1.1 Dimensional analysis  (Page 3/7)

Problem : Find the dimensional formula of thermal conductivity.

Solution : The thermal heat of conduction (Q) is given as :

$$Q=\frac{kA\left({\theta }_2-{\theta }_1\right)t}{d}$$

where “k” is thermal conductivity, “Q” is heat, “A” is area of cross-section, “θ” is temperature and “t” is time.

$$\Rightarrow \left[k\right]=\frac{\left[Qd\right]}{\left[A\left({\theta }_2-{\theta }_1\right)t\right]}$$

We make use of two facts (i) heat is energy and (ii) dimension of difference of temperature is same as that of temperature.

$$\Rightarrow \left[k\right]=\frac{\left[M{L}^2{T}^{-2}\right]\left[L\right]}{\left[L^{2}KT\right]}$$

$$\left[k\right]=\left[ML{T}^{-3}{K}^{-1}\right]$$

Problem : Find the dimension of expression,

$$\frac{CV}{{\epsilon }_0\rho }$$

where “C” is capacitance, “V” is potential difference, “ ${\epsilon }_0$ ” is permittivity of vacuum and “ρ” is specific resistance.

Solution :

By inspection of expression, we observe that capacitance of parallel plate capacitor is given by the expression containing “ ${\epsilon }_0$ ”. This may enable us to cancel “ ${\epsilon }_0$ ” from the ratio. The capacitance is given as :

$$C=\frac{{\epsilon }_{0}A}{d}$$

On the other hand, specific resistance,

$$\rho =\frac{RA}{L}$$

Hence,

$$\Rightarrow \left[\frac{CV}{{\epsilon }_0\rho }\right]=\frac{\left[{\epsilon }_{0}ALV\right]}{\left[{\epsilon }_{0}dRA\right]}$$

Applying Ohm’s law,

$$\Rightarrow \left[\frac{CV}{{\epsilon }_0\rho }\right]=\frac{\left[{\epsilon }_{0}ALV\right]}{\left[{\epsilon }_{0}LRA\right]}=\frac{\left[V\right]}{\left[R\right]}$$

Applying Ohm's law,

$$\Rightarrow \left[\frac{CV}{{\epsilon }_0\rho }\right]=\frac{\left[V\right]}{\left[R\right]}=\left[A\right]$$

Problem : Find the dimension of term given as :

$$\frac{E^2}{{\mu }_0}$$

where “E” is electric field and “ ${\mu }_0$ ” is absolute permeability of vacuum.

Solution : We can approach the problem by evaluating each of the quantities in the expression terms of basic quantities. Alternatively, however, we can evaluate the dimension of the given ratio by expressing the given expression in terms of physical quantities, whose dimensions can be easily found out. For illustration purpose, we shall use the second approach.

Here we can write given expression in the following manner :

$$\frac{E^2}{{\mu }_0}=\frac{{\epsilon }_0{E}^2}{{\epsilon }_0{\mu }_0}$$

where “ ${\epsilon }_0$ ” is absolute permittivity of vacuum.

However, we know that electric energy density of a medium is $\frac{1}{2}{\epsilon }_0{E}^2$ . It means that numerator has the same dimension as “energy/volume”. On the other hand, speed of light in vacuum is given as :

$$c=\frac{1}{\sqrt{\left({\epsilon }_0{\mu }_0\right)}}$$

Thus, dimension of denominator is same as that of $\frac{1}{c^2}$ .

Therefore,

$$\Rightarrow \frac{E^2}{{\mu }_0}=\frac{\left[\mathrm{energy/volume}\right]}{\left[\frac{1}{c^2}\right]}$$

$$\Rightarrow \frac{E^2}{{\mu }_0}=\left[\frac{M{L}^2{T}^{-2}}{L^3}\right][L{T}^{-1}{]}^2=[ML{T}^{-4}]$$