Question

The quantity $X = \dfrac{{{\varepsilon _0}LV}}{t}$ ; ${\varepsilon _0}$ is the permittivity of free space, $L$ is the length, $V$ is the potential difference and $t$ is the time. The dimensions of $X$ are the same as that of
A) resistance
B) charge
C) voltage
D) current

Answer 1

Hint:Any physical quantity can be expressed in terms of its dimensions. The three basic dimensions are mass $\left[ M \right]$, length $\left[ L \right]$ and time $\left[ T \right]$ . But for quantities like resistance, voltage, charge and current we make use of other dimensions like $\left[ I \right]$ in addition to the three basic dimensions to represent these quantities. The dimensional analysis can be employed to obtain the dimensions of the given unknown quantity.

Step by step solution.
Step 1: Express the dimensional formula for each term in the given quantity $X$ .
The given quantity is expressed as $X = \dfrac{{{\varepsilon _0}LV}}{t}$ ------- (1) where ${\varepsilon _0}$ is the permittivity of free space, $L$ is the length, $V$ is the potential difference and $t$ is the time.
The dimensional formula for the permittivity of free space is ${\varepsilon _0} \to \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{I^2}} \right]$ .
The dimensional formula for the length is $L \to \left[ L \right]$ .
The dimensional formula for the potential difference is $V \to \left[ {M{L^2}{T^{ - 3}}{I^{ - 1}}} \right]$ .
The dimensional formula for the time is $it \to \left[ T \right]$ .
Step 2: Express the equation (1) in terms of the dimensions of terms involved in it to obtain the dimensional formula for $X$ .
Equation (1) is given by, $X = \dfrac{{{\varepsilon _0}LV}}{t}$ .
Replacing ${\varepsilon _0}$ , $L$ , $V$ and $t$ in equation (1) by their respective dimensional formulas we get, $X \to \dfrac{{\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{I^2}} \right]\left[ L \right]\left[ {M{L^2}{T^{ - 3}}{I^{ - 1}}} \right]}}{{\left[ T \right]}}$
Simplifying the powers of the dimensions we get, $X \to \left[ {{M^{\left( { - 1 + 1} \right)}}{L^{\left( { - 3 + 1 + 2} \right)}}{T^{\left( {4 - 3 - 1} \right)}}{I^{\left( {2 - 1} \right)}}} \right] = \left[ I \right]$
Thus the dimension of the given quantity $X$ is the same as the dimension of current.

So the correct option is D.

Note:Alternate method
Given: $X = \dfrac{{{\varepsilon _0}LV}}{t}$ where ${\varepsilon _0}$ is the permittivity of free space, $L$ is the length, $V$ is the potential difference and $t$ is the time.
We can also find the dimension of the given by simplifying the above relation.
Now we know that capacitance $C = \dfrac{{{\varepsilon _0}A}}{d}$ where $A$ is the area of the capacitor plate and $d$ is the distance between the two plates.
And if the dimensions were considered then $C = \dfrac{{{\varepsilon _0}{L^2}}}{L} = {\varepsilon _0}L$ .
So the given relation for $X$ becomes, $X = \dfrac{{CV}}{t}$ .
Also, the relation between charge in a capacitor and potential difference across a capacitor is given by, $Q = CV$ .
$ \Rightarrow X = \dfrac{Q}{t}$
Now since current is defined as the rate of flow of charge, we have $X = I$ .
So the dimensions of $X$ must be the same as that of current and hence the correct option is D.