then \(\varepsilon_0 E^2\) has the dimensions is _______.
- A \(\left[\mathrm{M}^0 \mathrm{~L}^{-2} \mathrm{TA}\right]\)
- B \(\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\)
- C \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]\)
- D \(\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-2}\right]\)
Answer & Solution
Correct Answer
(B) \(\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\)
Step-by-step Solution
Detailed explanation
\(\mathrm{E}=\frac{\mathrm{K}}{\mathrm{R}^2}\) \(\mathrm{E}=\frac{\mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}^2}\) \(\varepsilon_0=\frac{\mathrm{Q}}{4 \pi \mathrm{R}^2 \mathrm{E}}\)…
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