The unit of electric permittivity is `(C^(2))/(Nm^(2))`. Find the dimesions of electric permittivity

Options

A. `[a^(2)M^(-1)L^(-3)T^(4)]`
B. `[A^(2)M^(-1)L^(-3)T^(0)]`
C. `[AM^(-1)L^(-3)T^(4)]`
D. `[A^(2)M^(0)L^(-3)T^(4)]`

Answer

To find the dimensions of electric permittivity given its unit as \((C^2)/(Nm^2)\), we will follow these steps:

### Step 1: Understand the units involved
The unit of electric permittivity is given as:
\[ \frac{C^2}{N \cdot m^2} \]
where:
- \(C\) is the unit of charge (Coulomb),
- \(N\) is the unit of force (Newton),
- \(m\) is the unit of length (meter).

### Step 2: Write the dimensions of each unit
1. **Coulomb (C)**:
The Coulomb is defined in terms of current (I) and time (T):
\[ C = A \cdot s \]
where \(A\) is the unit of current (Ampere) and \(s\) is the unit of time (second). Thus, the dimensions of Coulomb are:
\[ [C] = [A][T] = [I][T] \]

2. **Newton (N)**:
The Newton is defined as:
\[ N = kg \cdot m/s^2 \]
which can be expressed in dimensional form as:
\[ [N] = [M][L][T^{-2}] \]

3. **Meter (m)**:
The dimensions of meter are simply:
\[ [m] = [L] \]

### Step 3: Substitute the dimensions into the unit of electric permittivity
Now, substituting the dimensions into the expression for electric permittivity:
\[ \epsilon = \frac{C^2}{N \cdot m^2} \]
Substituting the dimensions we derived:
\[ \epsilon = \frac{(I \cdot T)^2}{(M \cdot L \cdot T^{-2}) \cdot (L^2)} \]

### Step 4: Simplify the expression
Now we can simplify this expression:
\[ \epsilon = \frac{I^2 \cdot T^2}{M \cdot L \cdot T^{-2} \cdot L^2} \]
This simplifies to:
\[ \epsilon = \frac{I^2 \cdot T^2}{M \cdot L^3 \cdot T^{-2}} = \frac{I^2 \cdot T^4}{M \cdot L^3} \]

### Step 5: Write the final dimensions of electric permittivity
Thus, the dimensions of electric permittivity \(\epsilon\) are:
\[ [\epsilon] = \frac{I^2 \cdot T^4}{M \cdot L^3} \]

### Final Answer:
The dimensions of electric permittivity are:
\[ [\epsilon] = M^{-1}L^{-3}T^{4}I^{2} \]