Capacitance has dimensions :

Options

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

Answer

To find the dimensions of capacitance, we start from the basic relationship involving charge (Q), capacitance (C), and voltage (V). The relationship is given by:

\[ Q = C \cdot V \]

### Step 1: Rearranging the Equation
From the equation above, we can express capacitance (C) in terms of charge (Q) and voltage (V):

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

### Step 2: Finding Dimensions of Charge (Q)
The dimension of charge (Q) can be derived from the definition of current (I), which is the charge per unit time:

\[ I = \frac{Q}{t} \]

Thus, we can express charge (Q) as:

\[ Q = I \cdot t \]

The dimensions of current (I) are:

\[ [I] = A \]

Where A is the dimension of electric current. Therefore, the dimensions of charge (Q) become:

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

### Step 3: Finding Dimensions of Voltage (V)
Voltage (V) can be defined as energy per unit charge. The dimension of energy (E) can be expressed as:

\[ E = F \cdot d \]

Where F is force and d is displacement. The dimensions of force (F) are:

\[ [F] = [M][a] = [M][L][T^{-2}] = MLT^{-2} \]

Thus, the dimensions of energy (E) are:

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

Now, substituting this into the dimension of voltage:

\[ [V] = \frac{[E]}{[Q]} = \frac{ML^2T^{-2}}{A \cdot T} = \frac{ML^2}{AT^3} \]

### Step 4: Finding Dimensions of Capacitance (C)
Now we can substitute the dimensions of charge (Q) and voltage (V) back into the equation for capacitance:

\[ [C] = \frac{[Q]}{[V]} = \frac{A \cdot T}{\frac{ML^2}{AT^3}} \]

This simplifies to:

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

### Step 5: Final Result
Thus, the dimensions of capacitance (C) are:

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

### Summary
The final dimensions of capacitance are:

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