Which of the following represents the dimension of capacitance ?

Options

A. `M^(-1)Li^(-2)A^(2)T^(4)`
B. `ML^(2)A^(-2)T^(-4)`
C. `MLA^(-1)T^(-4)`
D. `M^(-1)L^(-1)A^(2)T^(2)`

Explanation

To find the dimension of capacitance, we start with the relationship between charge (Q), capacitance (C), and potential difference (V). The formula is given by:

\[ Q = C \cdot V \]

From this equation, we can express capacitance as:

\[ C = \frac{Q}{V} \]

### Step 1: Identify the dimensions of charge (Q)
Charge (Q) is defined as the product of current (I) and time (T):

\[ Q = I \cdot T \]

The dimension of current (I) is represented as:

\[ [I] = A \] (where A is amperes)

Thus, the dimension of charge can be expressed as:

\[ [Q] = [I] \cdot [T] = A \cdot T \]

### Step 2: Identify the dimensions of potential difference (V)
Potential difference (V) is defined as the work done (W) per unit charge (Q):

\[ V = \frac{W}{Q} \]

The dimension of work (W) is given by:

\[ [W] = [F] \cdot [d] = [M \cdot L \cdot T^{-2}] \cdot [L] = M \cdot L^2 \cdot T^{-2} \]

Thus, the dimension of potential difference can be expressed as:

\[ [V] = \frac{[W]}{[Q]} = \frac{M \cdot L^2 \cdot T^{-2}}{A \cdot T} = M \cdot L^2 \cdot T^{-3} \cdot A^{-1} \]

### Step 3: Substitute the dimensions into the capacitance formula
Now, substituting the dimensions of charge and potential difference into the capacitance formula:

\[ [C] = \frac{[Q]}{[V]} = \frac{A \cdot T}{M \cdot L^2 \cdot T^{-3} \cdot A^{-1}} \]

### Step 4: Simplify the expression
Now, simplifying the expression:

\[ [C] = \frac{A^2 \cdot T^4}{M \cdot L^2} \]

This can be rewritten as:

\[ [C] = M^{-1} \cdot L^{-2} \cdot T^4 \cdot A^2 \]

### Final Answer
Thus, the dimension of capacitance is:

\[ [C] = M^{-1} \cdot L^{-2} \cdot T^4 \cdot A^2 \]