Question

The quantity $X$ is given by ${\epsilon }_{o}L\frac{\Delta V}{\Delta T}$, where ${\epsilon }_o$ is the permittivity of free space, $L$ is a length, $\Delta V$ is a potential difference and $\Delta T$ is a time interval. The dimensional formula for $X$ is the same as that of

• A. resistance
• B. charge
• C. voltage
• D. current

Answer 1

The correct option is D current

We know than, from coulomb's law

$E_0=\frac{1}{4\pi }\frac{Q_1{Q}_2}{F{R}^2}$

$\Rightarrow \left[E_0\right]=\frac{\left[Q_1\right]\left[Q_2\right]}{\left[F\right]\left[R^2\right]}$

$\Rightarrow \left[E_0\right]=\frac{A^2{T}^2}{mL{T}^{-2}{L}^2}=\left[\frac{A^2{T}^2}{mL{T}^3{L}^{-2}}\right]$

$V=-\int \overrightarrow{E}.\overrightarrow{d\pi }$

$\left[V\right]=\left[E\right]\left[L\right]$

$\left[V\right]=mL{T}^{-2}{A}^{-1}{T}^{-1}L=\left[m{L}^2{A}^{-1}{T}^{-3}\right]$

$\left[T\right]=T$

$\left[L\right]=L$

$X={ϵ}_{0}L\frac{\Delta V}{\Delta T}$

$\left[X\right]=\left[\frac{A^2{T}^2}{mL{T}^3{L}^{-2}}\right]\left[L\right]\left[\frac{m{L}^2{A}^{-1}{T}^{-3}}{T}\right]$

$\left[X\right]=\left[A\right]$

$\left[A\right]$ is the dimension of current.