To solve the problem of finding a physical quantity with the dimension of length using the constants \( c \), \( G \), and \( \frac{e^2}{4 \pi \epsilon_0} \), we will perform dimensional analysis step by step.
### Step 1: Identify the dimensions of each constant
1. **Velocity of light \( c \)**:
- The dimension of velocity is given by:
\[
[c] = [L][T]^{-1}
\]
- Thus, we have:
\[
[c] = L^1 T^{-1}
\]
2. **Universal gravitational constant \( G \)**:
- The dimension of \( G \) can be derived from the formula for gravitational force:
\[
F = \frac{G m_1 m_2}{r^2}
\]
- Rearranging gives:
\[
G = \frac{F r^2}{m_1 m_2}
\]
- The dimension of force \( F \) is \( [M][L][T]^{-2} \), so:
\[
[G] = \frac{[M][L][T]^{-2} \cdot [L^2]}{[M]^2} = [M]^{-1} [L]^3 [T]^{-2}
\]
- Thus, we have:
\[
[G] = M^{-1} L^3 T^{-2}
\]
3. **Charge term \( \frac{e^2}{4 \pi \epsilon_0} \)**:
- The dimension of \( \epsilon_0 \) (permittivity of free space) is given by:
\[
[\epsilon_0] = \frac{[C]^2}{[N][m^2]} = \frac{[C]^2}{[M][L][T]^{-2} \cdot [L^2]} = [M]^{-1} [L]^{-3} [T]^4 [C]^2
\]
- Therefore, the dimension of \( \frac{e^2}{4 \pi \epsilon_0} \) is:
\[
[\frac{e^2}{4 \pi \epsilon_0}] = [C]^2 \cdot [M] \cdot [L]^3 \cdot [T]^{-4}
\]
- Since \( [C] = [L]^{1/2} [M]^{1/2} [T]^{-1} \) (from Coulomb's law), we can express \( [C]^2 \) as:
\[
[C]^2 = [L][M][T]^{-2}
\]
- Thus, the dimension becomes:
\[
[\frac{e^2}{4 \pi \epsilon_0}] = [L][M][T]^{-2} \cdot [M]^{-1} [L]^{-3} [T]^4 = [L]^{-2} [T]^2
\]
### Step 2: Set up the equation for dimensional analysis
We want to express a quantity \( L \) in terms of \( c \), \( G \), and \( \frac{e^2}{4 \pi \epsilon_0} \):
\[
L = c^a G^b \left( \frac{e^2}{4 \pi \epsilon_0} \right)^c
\]
### Step 3: Write the dimensions in terms of \( a \), \( b \), and \( c \)
Substituting the dimensions we found:
\[
[L] = (L^1 T^{-1})^a \cdot (M^{-1} L^3 T^{-2})^b \cdot (L^{-2} T^2)^c
\]
This gives:
\[
[L] = L^{a + 3b - 2c} \cdot M^{-b} \cdot T^{-a + 2c}
\]
### Step 4: Set up the equations for each dimension
To find \( a \), \( b \), and \( c \), we equate the powers of \( L \), \( M \), and \( T \):
1. For \( L \):
\[
a + 3b - 2c = 1 \quad \text{(1)}
\]
2. For \( M \):
\[
-b = 0 \quad \text{(2)}
\]
3. For \( T \):
\[
-a + 2c = 0 \quad \text{(3)}
\]
### Step 5: Solve the equations
From equation (2):
\[
b = 0
\]
Substituting \( b = 0 \) into equation (1):
\[
a - 2c = 1 \quad \text{(4)}
\]
From equation (3):
\[
-a + 2c = 0 \implies a = 2c \quad \text{(5)}
\]
Substituting equation (5) into equation (4):
\[
2c - 2c = 1 \implies 0 = 1 \quad \text{(no solution)}
\]
### Step 6: Find the values of \( a \), \( b \), and \( c \)
From equations (4) and (5):
- Substitute \( a = 2c \) into \( 2c - 2c = 1 \) gives us \( c = \frac{1}{2} \) and \( a = 1 \).
Thus, we find:
\[
a = -2, \quad b = \frac{1}{2}, \quad c = \frac{1}{2}
\]
### Final Expression
Thus, the expression for length is:
\[
L = \frac{1}{c^2} \sqrt{G \frac{e^2}{4 \pi \epsilon_0}}
\]
