Q137 / 160 · AP EAMCET 2020 (17 Sep Shift 2) · Physics

Dimensions of \(\varepsilon_0\) are

  • Correct answer
    A
    \(\left[M^{-1} L^{-3} T^4 A^2\right]\)
  • B
    \(\left[M^0 L^{-3} T^3 A^3\right]\)
  • C
    \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^3 \mathrm{~A}\right]\)
  • D
    \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T} \mathrm{~A}^2\right]\)

Solution

We know that, force between two charges \(q_1\) and \(q_2\) placed at distance \(r\) is given as \(\begin{aligned} F & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2} \\ \Rightarrow \quad \varepsilon_0 & =\frac{q_1 q_2}{4 \pi F r^2} \Rightarrow\left[\varepsilon_0\right]=\frac{\left[q_1\right]\left[q_2\right]}{[F]\left[r^2\right]} \\ & =\frac{[\mathrm{AT}][\mathrm{AT}]}{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^2\right]}=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right] \end{aligned}\)

Hence, the correct answer is option: 'A'.

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