Question

If $C$ be the capacitance and $V$ be the electric potential, then the dimensional formula of $C{V}^2$ is

• A. $\left[M{L}^{-3}TA\right]$
• B. $\left[M^{0}L{T}^{-2}{A}^0\right]$
• C. $\left[M{L}^1{T}^{-2}{A}^{-1}\right]$
• D. $\left[M{L}^2{T}^{-2}{A}^0\right]$

Answer 1

The correct option is D

$\left[M{L}^2{T}^{-2}{A}^0\right]$

Step 1: Given data

Capacitance$\to C$

Electric potential$\to V$

We have to find the dimensional formula of $C{V}^2$.

Step 2: Formula used and solution

We know $C=\frac{Q}{V}$

$$\begin{gathered} C{V}^2=\left(\frac{Q}{V}\right)V^2\left(\because C=\frac{Q}{V}\right) \\ \Rightarrow C{V}^2=QV \\ WeknowthatdimensionalformulaofQ=\left[AT\right]andV=\left[M{L}^2{T}^{-3}{A}^{-1}\right] \\ Therefore,C{V}^2=\left[AT\right]\left[M{L}^2{T}^{-3}{A}^{-1}\right] \\ C{V}^2=\left[M{L}^2{T}^{-2}{A}^0\right] \end{gathered}$$

Dimensional formula of $C{V}^2$ is $\left[M{L}^2{T}^{-2}{A}^0\right]$

Hence, option D is correct.