Question

If ε, ϕ and t stands for permittivity, electric flux and time respectively, then dimension of $\varepsilon \dfrac{d\phi }{dt}$is same as that of
A) speed
B) current
C) charge
D) potential difference

Answer 1

Here we have given three physical quantities permittivity ε, electric flux ϕ and time t and we have to find the dimension of $\varepsilon \dfrac{d\phi }{dt}$. So we can use the dimensional analysis and find the dimension of the $\varepsilon \dfrac{d\phi }{dt}$. First we will write the dimension of the quantities given and by using it we will find the dimension of $\varepsilon \dfrac{d\phi }{dt}$.

Complete step by step answer:
Permittivity is a measure of polarizability of an electric field in a dielectric material and its SI unit is Farad per metre (F/m) and its dimension is $\left[ {{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}} \right]$. Here M is the dimension for mass, L is the dimension of length, T is dimension for time and A is dimension for electric current.
Now electric flux is the amount of electric field flowing across the area and its SI unit is weber (Wb). Its dimension is $\left[ {{M}^{1}}{{L}^{3}}{{T}^{-3}}{{A}^{-1}} \right]$
We know that time is one of the fundamental physical quantities and its dimension is [T].
Now dϕ is the difference in electric flux, therefore it will have the same dimension as the dimension of the electric flux. Similarly, dt is the difference in time and it will have the same dimension as of time.
By using the dimension of permittivity, electric flux and time the dimension of $\varepsilon \dfrac{d\phi }{dt}$ can be given as

& \varepsilon \dfrac{d\phi }{dt}=\left[ {{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}} \right]\dfrac{\left[ {{M}^{1}}{{L}^{3}}{{T}^{-3}}{{A}^{-1}} \right]}{T} \\\ & \Rightarrow \varepsilon \dfrac{d\phi }{dt}=\dfrac{\left[ {{M}^{0}}{{L}^{0}}{{T}^{1}}{{A}^{1}} \right]}{T} \\\ & \Rightarrow \varepsilon \dfrac{d\phi }{dt}=A \\\ \end{aligned}$$ As A is the dimension of electric current. Hence the dimension of $$\varepsilon \dfrac{d\phi }{dt}$$ is the same as the dimension of electric current. **Hence option B is correct.** **Note:** We can also use the formula of electric flux which is given by Gauss’ law as it will give as value of $$\varepsilon \dfrac{d\phi }{dt}$$ which will be equal to current. And we know that dimensions on both sides should be equal therefore it will be the same as the dimension of current. But this approach is better and appropriate, in case we don’t know the dimensions of the given quantities we can use Gauss’s law.