Dimensions of capcitance is

Options

A. `[M^(-1)L^(-2)T^(4)A^(2)]`
B. `[MLT^(-3)A^(-1)]`
C. `[ML^(2)T^(-3)A^(-1)]`
D. `[M^(-1)L^(-2)T^(4)A^(2)]`

Answer

To find the dimensions of capacitance, we start with the relationship between charge (Q), capacitance (C), and voltage (V) given by the formula:

\[ Q = C \cdot V \]

### Step 1: Rearranging the formula
We can rearrange this formula to express capacitance in terms of charge and voltage:

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

### Step 2: Finding the dimensions of charge (Q)
The charge (Q) can be expressed in terms of current (I) and time (T):

\[ Q = I \cdot T \]

The dimension of current (I) is represented as [A] (Ampere), so the dimensions of charge become:

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

### Step 3: Finding the dimensions of voltage (V)
Voltage (V) is defined as energy (E) per unit charge (Q). The dimension of energy can be expressed in terms of force (F) and distance (L):

\[ E = F \cdot L \]

The dimension of force (F) is given by Newton's second law:

\[ F = m \cdot a = m \cdot \frac{L}{T^2} \]

Thus, the dimensions of energy (E) become:

\[ [E] = [F] \cdot [L] = [m \cdot \frac{L}{T^2}] \cdot [L] = [m \cdot L^2 \cdot T^{-2}] \]

Now, substituting this into the expression for voltage:

\[ [V] = \frac{[E]}{[Q]} = \frac{[m \cdot L^2 \cdot T^{-2}]}{[A \cdot T]} \]

This simplifies to:

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

### Step 4: Substituting dimensions into capacitance formula
Now we can substitute the dimensions of charge and voltage back into the capacitance formula:

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

### Step 5: Simplifying the dimensions
This gives us:

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

Rearranging this, we find:

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

### Conclusion
Thus, the dimensions of capacitance are:

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