JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2024
- Medium
- MCQ
If \(\varepsilon_0\) is the permittivity of free space and \(E\) is the electric field, then \(\varepsilon_0 E^2\) has the dimensions is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\)
Step-by-step Solution
Detailed explanation
\(\mathrm{E}=\frac{\mathrm{K}}{\mathrm{R}^2}\)
\(\mathrm{E}=\frac{\mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}^2}\)
\(\varepsilon_0=\frac{\mathrm{Q}}{4 \pi \mathrm{R}^2 \mathrm{E}}\)
\(\text { Now, } \varepsilon_0 \mathrm{E}^2=\frac{\mathrm{Q}}{4 \pi \mathrm{R}^2\mathrm{E}} \cdot \mathrm{E}^2=\frac{\mathrm{Q}}{4 \pi \mathrm{R}^2} \cdot \mathrm{E}\)
\({\left[\varepsilon_0 \mathrm{E}^2\right]=\left[\frac{\mathrm{QE}}{\mathrm{R}^2}\right]=\frac{[\mathrm{Q}][\mathrm{E}]}{\left[\mathrm{R}^2\right]}=\frac{[\mathrm{Q}]}{\left[\mathrm{R}{ }^2\right][\mathrm{Q}][\mathrm{R}]}}\)
\(=\frac{[\mathrm{W}]}{\left[\mathrm{R}^3\right]}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~L}^3}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}\)
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