Match the following
`{:(,"Table - 1",,"Table - 2",),((A),sigma^(2)//epsi_(0),(P),C^(2)//J - m,),((B),epsi_(0),(Q),"Farad",),((C),("ampere - second")/("Volt"),(R),J//m^(3),),((D),(V)/(E),(S),"metre",):}`

Answer

To solve the matching question between Table 1 and Table 2, we will analyze the units provided in Table 1 and find their corresponding matches in Table 2.

### Step-by-Step Solution:

1. **Identify the Units in Table 1:**
- (A) \( \frac{\sigma^2}{\epsilon_0} \)
- (B) \( \epsilon_0 \)
- (C) \( \frac{\text{ampere} \cdot \text{second}}{\text{Volt}} \)
- (D) \( \frac{V}{E} \)

2. **Match (D) \( \frac{V}{E} \):**
- The unit \( \frac{V}{E} \) can be interpreted as \( \frac{\text{Voltage}}{\text{Electric Field}} \).
- From the relationship \( V = E \cdot L \), we can deduce that \( \frac{V}{E} = L \), where \( L \) is a length.
- Therefore, \( D \) matches with \( S \) (metre).

3. **Match (C) \( \frac{\text{ampere} \cdot \text{second}}{\text{Volt}} \):**
- The unit \( \frac{\text{ampere} \cdot \text{second}}{\text{Volt}} \) can be rewritten as \( \frac{I \cdot T}{V} \).
- This can be interpreted as charge \( Q \) (since \( Q = I \cdot T \)) divided by potential \( V \).
- The unit of capacitance is Farad, which is defined as \( \frac{C}{V} \).
- Thus, \( C \) matches with \( Q \) (Farad).

4. **Match (B) \( \epsilon_0 \):**
- The permittivity of free space \( \epsilon_0 \) can be derived from the equation \( F = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{R^2} \).
- Rearranging gives \( \epsilon_0 = \frac{Q^2}{4 \pi F R^2} \).
- The unit of \( \epsilon_0 \) can be expressed in terms of charge (Coulombs) and force (Newtons), leading to \( \text{Farad} \) (since \( \text{Farad} = \frac{C}{V} \)).
- Therefore, \( B \) matches with \( P \) (Farad).

5. **Match (A) \( \frac{\sigma^2}{\epsilon_0} \):**
- The surface charge density \( \sigma \) has units of \( \frac{C}{m^2} \).
- Thus, \( \sigma^2 \) has units of \( \frac{C^2}{m^4} \).
- Dividing by \( \epsilon_0 \) (which has units of \( \frac{C^2}{N \cdot m^2} \)), we get \( \frac{C^2/m^4}{C^2/(N \cdot m^2)} = \frac{N}{m^2} \).
- This is equivalent to \( J/m^3 \) (Joules per cubic meter).
- Therefore, \( A \) matches with \( R \) (J/m³).

### Final Matches:
- (A) \( \frac{\sigma^2}{\epsilon_0} \) → \( R \) (J/m³)
- (B) \( \epsilon_0 \) → \( P \) (Farad)
- (C) \( \frac{\text{ampere} \cdot \text{second}}{\text{Volt}} \) → \( Q \) (Farad)
- (D) \( \frac{V}{E} \) → \( S \) (metre)

### Summary of Matches:
- A → R
- B → P
- C → Q
- D → S