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Question: The quantity \(X = \dfrac{{{\varepsilon _0}LV}}{t}\) ; \({\varepsilon _0}\) is the permittivity of f...

The quantity X=ε0LVtX = \dfrac{{{\varepsilon _0}LV}}{t} ; ε0{\varepsilon _0} is the permittivity of free space, LL is the length, VV is the potential difference and tt is the time. The dimensions of XX are the same as that of
A) resistance
B) charge
C) voltage
D) current

Explanation

Solution

Any physical quantity can be expressed in terms of its dimensions. The three basic dimensions are mass [M]\left[ M \right], length [L]\left[ L \right] and time [T]\left[ T \right] . But for quantities like resistance, voltage, charge and current we make use of other dimensions like [I]\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 XX .
The given quantity is expressed as X=ε0LVtX = \dfrac{{{\varepsilon _0}LV}}{t} ------- (1) where ε0{\varepsilon _0} is the permittivity of free space, LL is the length, VV is the potential difference and tt is the time.
The dimensional formula for the permittivity of free space is ε0→[M−1L−3T4I2]{\varepsilon _0} \to \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{I^2}} \right] .
The dimensional formula for the length is L→[L]L \to \left[ L \right] .
The dimensional formula for the potential difference is V→[ML2T−3I−1]V \to \left[ {M{L^2}{T^{ - 3}}{I^{ - 1}}} \right] .
The dimensional formula for the time is it→[T]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 XX .
Equation (1) is given by, X=ε0LVtX = \dfrac{{{\varepsilon _0}LV}}{t} .
Replacing ε0{\varepsilon _0} , LL , VV and tt in equation (1) by their respective dimensional formulas we get, X→[M−1L−3T4I2][L][ML2T−3I−1][T]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→[M(−1+1)L(−3+1+2)T(4−3−1)I(2−1)]=[I]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 XX is the same as the dimension of current.

So the correct option is D.

Note: Alternate method
Given: X=ε0LVtX = \dfrac{{{\varepsilon _0}LV}}{t} where ε0{\varepsilon _0} is the permittivity of free space, LL is the length, VV is the potential difference and tt is the time.
We can also find the dimension of the given by simplifying the above relation.
Now we know that capacitance C=ε0AdC = \dfrac{{{\varepsilon _0}A}}{d} where AA is the area of the capacitor plate and dd is the distance between the two plates.
And if the dimensions were considered then C=ε0L2L=ε0LC = \dfrac{{{\varepsilon _0}{L^2}}}{L} = {\varepsilon _0}L .
So the given relation for XX becomes, X=CVtX = \dfrac{{CV}}{t} .
Also, the relation between charge in a capacitor and potential difference across a capacitor is given by, Q=CVQ = CV .
⇒X=Qt\Rightarrow X = \dfrac{Q}{t}
Now since current is defined as the rate of flow of charge, we have X=IX = I .
So the dimensions of XX must be the same as that of current and hence the correct option is D.