Question

The dimensions of $\frac{QV}{C}$ is,($Q$=charge,$V$=voltage,$C$=capacitance,take dimension of current as $k$)

• A. $M^2{L}^{-4}{T}^{-6}{k}^{-1}$
• B. $M^2{L}^4{T}^{-6}{k}^{-2}$
• C. $M^2{L}^{-4}{T}^6{k}^{-2}$
• D. $M^2{L}^4{T}^{-6}{k}^1$

Answer 1

The correct option is B $M^2{L}^4{T}^{-6}{k}^{-2}$

We know that, $Q=CV$ so $\frac{QV}{C}=\frac{\left(CV\right)V}{C}=V^2$

So we have to find out only dimension of $V^2$

Volatge $V=Ed=\frac{F}{Q}d$

We know that dimension of force, $\left[F\right]=\left[ML{T}^{-2}\right]$, distance $\left[d\right]=\left[L\right]$ and charge $\left[Q\right]=\left[kT\right]$ (as charge $=$ current$\times$ time)

Now, $\left[V\right]=\frac{\left[ML{T}^{-2}\right]}{\left[kT\right]}\left[L\right]=\left[M{L}^2{T}^{-3}{k}^{-1}\right]$

Thus, $\left[V^2\right]=[M^2{L}^4{T}^{-6}{k}^{-2}$]